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  • Zend Framework decorator subform add a class tag to DD wrapper tag

    - by Samuele
    I have this form: class Request_Form_Prova extends Zend_Form { public function init() { $this->setMethod('post'); $SubForm_Step = new Zend_Form_SubForm(); $SubForm_Step->setAttrib('class','Step'); $this->addSubform($SubForm_Step, 'Chicco'); $PrivacyCheck = $SubForm_Step->createElement('CheckBox', 'PrivacyCheck'); $PrivacyCheck->setLabel('I have read and I agre bla bla...') ->setRequired(true) ->setUncheckedValue(''); $PrivacyCheck->getDecorator('Label')->setOption('class', 'inline'); $SubForm_Step->addElement($PrivacyCheck); $SubForm_Step->addElement('submit', 'submit', array( 'ignore' => true, 'label' => 'OK', )); } } That generate this HTML: <form enctype="application/x-www-form-urlencoded" method="post" action=""> <dl class="zend_form"> <dt id="Chicco-label">&nbsp;</dt> <dd id="Chicco-element"> <fieldset id="fieldset-Chicco" class="Step"> <dl> <dt id="Chicco-PrivacyCheck-label"><label for="Chicco-PrivacyCheck" class="inline required">I have read and I agre bla bla...</label></dt> <dd id="Chicco-PrivacyCheck-element"> <input type="hidden" name="Chicco[PrivacyCheck]" value=""><input type="checkbox" name="Chicco[PrivacyCheck]" id="Chicco-PrivacyCheck" value="1"> </dd> <dt id="submit-label">&nbsp;</dt> <dd id="submit-element"> <input type="submit" name="Chicco[submit]" id="Chicco-submit" value="OK"> </dd> </dl> </fieldset> </dd> </dl> </form> How can I add a class="Test" to the <dd id="Chicco-element"> elemnt? In order to have it like that: <dd id="Chicco-element" class="Test"> I thought something like that but it don't work: $SubForm_Step->getDecorator('DdWrapper')->setOption('class', 'Test'); OR $SubForm_Step->getDecorator('DtDdWrapper')->setOption('class', 'Test'); How can I do it? And last question: How can I wrap that DD and DT element of a SubForm in another DL element? Like that: ( second line ) <dl class="zend_form"> <dl> <dt id="Chicco-label">&nbsp;</dt> <dd id="Chicco-element"> <fieldset id="fieldset-Chicco" class="Step"> <dl> .......

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  • Problem passing variables in php form.

    - by Joshxtothe4
    I have the following php form. I am trying to make it so that when the form is loaded, the values will be assigned the appropriate check- variable. This variable will contain either "checked or "". If it contains checked, the way it is displayed with the html should cause the relevant checkbox to be checked. As it is, the variables do not seem to be being passed. When I echo out $deleted or $notice from within the submitinfo branch, they are blank. Furthermore, nothing is being inserted into the database, and I am not getting any database error. How can I check this? <?php if (isset($_GET["cmd"])) $cmd = $_GET["cmd"]; else if (isset($_POST["cmd"])) $cmd = $_POST["cmd"]; else die("Invalid URL"); if (isset($_GET["pk"])) { $pk = $_GET["pk"]; } if (isset($_POST["deleted"])) { $deleted = $_POST["deleted"]; } if (isset($_POST["notice"])) { $notice = $_POST["notice"]; } $con = mysqli_connect("localhost","user","password", "db"); if (!$con) { echo "Can't connect to MySQL Server. Errorcode: %s\n". mysqli_connect_error(); exit; } $con->set_charset("utf8"); $getformdata = $con->query("select * from STATUS where ARTICLE_NO = '$pk'"); $checkDeleted = ""; $checkNotice = ""; while ($row = mysqli_fetch_assoc($getformdata)) { $checkDeleted = $row['deleted']; $checkNotice = $row['notice']; } if($cmd=="submitinfo") { $statusQuery = "INSERT INTO STATUS VALUES (?, ?)"; if ($statusInfo = $con->prepare($statusQuery)) { $statusInfo->bind_param("ss", $deleted, $notice); $statusInfo->execute(); $statusInfo->close(); echo "true"; } else { echo "false"; } print_r($con->error); } if($cmd=="EditStatusData") { echo "<form name=\"statusForm\" action=\"test.php\" method=\"post\" enctype=\"multipart/form-data\"> <h1>Editing information for auction: ".$pk."</h1> Löschung Ebay: <input type=\"checkbox\" name=\"deleted\" value=\"checked\" ".$checkDeleted." /> <br /> Abmahnung: <input type=\"checkbox\" name=\"notice\" value=\"checked\" ".$checkNotice." /> <br /> <input type=\"hidden\" name=\"cmd\" value=\"submitinfo\" /> <input name=\"Submit\" type=\"submit\" value=\"submit\" /> </form>"; } else { print_r($con->error); }

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  • jQuery Form plugin - no data from file upload?

    - by pojo
    I've been struggling a bit with the jQuery Form plugin. I want to create a file upload form that posts the data (JSON, from the chosen file) into a REST service exposed by a servlet. The URL for the POST is calculated from what the user chooses in a SELECT dropdown. When the upload is complete, I want to notify the user immediately, AJAX-style. The problem is that the POST header has a Content-Length of 0 and contains no data. I would appreciate any help! <html> <head> <script type="text/javascript" src="js/jquery-1.4.2.min.js">/* ppp */</script> <script type="text/javascript" src="js/jquery.form.js">/* ppp */</script> <script type="text/javascript"> function cb_beforesubmit (arr, $form, options) { // This should override the form's action attribute options.url = "/rest/services/" + $('#selectedaction')[0].value; return true; } function cb_success (rt, st, xhr, wf) { $('#response').html(rt + '<br>' + st + '<br>' + xhr); } $(document).ready(function () { var options = { beforeSubmit: cb_beforesubmit, success: cb_success, dataType: 'json', contentType: 'application/json', method: 'POST', }; $('#myform').ajaxForm(options); $.getJSON('/rest/services', function (data, ts) { for (var property in data) { if (typeof property == 'string') { $('#selectedaction').append('<option>' + property + '</option>'); } } }); }); </script> </head> <body> <form id="myform" action="/rest/services/foo1" method="POST" enctype="multipart/form-data"> <!-- The form does not seem to submit at all if I don't set action to a default value? !--> <select id="selectedaction"> <script type="text/javascript"> </script> </select> <input type="file" value="Choose"/> <input type="submit" value="Submit" /> </form> <div id="response"> </div> </body> </html>

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  • It is only uploading first row's file input

    - by user1304328
    I have a application which you can access here. If you open the application please click on the "Add" button a couple of times. This will add a new row into a table below. In each table row there is an AJAX file uploader. Now the problem is that if I click on the "Upload" button in any row except the first row, then the uploading only happens in the first row so it is only uploading the first file input only. Why is it doing this and how can I get it so that when then the user clicks the "Upload" button, the file input within that row of the "Upload" button is uploaded and not the first row being uploaded? Below is the full code where it appends the file AJAX file uploaded in each table row: function insertQuestion(form) { var $tbody = $('#qandatbl > tbody'); var $tr = $("<tr class='optionAndAnswer' align='center'></tr>"); var $image = $("<td class='image'></td>"); var $fileImage = $("<form action='upload.php' method='post' enctype='multipart/form-data' target='upload_target' onsubmit='startUpload();' >" + "<p id='f1_upload_process' align='center'>Loading...<br/><img src='Images/loader.gif' /><br/></p><p id='f1_upload_form' align='center'><br/><label>" + "File: <input name='fileImage' type='file' class='fileImage' /></label><br/><label><input type='submit' name='submitBtn' class='sbtn' value='Upload' /></label>" + "</p> <iframe id='upload_target' name='upload_target' src='#' style='width:0;height:0;border:0px solid #fff;'></iframe></form>"); $image.append($fileImage); $tr.append($image); $tbody.append($tr); } function startUpload(){ document.getElementById('f1_upload_process').style.visibility = 'visible'; document.getElementById('f1_upload_form').style.visibility = 'hidden'; return true; } function stopUpload(success){ var result = ''; if (success == 1){ result = '<span class="msg">The file was uploaded successfully!<\/span><br/><br/>'; } else { result = '<span class="emsg">There was an error during file upload!<\/span><br/><br/>'; } document.getElementById('f1_upload_process').style.visibility = 'hidden'; document.getElementById('f1_upload_form').innerHTML = result + '<label>File: <input name="fileImage" type="file"/><\/label><label><input type="submit" name="submitBtn" class="sbtn" value="Upload" /><\/label>'; document.getElementById('f1_upload_form').style.visibility = 'visible'; return true; }

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  • Looping through array in PHP to post several multipart form-data

    - by Léon Pelletier
    I'm trying in an asp web application to code a function that would loop through a list of files in a multiple upload form and send them one by one. Is this something that can be done in ASP? Because I've read some posts about how to attach several files together, but saw nothing about looping through the files. I can easily imagine it in C# via HttpWebRequest or with socket, but in php, I guess there are already function designed to handle it? // This is false/pseudo-code :) for (int index = 0; index < number_of_files; index++) { postfile(file[index]); } And in each iteration, it should send a multipart form-data POST. postfile(TheFileInfos) should make a POST like it: POST /afs.aspx?fn=upload HTTP/1.1 [Header stuff] Content-Type: multipart/form-data; boundary=----------Ef1Ef1cH2Ij5GI3ae0gL6KM7GI3GI3 [Header stuff] ------------Ef1Ef1cH2Ij5GI3ae0gL6KM7GI3GI3 Content-Disposition: form-data; name="Filename" myimage1.png ------------Ef1Ef1cH2Ij5GI3ae0gL6KM7GI3GI3 Content-Disposition: form-data; name="fileid" 58e21ede4ead43a5201206101806420000007667212251 ------------Ef1Ef1cH2Ij5GI3ae0gL6KM7GI3GI3 Content-Disposition: form-data; name="Filedata"; filename="myimage1.png" Content-Type: application/octet-stream [Octet Stream] [Edit] I'll try it: <html> <head> <title>Untitled Document</title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"> </head> <body> <form name="form1" enctype="multipart/form-data" method="post" action="processFiles.php"> <p> <? // start of dynamic form $uploadNeed = $_POST['uploadNeed']; for($x=0;$x<$uploadNeed;$x++){ ?> <input name="uploadFile<? echo $x;?>" type="file" id="uploadFile<? echo $x;?>"> </p> <? // end of for loop } ?> <p><input name="uploadNeed" type="hidden" value="<? echo $uploadNeed;?>"> <input type="submit" name="Submit" value="Submit"> </p> </form> </body> </html>

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  • how to upload a audio file using REST webservice in Google App Engine for Java

    - by sathya
    Am using google app engine with eclipse IDE and trying to upload a audio file. I used the File Upload in Google App Engine For Java and can able to upload the file successfully. Now am planning to use REST web service for it. I had analyzed in developers.google but i failed. Can anyone suggest me how to implement REST Web services in google app engine using Eclipse. The code google provided is shown below, // file Upload.java public class Upload extends HttpServlet { private BlobstoreService blobstoreService = BlobstoreServiceFactory.getBlobstoreService(); public void doPost(HttpServletRequest req, HttpServletResponse res) throws ServletException, IOException { Map<String, BlobKey> blobs = blobstoreService.getUploadedBlobs(req); BlobKey blobKey = blobs.get("myFile"); if (blobKey == null) { res.sendRedirect("/"); } else { res.sendRedirect("/serve?blob-key=" + blobKey.getKeyString()); }}} // file Serve.java public class Serve extends HttpServlet { private BlobstoreService blobstoreService = BlobstoreServiceFactory.getBlobstoreService(); public void doGet(HttpServletRequest req, HttpServletResponse res) throws IOException { BlobKey blobKey = new BlobKey(req.getParameter("blob-key")); blobstoreService.serve(blobKey, res); }} // file index.jsp <%@ page import="com.google.appengine.api.blobstore.BlobstoreServiceFactory" %> <%@ page import="com.google.appengine.api.blobstore.BlobstoreService" %> <% BlobstoreService blobstoreService = BlobstoreServiceFactory.getBlobstoreService(); %> <form action="<%= blobstoreService.createUploadUrl("/upload") %>" method="post" enctype="multipart/form-data"> <input type="file" name="myFile"> <input type="submit" value="Submit"> </form> // web.xml <servlet> <servlet-name>Upload</servlet-name> <servlet-class>Upload</servlet-class> </servlet> <servlet> <servlet-name>Serve</servlet-name> <servlet-class>Serve</servlet-class> </servlet> <servlet-mapping> <servlet-name>Upload</servlet-name> <url-pattern>/upload</url-pattern> </servlet-mapping> <servlet-mapping> <servlet-name>Serve</servlet-name> <url-pattern>/serve</url-pattern> </servlet-mapping> Now how to provide a rest web service for the above code. Kindly suggest me an idea.

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  • jquery ajax form success callback not being called

    - by Michael Merchant
    I'm trying to upload a file using "AJAX", process data in the file and then return some of that data to the UI so I can dynamically update the screen. I'm using the JQuery Ajax Form Plugin, jquery.form.js found at http://jquery.malsup.com/form/ for the javascript and using Django on the back end. The form is being submitted and the processing on the back end is going through without a problem, but when a response is received from the server, my Firefox browser prompts me to download/open a file of type "application/json". The file has the json content that I've been trying to send to the browser. I don't believe this is an issue with how I'm sending the json as I have a modularized json_wrapper() function that I'm using in multiple places in this same application. Here is what my form looks after Django templates are applied: <form method="POST" enctype="multipart/form-data" action="/test_suites/active/upload_results/805/"> <p> <label for="id_resultfile">Upload File:</label> <input type="file" id="id_resultfile" name="resultfile"> </p> </form> You won't see any submit buttons because I'm calling submit with a button else where and am using ajaxSubmit() from the jquery.form.js plugin. Here is the controlling javascript code: function upload_results($dialog_box){ $form = $dialog_box.find("form"); var options = { type: "POST", success: function(data){ alert("Hello!!"); }, dataType: "json", error: function(){ console.log("errors"); }, beforeSubmit: function(formData, jqForm, options){ console.log(formData, jqForm, options); }, } $form.submit(function(){ $(this).ajaxSubmit(options); return false; }); $form.ajaxSubmit(options); } As you can see, I've gotten desperate to see the success callback function work and simply have an alert message created on success. However, we never reach that call. Also, the error function is not called and the beforeSubmit function is executed. The file that I get back has the following contents: {"count": 18, "failed": 0, "completed": 18, "success": true, "trasaction_id": "SQEID0.231"} I use 'success' here to denote whether or not the server was able to run the post command adequately. If it failed the result would look something like: {"success": false, "message":"<error_message>"} Your time and help is greatly appreciated. I've spent a few days on this now and would love to move on.

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  • why resubmit after refresh php page

    - by user2719452
    why resubmit after refresh php page? try it, go to: http://qass.im/message-envelope/ and upload any image now try click F5, after refresh page "resubmit" Why? I don't want resubmit after refresh page What is the solution? See this is my form code: <form id="uploadedfile" name="uploadedfile" enctype="multipart/form-data" action="upload.php" method="POST"> <input name="uploadedfile" type="file" /> <input type="submit" value="upload" /> </form> See this is php code upload.php file: <?php $allowedExts = array("gif", "jpeg", "jpg", "png", "zip", "pdf", "docx", "rar", "txt", "doc"); $temp = explode(".", $_FILES["uploadedfile"]["name"]); $extension = end($temp); $newname = $extension.'_'.substr(str_shuffle(str_repeat("0123456789abcdefghijklmnopqrstuvwxyz", 7)), 4, 7); $imglink = 'attachment/attachment_file_'; $uploaded = $imglink .$newname.'.'.$extension; if ((($_FILES["uploadedfile"]["type"] == "image/jpeg") || ($_FILES["uploadedfile"]["type"] == "image/jpeg") || ($_FILES["uploadedfile"]["type"] == "image/jpg") || ($_FILES["uploadedfile"]["type"] == "image/pjpeg") || ($_FILES["uploadedfile"]["type"] == "image/x-png") || ($_FILES["uploadedfile"]["type"] == "image/gif") || ($_FILES["uploadedfile"]["type"] == "image/png") || ($_FILES["uploadedfile"]["type"] == "application/msword") || ($_FILES["uploadedfile"]["type"] == "text/plain") || ($_FILES["uploadedfile"]["type"] == "application/vnd.openxmlformats-officedocument.wordprocessingml.document") || ($_FILES["uploadedfile"]["type"] == "application/pdf") || ($_FILES["uploadedfile"]["type"] == "application/x-rar-compressed") || ($_FILES["uploadedfile"]["type"] == "application/x-zip-compressed") || ($_FILES["uploadedfile"]["type"] == "application/zip") || ($_FILES["uploadedfile"]["type"] == "multipart/x-zip") || ($_FILES["uploadedfile"]["type"] == "application/x-compressed") || ($_FILES["uploadedfile"]["type"] == "application/octet-stream")) && ($_FILES["uploadedfile"]["size"] < 5242880) // Max size is 5MB && in_array($extension, $allowedExts)) { move_uploaded_file($_FILES["uploadedfile"]["tmp_name"], $uploaded ); echo '<a target="_blank" href="'.$uploaded.'">click</a>'; // If has been uploaded file echo '<h3>'.$uploaded.'</h3>'; } if($_FILES["uploadedfile"]["error"] > 0){ echo '<h3>Please choose file to upload it!</h3>'; // If you don't choose file } elseif(!in_array($extension, $allowedExts)){ echo '<h3>This extension is not allowed!</h3>'; // If you choose file not allowed } elseif($_FILES["uploadedfile"]["size"] > 5242880){ echo "Big size!"; // If you choose big file } ?> if you have solution, please edit my php code and paste your solution code! Thanks.

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  • jQuery slider onChange auto submit form

    - by bikey77
    I'm using a jQuery slider as a price range selector in a form. I'd like to have the form submit automatically when one of the values has been changed. I used a couple of examples I found on SO but they didn't work with my code. <form action="itemlist.php" method="post" enctype="application/x-www-form-urlencoded" name="priceform" id="priceform" target="_self"> <div id="slider-holder"> Prices: From <span id="pricefromlabel">100 &#8364;</span> To <span id="pricetolabel">500 &#8364;</span> <input type="hidden" id="pricefrom" name="pricefrom" value="100" /> <input type="hidden" id="priceto" name="priceto" value="500" /> <div id="slider-range"></div> <input name="Search" type="submit" value="Search" /> </div> </form> This is the code that displays the values of the slider and updates 2 hidden form fields I use to store the prices in order to submit: <script> $(function() { $("#slider-range" ).slider({ range: true, min: 0, max: 1000, values: [ <?=$minprice?>, <?=$maxprice?> ], start: function (event, ui) { event.stopPropagation(); }, slide: function( event, ui ) { $( "#pricefrom" ).val(ui.values[0]); $( "#priceto" ).val(ui.values[1]); $( "#pricefromlabel" ).html(ui.values[0] + ' &euro;'); $( "#pricetolabel" ).html(ui.values[1] + ' &euro;'); } }); return false; }); </script> I've tried adding this code as well as an data-autosubmit="true" attribute to the div but no result. $(function() { $('[data-autosubmit="true"]').change(function() { parentForm = $(this).('#priceform'); clearTimeout(submitTimeout); submitTimeout = setTimeout(function() { parentForm.submit() }, 100); }); I've also tried adding a $.post() event to the slider but I'm not very good with jQuery so I'm probably doing it wrong. Any help will be appreciated.

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  • Yii file upload issue

    - by user1289853
    Couple of months ago,I developed a simple app using YII,one of the feature was to upload file. The feature was working well in my dev machine,couple of days ago client found the file upload feature is not working in his server since deployment. And after that I test my dev machine that was not working too. My controller looks: public function actionEntry() { if (!Yii::app()->user->isGuest) { $model = new TrackForm; if (isset($_POST['TrackForm'])) { $entry = new Track; try { $entry->product_image = $_POST['TrackForm']['product_image']; $entry->product_image = CUploadedFile::getInstance($model, 'product_image'); if ($entry->save()) { if ($entry->product_image) { $entry->product_image->saveAs($entry->product_image->name, '/trackshirt/uploads'); } } $this->render('success', array('model' => $model)); // redirect to success page } } catch (Exception $e) { echo 'Caught exception: ', $e->getMessage(), "\n"; } } else { $this->render('entry', array('model' => $model)); } } } Model is like below: <?php class Track extends CActiveRecord { public static function model($className=__CLASS__) { return parent::model($className); } public function tableName() { return 'product_details'; } } My view looks: <?php $form = $this->beginWidget('CActiveForm', array( 'id' => 'hide-form', 'enableClientValidation' => true, 'clientOptions' => array( 'validateOnSubmit' => true, ), 'htmlOptions' => array('enctype' => 'multipart/form-data'), )); ?> <p class="auto-style2"><strong>Administration - Add New Product</strong></p> <table align="center" style="width: 650px"><td class="auto-style3" style="width: 250px">Product Image</td> <td> <?php echo $form->activeFileField($model, 'product_image'); ?> </td> </tr> </table> <p class="auto-style1"> <div style="margin-leftL:-100px;"> <?php echo CHtml::submitButton('Submit New Product Form'); ?> </div> <?php $this->endWidget(); ?> Any idea where is the problem?I tried to debug it but every time it returns Null. Thanks.

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  • upload form only works in Firefox when uploading ASCII .stl 3D files

    - by NathanPDX
    uploadform.html and upload_file.php (below) works fine in Firefox but fails in Chrome, IE, and Safari when uploading ASCII .stl 3D files. Error message is "Invalid file" and problem occurs with multiple computers and multiple .stl files. When I modify the code to support other file types like JPG and PDF it allows those file types in all three web browsers. Also, Firefox only allows the .stl upload if I include application/octet-stream in the mime types section. Why doesn't this work outside of Firefox? uploadform.html: <!doctype html> <html> <body> <form action="upload_file.php" method="post" enctype="multipart/form-data"> <label for="file">Filename:</label> <input type="file" name="file" id="file" /> <br /> <input type="submit" name="submit" value="Submit" /> </form> </body> </html> upload_file.php: <!doctype html> <html> <body> <?php $allowedExts = array("stl"); $extension = end(explode(".", $_FILES["file"]["name"])); if ( ( ($_FILES["file"]["type"] == "application/sla") || ($_FILES["file"]["type"] == "application/octet-stream") || ($_FILES["file"]["type"] == "text/plain") || ($_FILES["file"]["type"] == "application/unknown") ) && ($_FILES["file"]["size"] < 2000000) && in_array($extension, $allowedExts) ) { if ($_FILES["file"]["error"] > 0) { echo "Return Code: " . $_FILES["file"]["error"] . "<br />"; } else { echo "Upload: " . $_FILES["file"]["name"] . "<br />"; echo "Size: " . ($_FILES["file"]["size"] /1024) . " KB<br />"; if (file_exists("upload/" . $_FILES["file"]["name"])) { echo $_FILES["file"]["name"] . " already exists. "; } else { move_uploaded_file($_FILES["file"]["tmp_name"], "upload/" . $_FILES["file"]["name"]); echo "successful upload"; } } } else { echo "Invalid file"; } ?> </body> </html>

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  • Not Inserted into database in PHP script

    - by Aruna
    Hi, i am having an form for uploading an excel file like <form enctype="multipart/form-data" action="http://localhost/joomla/Joomla_1.5.7/index.php?option=com_jumi&fileid=7" method="POST"> Please choose a file: <input name="file" type="file" id="file" /><br /> <input type="submit" value="Upload" /> </form> And in the FIle http://localhost/joomla/Joomla_1.5.7/index.php?option=com_jumi&fileid=7 i have retrived the uploaded file contents by <?php echo "Name". $_FILES['file']['name'] . "<br />"; echo "Type: " . $_FILES["file"]["type"] . "<br />"; echo "Size: " . ($_FILES["file"]["size"] / 1024) . " Kb<br />"; echo "Stored in: " . $_FILES["file"]["tmp_name"] . "<br />"; $string = file_get_contents( $_FILES["file"]["tmp_name"] ); foreach ( explode( "\n", $string ) as $userString ) { $userString = trim( $userString ); echo $userString; $array = explode( ';', $userString ); $userdata = array ( 'cf_id'=>trim( $array[0] ), 'text_0'=>trim( $array[1] ), 'text_1'=>trim( $array[2] ), 'text_2'=>trim( $array[3] ), 'text_3'=>trim($array[4]), 'text_6'=>trim($array[5]), 'text_7'=>trim($array[6]), 'text_9'=>trim($array[7]), 'text_12'=>trim($array[8])); global $db; $db =& JFactory::getDBO(); $query = 'INSERT INTO #__chronoforms_UploadAuthor VALUES ("'.$userdata['cf_id'].'","'.$userdata['text_0'].'","'.$userdata['text_1'].'","'.$userdata['text_2'].'","'.$userdata['text_3'].'","'.$userdata['text_6'].'","'.$userdata['text_7'].'","'.$userdata['text_9'].'")'; $db->Execute($query); echo "<br>"; } i am trying to insert the contents into the table #__chronoforms_UploadAuthor in Joomla database . it doent shows any error but it is not inserted into the database.. Please help me.. Y the contents are not inserted into the database...And how to make it inserted into the database..

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  • how to list out the submited data in same page where form submitted?

    - by OM The Eternity
    I have a form with 3 text values and one image.. I want to save these values such that i can display these records in the list below.. how can i do that... I am using osCommerce For example: <form method="post" id="fm-form" action ="" enctype="multipart/form-data"> <label>Name:</label> <input type="text" id="fm-name" name="fm-name" value="" /> <label>Email:</label> <input type="text" id="fm-email" name="fm-email" value="" /> <label>Birthdate:</label> <input type="text" id="fm-birthdate" name="fm-birthdate" value="" /> <input type="file" id="fm-image" name="fm-image"/> <input type="submit" id="fm-submit" value="Save it"> </form> <table width="100%" border="0" cellspacing="0" cellpadding="0"> <tr > <td align="center" class="productListing-heading">Product(s)</td> <td align="center" class="productListing-heading">Edit</td> <td align="center" class="productListing-heading">Delete</td> </tr> <?php for($i=0;$i<$count_image;$i++){?> <tr> <td align="left" class="productListing-data1"> <?php echo tep_image(DIR_WS_IMAGES . $file_realname, $save_image[$i], '110', '110');?> </td> <td align="center" class="productListing-data1">Edit</td> <td align="center" class="productListing-data1">Delete</td> </tr> <tr><td>&nbsp;</td></tr> <?php }?> </table> In the above format as the form is submitted the image has to be stored in a count_image array variable... and the on its count, the list below the form is displayed.. but i cannot get it worked.. could u pls help in doing this...

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  • Using the HTML5 &lt;input type=&quot;file&quot; multiple=&quot;multiple&quot;&gt; Tag in ASP.NET

    - by Rick Strahl
    Per HTML5 spec the <input type="file" /> tag allows for multiple files to be picked from a single File upload button. This is actually a very subtle change that's very useful as it makes it much easier to send multiple files to the server without using complex uploader controls. Please understand though, that even though you can send multiple files using the <input type="file" /> tag, the process of how those files are sent hasn't really changed - there's still no progress information or other hooks that allow you to automatically make for a nicer upload experience without additional libraries or code. For that you will still need some sort of library (I'll post an example in my next blog post using plUpload). All the new features allow for is to make it easier to select multiple images from disk in one operation. Where you might have required many file upload controls before to upload several files, one File control can potentially do the job. How it works To create a file input box that allows with multiple file support you can simply do:<form method="post" enctype="multipart/form-data"> <label>Upload Images:</label> <input type="file" multiple="multiple" name="File1" id="File1" accept="image/*" /> <hr /> <input type="submit" id="btnUpload" value="Upload Images" /> </form> Now when the file open dialog pops up - depending on the browser and whether the browser supports it - you can pick multiple files. Here I'm using Firefox using the thumbnail preview I can easily pick images to upload on a form: Note that I can select multiple images in the dialog all of which get stored in the file textbox. The UI for this can be different in some browsers. For example Chrome displays 3 files selected as text next to the Browse… button when I choose three rather than showing any files in the textbox. Most other browsers display the standard file input box and display the multiple filenames as a comma delimited list in the textbox. Note that you can also specify the accept attribute in the <input> tag, which specifies a mime-type to specify what type of content to allow.Here I'm only allowing images (image/*) and the browser complies by just showing me image files to display. Likewise I could use text/* for all text formats registered on the machine or text/xml to only show XML files (which would include xml,xst,xsd etc.). Capturing Files on the Server with ASP.NET When you upload files to an ASP.NET server there are a couple of things to be aware of. When multiple files are uploaded from a single file control, they are assigned the same name. In other words if I select 3 files to upload on the File1 control shown above I get three file form variables named File1. This means I can't easily retrieve files by their name:HttpPostedFileBase file = Request.Files["File1"]; because there will be multiple files for a given name. The above only selects the first file. Instead you can only reliably retrieve files by their index. Below is an example I use in app to capture a number of images uploaded and store them into a database using a business object and EF 4.2.for (int i = 0; i < Request.Files.Count; i++) { HttpPostedFileBase file = Request.Files[i]; if (file.ContentLength == 0) continue; if (file.ContentLength > App.Configuration.MaxImageUploadSize) { ErrorDisplay.ShowError("File " + file.FileName + " is too large. Max upload size is: " + App.Configuration.MaxImageUploadSize); return View("UploadClassic",model); } var image = new ClassifiedsBusiness.Image(); var ms = new MemoryStream(16498); file.InputStream.CopyTo(ms); image.Entered = DateTime.Now; image.EntryId = model.Entry.Id; image.ContentType = "image/jpeg"; image.ImageData = ms.ToArray(); ms.Seek(0, SeekOrigin.Begin); // resize image if necessary and turn into jpeg Bitmap bmp = Imaging.ResizeImage(ms.ToArray(), App.Configuration.MaxImageWidth, App.Configuration.MaxImageHeight); ms.Close(); ms = new MemoryStream(); bmp.Save(ms,ImageFormat.Jpeg); image.ImageData = ms.ToArray(); bmp.Dispose(); ms.Close(); model.Entry.Images.Add(image); } This works great and also allows you to capture input from multiple input controls if you are dealing with browsers that don't support multiple file selections in the file upload control. The important thing here is that I iterate over the files by index, rather than using a foreach loop over the Request.Files collection. The files collection returns key name strings, rather than the actual files (who thought that was good idea at Microsoft?), and so that isn't going to work since you end up getting multiple keys with the same name. Instead a plain for loop has to be used to loop over all files. Another Option in ASP.NET MVC If you're using ASP.NET MVC you can use the code above as well, but you have yet another option to capture multiple uploaded files by using a parameter for your post action method.public ActionResult Save(HttpPostedFileBase[] file1) { foreach (var file in file1) { if (file.ContentLength < 0) continue; // do something with the file }} Note that in order for this to work you have to specify each posted file variable individually in the parameter list. This works great if you have a single file upload to deal with. You can also pass this in addition to your main model to separate out a ViewModel and a set of uploaded files:public ActionResult Edit(EntryViewModel model,HttpPostedFileBase[] uploadedFile) You can also make the uploaded files part of the ViewModel itself - just make sure you use the appropriate naming for the variable name in the HTML document (since there's Html.FileFor() extension). Browser Support You knew this was coming, right? The feature is really nice, but unfortunately not supported universally yet. Once again Internet Explorer is the problem: No shipping version of Internet Explorer supports multiple file uploads. IE10 supposedly will, but even IE9 does not. All other major browsers - Chrome, Firefox, Safari and Opera - support multi-file uploads in their latest versions. So how can you handle this? If you need to provide multiple file uploads you can simply add multiple file selection boxes and let people either select multiple files with a single upload file box or use multiples. Alternately you can do some browser detection and if IE is used simply show the extra file upload boxes. It's not ideal, but either one of these approaches makes life easier for folks that use a decent browser and leaves you with a functional interface for those that don't. Here's a UI I recently built as an alternate uploader with multiple file upload buttons: I say this is my 'alternate' uploader - for my primary uploader I continue to use an add-in solution. Specifically I use plUpload and I'll discuss how that's implemented in my next post. Although I think that plUpload (and many of the other packaged JavaScript upload solutions) are a better choice especially for large uploads, for simple one file uploads input boxes work well enough. The advantage of this solution is that it's very easy to handle on the server side. Any of the JavaScript controls require special handling for uploads which I'll also discuss in my next post.© Rick Strahl, West Wind Technologies, 2005-2012Posted in HTML5  ASP.NET  MVC   Tweet !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0];if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src="//platform.twitter.com/widgets.js";fjs.parentNode.insertBefore(js,fjs);}}(document,"script","twitter-wjs"); (function() { var po = document.createElement('script'); po.type = 'text/javascript'; po.async = true; po.src = 'https://apis.google.com/js/plusone.js'; var s = document.getElementsByTagName('script')[0]; s.parentNode.insertBefore(po, s); })();

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  • How to use Azure storage for uploading and displaying pictures.

    - by Magnus Karlsson
    Basic set up of Azure storage for local development and production. This is a somewhat completion of the following guide from http://www.windowsazure.com/en-us/develop/net/how-to-guides/blob-storage/ that also involves a practical example that I believe is commonly used, i.e. upload and present an image from a user.   First we set up for local storage and then we configure for them to work on a web role. Steps: 1. Configure connection string locally. 2. Configure model, controllers and razor views.   1. Setup connectionsstring 1.1 Right click your web role and choose “Properties”. 1.2 Click Settings. 1.3 Add setting. 1.4 Name your setting. This will be the name of the connectionstring. 1.5 Click the ellipsis to the right. (the ellipsis appear when you mark the area. 1.6 The following window appears- Select “Windows Azure storage emulator” and click ok.   Now we have a connection string to use. To be able to use it we need to make sure we have windows azure tools for storage. 2.1 Click Tools –> Library Package manager –> Manage Nuget packages for solution. 2.2 This is what it looks like after it has been added.   Now on to what the code should look like. 3.1 First we need a view which collects images to upload. Here Index.cshtml. 1: @model List<string> 2:  3: @{ 4: ViewBag.Title = "Index"; 5: } 6:  7: <h2>Index</h2> 8: <form action="@Url.Action("Upload")" method="post" enctype="multipart/form-data"> 9:  10: <label for="file">Filename:</label> 11: <input type="file" name="file" id="file1" /> 12: <br /> 13: <label for="file">Filename:</label> 14: <input type="file" name="file" id="file2" /> 15: <br /> 16: <label for="file">Filename:</label> 17: <input type="file" name="file" id="file3" /> 18: <br /> 19: <label for="file">Filename:</label> 20: <input type="file" name="file" id="file4" /> 21: <br /> 22: <input type="submit" value="Submit" /> 23: 24: </form> 25:  26: @foreach (var item in Model) { 27:  28: <img src="@item" alt="Alternate text"/> 29: } 3.2 We need a controller to receive the post. Notice the “containername” string I send to the blobhandler. I use this as a folder for the pictures for each user. If this is not a requirement you could just call it container or anything with small characters directly when creating the container. 1: public ActionResult Upload(IEnumerable<HttpPostedFileBase> file) 2: { 3: BlobHandler bh = new BlobHandler("containername"); 4: bh.Upload(file); 5: var blobUris=bh.GetBlobs(); 6: 7: return RedirectToAction("Index",blobUris); 8: } 3.3 The handler model. I’ll let the comments speak for themselves. 1: public class BlobHandler 2: { 3: // Retrieve storage account from connection string. 4: CloudStorageAccount storageAccount = CloudStorageAccount.Parse( 5: CloudConfigurationManager.GetSetting("StorageConnectionString")); 6: 7: private string imageDirecoryUrl; 8: 9: /// <summary> 10: /// Receives the users Id for where the pictures are and creates 11: /// a blob storage with that name if it does not exist. 12: /// </summary> 13: /// <param name="imageDirecoryUrl"></param> 14: public BlobHandler(string imageDirecoryUrl) 15: { 16: this.imageDirecoryUrl = imageDirecoryUrl; 17: // Create the blob client. 18: CloudBlobClient blobClient = storageAccount.CreateCloudBlobClient(); 19: 20: // Retrieve a reference to a container. 21: CloudBlobContainer container = blobClient.GetContainerReference(imageDirecoryUrl); 22: 23: // Create the container if it doesn't already exist. 24: container.CreateIfNotExists(); 25: 26: //Make available to everyone 27: container.SetPermissions( 28: new BlobContainerPermissions 29: { 30: PublicAccess = BlobContainerPublicAccessType.Blob 31: }); 32: } 33: 34: public void Upload(IEnumerable<HttpPostedFileBase> file) 35: { 36: // Create the blob client. 37: CloudBlobClient blobClient = storageAccount.CreateCloudBlobClient(); 38: 39: // Retrieve a reference to a container. 40: CloudBlobContainer container = blobClient.GetContainerReference(imageDirecoryUrl); 41: 42: if (file != null) 43: { 44: foreach (var f in file) 45: { 46: if (f != null) 47: { 48: CloudBlockBlob blockBlob = container.GetBlockBlobReference(f.FileName); 49: blockBlob.UploadFromStream(f.InputStream); 50: } 51: } 52: } 53: } 54: 55: public List<string> GetBlobs() 56: { 57: // Create the blob client. 58: CloudBlobClient blobClient = storageAccount.CreateCloudBlobClient(); 59: 60: // Retrieve reference to a previously created container. 61: CloudBlobContainer container = blobClient.GetContainerReference(imageDirecoryUrl); 62: 63: List<string> blobs = new List<string>(); 64: 65: // Loop over blobs within the container and output the URI to each of them 66: foreach (var blobItem in container.ListBlobs()) 67: blobs.Add(blobItem.Uri.ToString()); 68: 69: return blobs; 70: } 71: } 3.4 So, when the files have been uploaded we will get them to present them to out user in the index page. Pretty straight forward. In this example we only present the image by sending the Uri’s to the view. A better way would be to save them up in a view model containing URI, metadata, alternate text, and other relevant information but for this example this is all we need.   4. Now press F5 in your solution to try it out. You can see the storage emulator UI here:     4.1 If you get any exceptions or errors I suggest to first check if the service Is running correctly. I had problem with this and they seemed related to the installation and a reboot fixed my problems.     5. Set up for Cloud storage. To do this we need to add configuration for cloud just as we did for local in step one. 5.1 We need our keys to do this. Go to the windows Azure menagement portal, select storage icon to the right and click “Manage keys”. (Image from a different blog post though).   5.2 Do as in step 1.but replace step 1.6 with: 1.6 Choose “Manually entered credentials”. Enter your account name. 1.7 Paste your Account Key from step 5.1. and click ok.   5.3. Save, publish and run! Please feel free to ask any questions using the comments form at the bottom of this page. I will get back to you to help you solve any questions. Our consultancy agency also provides services in the Nordic regions if you would like any further support.

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  • Image Preview in ASP.NET MVC

    - by imran_ku07
      Introduction :         Previewing an image is a great way to improve the UI of your site. Also it is always best to check the file type, size and see a preview before submitting the whole form. There are some ways to do this using simple JavaScript but not work in all browsers (like FF3).In this Article I will show you how do this using ASP.NET MVC application. You also see how this will work in case of nested form.   Description :          Create a new ASP.NET MVC project and then add a file upload and image control into your View. <form id="form1" method="post" action="NerdDinner/ImagePreview/AjaxSubmit">            <table>                <tr>                    <td>                        <input type="file" name="imageLoad1" id="imageLoad1"  onchange="ChangeImage(this,'#imgThumbnail')" />                    </td>                </tr>                <tr>                    <td align="center">                        <img src="images/TempImage.gif" id="imgThumbnail" height="200px" width="200px">                     </td>                </tr>            </table>        </form>           Note that here NerdDinner is refers to the virtual directory name, ImagePreview is the Controller and ImageLoad is the action name which you will see shortly          I will use the most popular jQuery form plug-in, that turns a form into an AJAX form with very little code. Therefore you must get these from Jquery site and then add these files into your page.          <script src="NerdDinner/Scripts/jquery-1.3.2.js" type="text/javascript"></script>        <script src="NerdDinner/Scripts/jquery.form.js" type="text/javascript"></script>            Then add the javascript function. <script type="text/javascript">function ChangeImage(fileId,imageId){ $("#form1").ajaxSubmit({success: function(responseText){ var d=new Date(); $(imageId)[0].src="NerdDinner/ImagePreview/ImageLoad?a="+d.getTime(); } });}</script>             This function simply submit the form named form1 asynchronously to ImagePreviewController's method AjaxSubmit and after successfully receiving the response, it will set the image src property to the action method ImageLoad. Here I am also adding querystring, preventing the browser to serve the cached image.           Now I will create a new Controller named ImagePreviewController. public class ImagePreviewController : Controller { [AcceptVerbs(HttpVerbs.Post)] public ActionResult AjaxSubmit(int? id) { Session["ContentLength"] = Request.Files[0].ContentLength; Session["ContentType"] = Request.Files[0].ContentType; byte[] b = new byte[Request.Files[0].ContentLength]; Request.Files[0].InputStream.Read(b, 0, Request.Files[0].ContentLength); Session["ContentStream"] = b; return Content( Request.Files[0].ContentType+";"+ Request.Files[0].ContentLength ); } public ActionResult ImageLoad(int? id) { byte[] b = (byte[])Session["ContentStream"]; int length = (int)Session["ContentLength"]; string type = (string)Session["ContentType"]; Response.Buffer = true; Response.Charset = ""; Response.Cache.SetCacheability(HttpCacheability.NoCache); Response.ContentType = type; Response.BinaryWrite(b); Response.Flush(); Session["ContentLength"] = null; Session["ContentType"] = null; Session["ContentStream"] = null; Response.End(); return Content(""); } }             The AjaxSubmit action method will save the image in Session and return content type and content length in response. ImageLoad action method will return the contents of image in response.Then clear these Sessions.           Just run your application and see the effect.   Checking Size and Content Type of File:          You may notice that AjaxSubmit action method is returning both content type and content length. You can check both properties before submitting your complete form.     $(myform).ajaxSubmit({success: function(responseText)            {                                var contentType=responseText.substring(0,responseText.indexOf(';'));                var contentLength=responseText.substring(responseText.indexOf(';')+1);                // Here you can do your validation                var d=new Date();                $(imageId)[0].src="http://weblogs.asp.net/MoneypingAPP/ImagePreview/ImageLoad?a="+d.getTime();            }        });  Handling Nested Form Case:          The above code will work if you have only one form. But this is not the case always.You may have a form control which wraps all the controls and you do not want to submit the whole form, just for getting a preview effect.           In this case you need to create a dynamic form control using JavaScript, and then add file upload control to this form and submit the form asynchronously  function ChangeImage(fileId,imageId)         {            var myform=document.createElement("form");                    myform.action="NerdDinner/ImagePreview/AjaxSubmit";            myform.enctype="multipart/form-data";            myform.method="post";            var imageLoad=document.getElementById(fileId).cloneNode(true);            myform.appendChild(imageLoad);            document.body.appendChild(myform);            $(myform).ajaxSubmit({success: function(responseText)                {                                    var contentType=responseText.substring(0,responseText.indexOf(';'));                    var contentLength=responseText.substring(responseText.indexOf(';')+1);                    var d=new Date();                    $(imageId)[0].src="http://weblogs.asp.net/MoneypingAPP/ImagePreview/ImageLoad?a="+d.getTime();                    document.body.removeChild(myform);                }            });        }            You also need append the child in order to send request and remove them after receiving response.

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  • ASP.NET MVC File Upload Error - "The input is not a valid Base-64 string"

    - by Justin
    Hey all, I'm trying to add a file upload control to my ASP.NET MVC 2 form but after I select a jpg and click Save, it gives the following error: The input is not a valid Base-64 string as it contains a non-base 64 character, more than two padding characters, or a non-white space character among the padding characters. Here's the view: <% using (Html.BeginForm("Save", "Developers", FormMethod.Post, new {enctype = "multipart/form-data"})) { %> <%: Html.ValidationSummary(true) %> <fieldset> <legend>Fields</legend> <div class="editor-label"> Login Name </div> <div class="editor-field"> <%: Html.TextBoxFor(model => model.LoginName) %> <%: Html.ValidationMessageFor(model => model.LoginName) %> </div> <div class="editor-label"> Password </div> <div class="editor-field"> <%: Html.Password("Password") %> <%: Html.ValidationMessageFor(model => model.Password) %> </div> <div class="editor-label"> First Name </div> <div class="editor-field"> <%: Html.TextBoxFor(model => model.FirstName) %> <%: Html.ValidationMessageFor(model => model.FirstName) %> </div> <div class="editor-label"> Last Name </div> <div class="editor-field"> <%: Html.TextBoxFor(model => model.LastName) %> <%: Html.ValidationMessageFor(model => model.LastName) %> </div> <div class="editor-label"> Photo </div> <div class="editor-field"> <input id="Photo" name="Photo" type="file" /> </div> <p> <%: Html.Hidden("DeveloperID") %> <%: Html.Hidden("CreateDate") %> <input type="submit" value="Save" /> </p> </fieldset> <% } %> And the controller: //POST: /Secure/Developers/Save/ [AcceptVerbs(HttpVerbs.Post)] public ActionResult Save(Developer developer) { //get profile photo. var upload = Request.Files["Photo"]; if (upload.ContentLength > 0) { string savedFileName = Path.Combine( ConfigurationManager.AppSettings["FileUploadDirectory"], "Developer_" + developer.FirstName + "_" + developer.LastName + ".jpg"); upload.SaveAs(savedFileName); } developer.UpdateDate = DateTime.Now; if (developer.DeveloperID == 0) {//inserting new developer. DataContext.DeveloperData.Insert(developer); } else {//attaching existing developer. DataContext.DeveloperData.Attach(developer); } //save changes. DataContext.SaveChanges(); //redirect to developer list. return RedirectToAction("Index"); } Thanks, Justin

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  • Post data to aspx page from iphone application

    - by Dipen
    Hi, I am developing a application. In which i am posting a image to .aspx page.The HTML for the page is as below. <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd>"> <html xmlns="http://www.w3.org/1999/xhtml>"> <head><title> Untitled Page </title><link href="App_Themes/XXX/XXX.css" type="text/css" rel="stylesheet" /></head> <body> <form name="form1" method="post" action="Default16.aspx" id="form1" enctype="multipart/form-data"> <div> <input type="hidden" name="__VIEWSTATE" id="__VIEWSTATE" value="/wEPDwUKLTQwMjY2MDA0Mw9kFgICAw8WAh4HZW5jdHlwZQUTbXVsdGlwYXJ0L2Zvcm0tZGF0YWRktr+hG1VVXZsO01PCyj61d6Ulqy8=" /> </div> <div> <div style="float:left;margin:10px"> <input type="file" name="fuImage" id="fuImage" /> </div> <div style="float:right"> <input type="submit" name="btnPost" value="Post Image" id="btnPost" /> </div> </div> </form> </body> </html> Now i am sending a request from my application then i am getting " Error Domain=kCFErrorDomainCFNetwork Code=303 UserInfo=0xf541c0 "Operation could not be completed. (kCFErrorDomainCFNetwork error 303.)"" I have tried using SynchronousRequest and aSynchronousRequest but both are not working. I have also used apple sample code. Here is the code for iPhone app UIImage *image=[UIImage imageNamed:@"photo2.jpg"]; NSData *imageData = UIImageJPEGRepresentation(image, 90); NSString *urlString = @"http://XXXXXXXX.com/Post.aspx"; NSMutableURLRequest *request = [[[NSMutableURLRequest alloc] init] autorelease]; [request setURL:[NSURL URLWithString:urlString]]; [request setHTTPMethod:@"POST"]; NSString *boundary = [NSString stringWithString:@"----WebKitFormBoundarylU9pAl5wPrF+Tk52"]; NSString *contentType = [NSString stringWithFormat:@"multipart/form-data; boundary=%@", boundary]; [request addValue:contentType forHTTPHeaderField:@"Content-Type"]; NSMutableData *body = [NSMutableData data]; [body appendData:[[NSString stringWithFormat:@"\r\n--%@\r\n", boundary] dataUsingEncoding:NSUTF8StringEncoding]]; [body appendData:[[NSString stringWithString:@"Content-Disposition: form-data; name=\"fuimage\"; filename=\"asd.jpg\"\r\n"] dataUsingEncoding:NSUTF8StringEncoding]]; [body appendData:[[NSString stringWithString:@"Content-Type: image/jpeg\r\n\r\n"] dataUsingEncoding:NSUTF8StringEncoding]]; [body appendData:[NSData dataWithData:imageData]]; [body appendData:[[NSString stringWithFormat:@"\r\n--%@--\r\n", boundary] dataUsingEncoding:NSUTF8StringEncoding]]; [request setHTTPBody:body]; // NSData *returnData = [NSURLConnection sendSynchronousRequest:request returningResponse:nil error:nil]; // NSString *returnString = [[NSString alloc] initWithData:returnData encoding:NSUTF8StringEncoding]; // NSLog(returnString); NSURLConnection *theConnection = [[NSURLConnection alloc] initWithRequest:request delegate:self]; if( theConnection ) { webData = [[NSMutableData data] retain]; } else { NSLog(@"theConnection is NULL"); } Thanks in Advance.

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  • HttpWebRequest: How to find a postal code at Canada Post through a WebRequest with x-www-form-enclos

    - by Will Marcouiller
    I'm currently writing some tests so that I may improve my skills with the Internet interaction through Windows Forms. One of those tests is to find a postal code which should be returned by Canada Post website. My default URL setting is set to: http://www.canadapost.ca/cpotools/apps/fpc/personal/findByCity?execution=e4s1 The required form fields are: streetNumber, streetName, city, province The contentType is "application/x-www-form-enclosed" EDIT: Please consider the value "application/x-www-form-encoded" instead of point 3 value as the contentType. (Thanks EricLaw-MSFT!) The result I get is not the result expected. I get the HTML source code of the page where I could manually enter the information to find the postal code, but not the HTML source code with the found postal code. Any idea of what I'm doing wrong? Shall I consider going the XML way? Is it first of all possible to search on Canada Post anonymously? Here's a code sample for better description: public static string FindPostalCode(ICanadadianAddress address) { var postData = string.Concat(string.Format("&streetNumber={0}", address.StreetNumber) , string.Format("&streetName={0}", address.StreetName) , string.Format("&city={0}", address.City) , string.Format("&province={0}", address.Province)); var encoding = new ASCIIEncoding(); byte[] postDataBytes = encoding.GetBytes(postData); request = (HttpWebRequest)WebRequest.Create(DefaultUrlSettings); request.ImpersonationLevel = System.Security.Principal.TokenImpersonationLevel.Anonymous; request.Container = new CookieContainer(); request.Timeout = 10000; request.ContentType = contentType; request.ContentLength = postDataBytes.LongLength; request.Method = @"post"; var senderStream = new StreamWriter(request.GetRequestStream()); senderStream.Write(postDataBytes, 0, postDataBytes.Length); senderStream.Close(); string htmlResponse = new StreamReader(request.GetResponse().GetResponseStream()).ReadToEnd(); return processedResult(htmlResponse); // Processing the HTML source code parsing, etc. } I seem stuck in a bottle neck in my point of view. I find no way out to the desired result. EDIT: There seems to have to parameters as for the ContentType of this site. Let me explain. There's one with the "meta"-variables which stipulates the following: meta http-equiv="Content-Type" content="application/xhtml+xml, text/xml, text/html; charset=utf-8" And another one later down the code that is read as: form id="fpcByAdvancedSearch:fpcSearch" name="fpcByAdvancedSearch:fpcSearch" method="post" action="/cpotools/apps/fpc/personal/findByCity?execution=e1s1" enctype="application/x-www-form-urlencoded" My question is the following: With which one do I have to stick? Let me guess, the first ContentType is to be considered as the second is only for another request to a function or so when the data is posted? EDIT: As per request, the closer to the solution I am is listed under this question: WebRequest: How to find a postal code using a WebRequest against this ContentType=”application/xhtml+xml, text/xml, text/html; charset=utf-8”? Thanks for any help! :-)

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  • Animated background image in a hidden <div> doesn't load or loads not animated

    - by Guanche
    Hello, I have spent the whole day trying to make a script which on "submit" hides the form and shows hidden with animated progress bar. The problem is that Internet Explorer doesn't show animated gif images in hidden divs. The images are static. I visited many websites and found a script which uses: document.getElementById(id).style.backgroundImage = 'url(/images/load.gif)'; Finally, my script works in Internet Explorer, Firefox, Opera but... Google Chrome doesn't display the image at all. I see only div text. After many tests I discovered the following: the only way to see the background image in Google Chrome is to include the same image somewhere in the page (outside of hidden div) with 1px dimensions: <img src="/images/load.gif" width="1" heigh="1" /> This did the trick but... after this dirty solution Microsoft Explorer for some reason shows the image as static again. So, my question is: is there any way how to force Gogle Chrome to show the image? Thanks. This is my script: <script language="JavaScript" type="text/javascript"> function ver (id, elementId){ if (document.getElementById('espera').style.visibility == "visible") { return false; }else{ var esplit = document.forms[0]['userfile'].value.split("."); ext = esplit[esplit.length-1]; if (document.forms[0]['userfile'].value == '') { alert('Please select a file'); return false; }else{ if ((ext.toLowerCase() == 'jpg')) { document.getElementById(id).style.position = 'absolute'; document.getElementById(id).style.display = 'inline'; document.getElementById(id).style.visibility = "visible"; document.getElementById(id).style.backgroundImage = 'url(/images/load.gif)'; document.getElementById(id).style.height = "100px"; document.getElementById(id).style.backgroundColor = '#f3f3f3'; document.getElementById(id).style.backgroundRepeat = "no-repeat"; document.getElementById(id).style.backgroundPosition = "50% 50%"; var element; if (document.all) element = document.all[elementId]; else if (document.getElementById) element = document.getElementById(elementId); if (element && element.style) element.style.display = 'none'; return true; }else{ alert('This is not a jpg file'); return false; } } } } </script> <div id="frmDiv"> <form enctype="multipart/form-data" action="/upload.php" method="post" name="upload3" onsubmit="return ver('espera','frmDiv');"> <input type="hidden" name="max_file_size" value="4194304" /> <table border="0" cellspacing="1" cellpadding="2" width="100%"> <tr bgcolor="#f5f5f5"> <td>File (jpg)</td> <td> <input type="file" name="userfile" class="upf" /></td></tr> <tr bgcolor="#f5f5f5"> <td>&nbsp;</td> <td> <input class="upf2" type="submit" name="add" value="Upload" /> </td></tr></table></form> </div> <div id="espera" style="display:none;text-align:center;float:left;width:753px;">&nbsp;<br />&nbsp;<br />&nbsp;<br />&nbsp;<br /> &nbsp;<br />Please wait...<br />&nbsp; </div>

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  • How to use metaWeblog.newPost (xmlrpc api) properly with PHP ?

    - by Sjne
    Update: solved this problem see answer I want to make new posts on my blog remotely with XMLRPC api and i m trying to use metaWeblog.newPost function, because it provides much features. I successfully added new posts into wordpress but i failed to post it in a defined category(categories) i tried lots of various things but failed at end now i m using code from this site http://www.samuelfolkes.com/2009/08/posting-to-wordpress-with-php-and-xml-rpc/ after stripping down the code for my needs here's what i got and its working fine remotepost.class.php <?php class remotePost { private $client; private $wpURL = 'http://localhost/wp/xmlrpc.php '; private $ixrPath = '/wp-includes/class-IXR.php'; private $uname = 'zxc'; private $pass = 'zxc'; public $postID; function __construct($content) { if(!is_array($content)) throw new Exception('Invalid Argument'); include $this->ixrPath; $this->client = new IXR_Client($this->wpURL); $this->postID = $this->postContent($content); } private function postContent($content) { $content['description'] = $content['description']; if(!$this->client->query('metaWeblog.newPost','',$this->uname,$this->pass,$content,true)) throw new Exception($this->client->getErrorMessage()); return $this->client->getResponse(); } } ?> post.php ( you can name it whatever you want ) <?php if(isset($_POST['submit'])) { include "remotepost.class.php"; $content['title'] = $_POST['title']; $content['categories'] = $_POST['category']; $content['description'] = $_POST['description']; try { $posted = new remotePost($content); $pid = $posted->postID; } catch(Exception $e) { echo $e->getMessage(); } } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>WordPress Poster</title> </head> <body> <?php if(isset($_POST['submit'])) echo "Posted! <a href=\"http://localhost/wp/?p=$pid\">View Post</a><br /><br />"; ?> <form enctype="multipart/form-data" method="post" action="#"> Title <input type="text" name="title" /> <br /> Category <input type="text" name="category" /> <br /> Description <input type="text" name="description" /> <br /> <input type="submit" value="Submit" name="submit" /> </form> </body> </html> dont know whats wrong in this code :( ,failing to post in right directory

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  • Simple PHP upload form not working

    - by Lea
    Hi all, I seem to run into so many issues when dealing with writing to directories. If someone could please look over this script, and tell me why it's not working, I'd be so appreciative. Upon uploading the file from the form, I don't get anything.. It doesnt output any errors, it simply just refreshes. Thanks, Lea <?php include ('config.php'); if( isset($_POST['submit']) ) { $target = ''.$_SERVER['DOCUMENT_ROOT'].'/images/'; $target = $target . basename( $_FILES['photo']['name']) ; $url = basename( $_FILES['photo']['name']) ; if(move_uploaded_file($_FILES['photo']['tmp_name'], $target)) { $moved = "File has been moved to location $target"; $name = mysql_real_escape_string($_POST['photoname']); mysql_query("INSERT INTO photos (photo_name, photo_image) VALUES ('$name', '$url' )") or die(mysql_error()); $success = "Photo has been added!"; } else { $moved = "File has not been moved to $target"; $success = "Failed to upload:("; } } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Photo Upload</title> <meta name="robots" content="index, follow" /> <meta name="keywords" content="" /> <meta name="description" content="" /> <link href="<?php echo $globalurl; ?>styles.css" rel="stylesheet" type="text/css" /> </head> <body> <div class="holder"> <br /><br /> <b>Add a new photo</b> <hr /> <br /> <b><?php echo "$success<br />"; ?> <?php echo "$moved<br />"; ?></b> <form enctype="multipart/form-data" method="post" action="<?php echo $PHP_SELF; ?>"> <table cellspacing="0" cellpadding="0" width="500px"> <tbody> <tr> <td valign="top">Photo Name:</td> <td valign="top"> <input type="text" name="photoname" /><br /> <br /> </td> </tr> <tr> <td valign="top">Photo:</td> <td valign="top"> <input type="file" name="photo"><br /> <br /> </td> </tr> </tbody> </table> <input type="submit" value="submit" /> </form> </div> </body> </html>

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  • Insert HTML into a page with AJAX

    - by Silvio Iannone
    Hi there, i'm currently developing a website and i need that the pages loads dinamicallly based on what actions the user does. Example: if the user clicks on the button 'Settings' an ajax funcion will load from an external page the code and will put into the div with tag 'settings'. This is the code i use to make the Ajax request: function get_page_content(page, target_id) { xmlhttp = new XMLHttpRequest(); xmlhttp.onreadystatechange = function() { if(xmlhttp.readyState == 4 && xmlhttp.status == 200) { document.getElementById(target_id).innerHTML = xmlhttp.responseText; // After getting the response we have to re-apply ui effects or they // won't be available on new elements coming from request. $('button').sb_animateButton(); $('input').sb_animateInput(); } } xmlhttp.open('GET', 'engine/ajax/get_page_content.php?page=' + page, true); xmlhttp.send(); } And this is where the ajax results will be put by first snippet: <div id="settings_appearance"> </div> The code is called from a function here: <div class="left_menu_item" id="left_menu_settings_appearance" onclick="show_settings_appearance()"> Appearance </div> And this is the html that the ajax function will put into the settings_appearance div: <script type="text/javascript"> $(function() { $('#upload_hidden_frame').hide(); show_mybrain(); document.getElementById('avatar_upload_form').onsubmit = function() { document.getElementById('avatar_upload_form').target = 'upload_hidden_frame'; upload_avatar(); } }); </script> <div class="title">Appearance</div> <iframe id="upload_hidden_frame" name="upload_hidden_frame" src="" class="error_message"></iframe> <table class="sub_container" id="avatar_upload_form" method="post" enctype="multipart/form-data" action="engine/ajax/upload_avatar.php"> <tr> <td><label for="file">Avatar</label></td> <td><input type="file" name="file" id="file" class="file_upload" /></td> <td><button type="submit" name="button_upload">Upload</button></td> </tr> <tr> <td><div class="hint">The image must be in PNG, JPEG or GIF format.</div></td> </tr> </table> I would like to know if there's a way to execute also the javascript code that's returned by the ajax function and if it's possible to apply some customized ui effects i build that are loaded with the main page. Thanks for helping. P.S. This is the script that applies the ui effects: <script type="text/javascript"> // UI effects $(document).ready(function() { $('button').sb_animateButton(); $('input').sb_animateInput(); $('.top_menu_item').sb_animateMenuItem(); $('.top_menu_item_right').sb_animateMenuItem(); $('.left_menu_item').sb_animateMenuItem(); }); </script>

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  • php, mySQL & AJAX: Unable to use sessions across the scripts in the same domain

    - by Devner
    Hi all, I have the following pages: page1.php, page2.php and page3.php. Code in each of them is as below CODE: page1.php <script type="text/javascript"> $(function(){ $('#imgID').upload({ submit_to_url: "page2.php", file_name: 'myfile1', description : "Image", limit : 1, file_types : "*.jpg", }) }); </script> <body> <form action="page3.php" method="post" enctype="multipart/form-data" name="frm1" id="frm1"> //Some other text fields <input type="submit" name="submit" id="submit" value="Submit" /> </form> </body> page2.php <?php session_start(); $a = $_SESSION['a']; $b = $_SESSION['b']; $c = $_SESSION['c']; $res = mysql_query("SELECT col FROM table WHERE col1 = $a AND col2 = $b AND col3 = $c LIMIT 1"); $num_rows = mysql_num_rows($res); echo $num_rows; //echos 0 when in fact it should have been 1 because the data in the Session exists. //Ok let's proceed further //... Do some stuff... //Store some more values and create new session variables (and assume that page1.php is going to be able to use it) $_SESSION['d'] = 'd'; $_SESSION['e'] = 'e'; $_SESSION['f'] = 'f'; if (move_uploaded_file($_FILES['file']['tmp_name'], $file)) { echo "success"; } else { echo "error ".$_FILES['file']['error']; } ?> page3.php <?php session_start(); if( isset($_POST['submit']) ) { //These sessions are non-existent although the AJAX request //to page2.php may have created them when called via AJAX from within page1.php echo $_SESSION['d'].$_SESSION['e'].$_SESSION['f']; ?> } ?> As the code says it I am posting some info via AJAX call from page1.php to page2.php. page2.php is supposed to be able to use the session values from page1.php i.e. $_SESSION['a'], $_SESSION['b'] and $_SESSION['c'] but it does not. Why? How can I fix this? page2.php is creating some more sessions after some processing is done and a response is sent back to page1.php. The submit button of the form on page1.php is hit and the page gets POST'ed to page3.php. But when the SESSION info that gets created in page2.php is echoed, it's blank signifying that SESSIONS from page2.php are not used. How can I fix this? I looked over a lot of information and have spent about 50 hours trying to do different things with my scripts before arriving at the above conclusions. My app. is custom made using function (not OOPS) and does not use any PHP frameworks & I am not even about to use any as my knowledge of OOP concepts is limited any many frameworks are object oriented. I came across race conditions, but the solutions provided don't help too much. One more solution of using DB to hold sessions and seek and retrieve from DB is the last thing on my mind and I really want to avoid creating table, coding and maintaining code for a task as simple as just keeping sessions across pages in the same domain. So my request is: Is there a way that I can solve the above problem(s) via simple coding in present conditions? Any help is appreciated. Thank you.

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  • CSS: Horizontal UL: Getting it centered

    - by Steve
    I'm trying to make a horizontal menu/list. It has a mix of independent buttons and buttons that are wrapped in their own individual forms. With much hacking I got all of the buttons, in forms and not in forms to align horizontally. I haven't been able to get the whole thing to center on the page though. Could someone point out to me what I am not seeing? <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"><head> <meta http-equiv="content-type" content="text/html; charset=ISO-8859-1"> <link rel="shortcut icon" href="http://localhost:7001/nsd/images/favicon.ico"> <link rel="StyleSheet" href="veci_files/nsd.css" type="text/css"> <style type = "text/css"> #horizontal_li_menu_container ul { margin-left: auto; margin-right:auto; text-align:center; border: 1px solid green; width:1000px; } #horizontal_li_menu_container_ul { list-style-type: none; text-decoration: none; border: 1px solid red; } #horizontal_li_menu_container li { display: inline;float:left; } </style> </head> <body> <div id = "horizontal_li_menu_container"> <ul id = "horizontal_li_menu_container_ul"> <li> <input value="Update" onclick="location.href='#'" name="button" type="button"/> </li> <li> <form name="formExportVECI" method="post" action="exportveci"> <input name="person_id" value="661774" type="hidden"> <input name="submitExport" value="Export To Microsoft Excel" type="submit"> </form> </li> <li> <form id="ufdh" name="formImportVECI" action="importveci" method="post" enctype="multipart/form-data"> <input name="person_id" value="661774" type="hidden"> <input value="Import From Microsoft Excel" path="Upload" type="submit"> <input id="fileData" name="fileData" value="" type="file"> </form> </li> <li> <input value="Search/Home" onclick="location.href='search'" name="buttonHome" type="button"/> </li> </ul> </div> </body></html>

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