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  • Cocoa WebView won't render all images on OSX 10.8

    - by user2906962
    I'm currently developing an application for OS X, backwards compatible with OS X 10.6. At some point I create a WebView in which I load html content that I create dynamically. The html content is formed only of image links <img src= and text, there is no javascript or anything of that kind. All the images (there are only 5 png images) are stored locally and their size is 4 KB. The problem I have is that some images (those that are not on the visible side of the "scroll"), the very first time I run the application,the images are not shown unless I drag the window to another screen or load again the view controller that contains the WebView. In those cases the images appear on the "scroll" even if they are offsite. I've tried creating the WebView both with IB and programatically, I've used WebPreferences like Autosaves, AllowsAnimatedImages … I've tried using NSURLCache to load each image so that the WebView will get access to them easier ... same result. Taking into account that my code is quite extensive I'm gonna post only the bits that I think are relevant: NSString *finalHtml ... //contains the complete html CGRect screenRect = [self.fixedView bounds]; CGRect webFrame = CGRectMake(0.0f, 0.0f, screenRect.size.width, screenRect.size.height); self.miwebView=[[WebView alloc] initWithFrame:webFrame]; [self.miwebView setEditable:NO]; [self.miwebView setUIDelegate:self]; ... NSURLCache *URLCache = [[NSURLCache alloc] initWithMemoryCapacity:4 * 1024 * 1024 diskCapacity:20 * 1024 * 1024 diskPath:nil]; [NSURLCache setSharedURLCache:URLCache]; NSString *imagePath = [[NSBundle mainBundle] pathForResource:@"line" ofType:@"png"]; NSURL *resourceUrl = [NSURL URLWithString:imagePath]; NSURLRequest *request = [NSURLRequest requestWithURL:resourceUrl cachePolicy:NSURLRequestUseProtocolCachePolicy timeoutInterval:10.0f]; [URLCache cachedResponseForRequest:request]; ... [self.miwebView setResourceLoadDelegate:self]; WebPreferences *webPref = [[WebPreferences alloc]init]; [webPref setAutosaves:YES]; [webPref setAllowsAnimatedImages:YES]; [webPref setAllowsAnimatedImageLooping:YES]; [self.miwebView setPreferences:webPref]; NSString *pathResult = [[NSBundle mainBundle] bundlePath]; NSURL *baseURLRes = [NSURL fileURLWithPath:pathResult]; [[self.miwebView mainFrame] loadHTMLString:finalHtml baseURL:baseURLRes]; [self.fixedView addSubview:self.miwebView]; I should also mention that if an image is caught somewhere in between the visible and non visible side of the "scroll" only the visible bit of the image is going to be rendered even if the page gets scrolled up ... so I think all this is some rendering issue ... I appreciate your help, thank you!

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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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  • Defined variables and arrays vs functions in php

    - by Frank Presencia Fandos
    Introduction I have some sort of values that I might want to access several times each page is loaded. I can take two different approaches for accessing them but I'm not sure which one is 'better'. Three already implemented examples are several options for the Language, URI and displaying text that I describe here: Language Right now it is configured in this way: lang() is a function that returns different values depending on the argument. Example: lang("full") returns the current language, "English", while lang() returns the abbreviation of the current language, "en". There are many more options, like lang("select"), lang("selectact"), etc that return different things. The code is too long and irrelevant for the case so if anyone wants it just ask for it. Url The $Url array also returns different values depending on the request. The whole array is fully defined in the beginning of the page and used to get shorter but accurate links of the current page. Example: $Url['full'] would return "http://mypage.org/path/to/file.php?page=1" and $Url['file'] would return "file.php". It's useful for action="" within the forms and many other things. There are more values for $Url['folder'], $Url['file'], etc. Same thing about the code, if wanted, just request it. Text [You can skip this section] There's another array called $Text that is defined in the same way than $Url. The whole array is defined at the beginning, making a mysql call and defining all $Text[$i] for current page with a while loop. I'm not sure if this is more efficient than multiple calls for a single mysql cell. Example: $Text['54'] returns "This is just a test array!" which this could perfectly be implemented with a function like text(54). Question With the 3 examples you can see that I use different methods to do almost the same function (no pun intended), but I'm not sure which one should become the standard one for my code. I could create a function called url() and other called text() to output what I want. I think that working with functions in those cases is better, but I'm not sure why. So I'd really appreciate your opinions and advice. Should I mix arrays and functions in the way I described or should I just use funcions? Please, base your answer in this: The source needs to be readable and reusable by other developers Resource consumption (processing, time and memory). The shorter the code the better. The more you explain the reasons the better. Thank you PS, now I know the differences between $Url and $Uri.

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  • PHP & MySQL - saving and looping problems.

    - by R.I.P.coalMINERS
    I'm new to PHP and MySQL I want a user to be able to store multiple names and there meanings in a MySQL database tables named names using PHP I will dynamically create form fields with JQuery every time a user clicks on a link so a user can enter 1 to 1,000,000 different names and there meanings which will be stored in a table called names. Since I asked my last question I figured out how to store my values from my form using the for loop but every time I loop my values when I add one or more dynamic fields the second form field named meaning will not save the value entered also my dynamic form fields keep looping doubling, tripling and so on the entered values into the database it all depends on how many form fields are added dynamically. I was wondering how can I fix these problems? On a side note I replaced the query with echo's to see the values that are being entered. Here is the PHP code. <?php if(isset($_POST['submit'])) { $mysqli = mysqli_connect("localhost", "root", "", "site"); $dbc = mysqli_query($mysqli,"SELECT * FROM names WHERE userID='$userID'"); $name = $_POST['name']; $meaning = $_POST['meaning']; if(isset($name['0']) && mysqli_num_rows($dbc) == 0 && trim($name['0'])!=='' && trim($meaning['0'])!=='') { for($n = 0; $n < count($name); $n++) { for($m = 0; $m < count($meaning); $m++) { echo $name[$n] . '<br />'; echo $meaning[$m] . '<br /><br />'; break; } } } } ?> And here is the HTML code. <form method="post" action="index.php"> <ul> <li><label for="name">Name: </label><input type="text" name="name[]" id="name" /></li> <li><label for="meaning">Meaning: </label><input type="text" name="meaning[]" id="meaning" /></li> <li><input type="submit" name="submit" value="Save" /></li> </ul> </form> If needed I will place the JQuery code.

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  • Not sure what happens to my apps objects when using NSURLSession in background - what state is my app in?

    - by Avner Barr
    More of a general question - I don't understand the workings of NSURLSession when using it in "background session mode". I will supply some simple contrived example code. I have a database which holds objects - such that portions of this data can be uploaded to a remote server. It is important to know which data/objects were uploaded in order to accurately display information to the user. It is also important to be able to upload to the server in a background task because the app can be killed at any point. for instance a simple profile picture object: @interface ProfilePicture : NSObject @property int userId; @property UIImage *profilePicture; @property BOOL successfullyUploaded; // we want to know if the image was uploaded to out server - this could also be a property that is queryable but lets assume this is attached to this object @end Now Lets say I want to upload the profile picture to a remote server - I could do something like: @implementation ProfilePictureUploader -(void)uploadProfilePicture:(ProfilePicture *)profilePicture completion:(void(^)(BOOL successInUploading))completion { NSUrlSession *uploadImageSession = ..... // code to setup uploading the image - and calling the completion handler; [uploadImageSession resume]; } @end Now somewhere else in my code I want to upload the profile picture - and if it was successful update the UI and the database that this action happened: ProfilePicture *aNewProfilePicture = ...; aNewProfilePicture.profilePicture = aImage; aNewProfilePicture.userId = 123; aNewProfilePicture.successfullyUploaded = NO; // write the change to disk [MyDatabase write:aNewProfilePicture]; // upload the image to the server ProfilePictureUploader *uploader = [ProfilePictureUploader ....]; [uploader uploadProfilePicture:aNewProfilePicture completion:^(BOOL successInUploading) { if (successInUploading) { // persist the change to my db. aNewProfilePicture.successfullyUploaded = YES; [Mydase update:aNewProfilePicture]; // persist the change } }]; Now obviously if my app is running then this "ProfilePicture" object is successfully uploaded and all is well - the database object has its own internal workings with data structures/caches and what not. All callbacks that may exist are maintained and the app state is straightforward. But I'm not clear what happens if the app "dies" at some point during the upload. It seems that any callbacks/notifications are dead. According to the API documentation- the uploading is handled by a separate process. Therefor the upload will continue and my app will be awakened at some point in the future to handle completion. But the object "aNewProfilePicture" is non existant at that point and all callbacks/objects are gone. I don't understand what context exists at this point. How am I supposed to ensure consistency in my DB and UI (For instance update the "successfullyUploaded" property for that user)? Do I need to re-work everything touching the DB or UI to correspond with the new API and work in a context free environment?

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  • I am having trouble using jquery to submit a form. It was working before

    - by noah
    When a user clicks a link it uses jquery ajax to submit a form to go to paypal. Not working for some reason. Really appreciate any help... LINK TO CLICK I put this in an href for onClick: javascript:go_paypal(); CODE TO EXECUTE ON CLICK function go_paypal() { data = 'req_paypal=1'; $.blockUI({ message: '<h1> Going to Paypal...</h1>',css:{background:'#000'} }); $.ajax({ type: "POST", url: "index.php", data: data, success: function(data) { $("#paypal_form").html(data); $("#payPalForm").submit(); } , error: function() {$.unblockUI(); alert('Unable to communicate to server.'); } }); return false; } CODE TO GO ON SUBMIT if(isset($_POST['req_paypal']) && $_POST['req_paypal'] == 1 ) { $sql = 'INSERT INTO `transactions` (id,type,ip,time,ammount,status) VALUES (NULL,1,\''.$_SERVER['REMOTE_ADDR'].'\',\''.time().'\',\''.$global['paypal_prod_amount'].'\',0) '; echo $sql; mysql_query($sql); $id = mysql_insert_id(); $html = ' <form action="https://www.sandbox.paypal.com/cgi-bin/webscr" method="post" id="payPalForm"> <input type="hidden" name="item_number" value="One Year of Imgur Pro"> <input type="hidden" name="cmd" value="_xclick"> <input type="hidden" name="no_note" value="1"> <input type="hidden" name="business" value="'.$global['paypal_email'].'"> <input type="hidden" name="custom" value="'.base64_encode($id).'"> <input type="hidden" name="currency_code" value="USD"> <input type="hidden" name="return" value="'.$global['paypal_return'].'"> <input name="item_name" type="hidden" id="item_name" value="One Year of Imgur Pro" > <input name="amount" type="hidden" id="amount" value="'.$global['paypal_prod_amount'].'" > </form> '; echo $html;exit; }

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  • How to create multiple Repository object inside a Repository class using Unit Of Work?

    - by Santosh
    I am newbie to MVC3 application development, currently, we need following Application technologies as requirement MVC3 framework IOC framework – Autofac to manage object creation dynamically Moq – Unit testing Entity Framework Repository and Unit Of Work Pattern of Model class I have gone through many article to explore an basic idea about the above points but still I am little bit confused on the “Repository and Unit Of Work Pattern “. Basically what I understand Unit Of Work is a pattern which will be followed along with Repository Pattern in order to share the single DB Context among all Repository object, So here is my design : IUnitOfWork.cs public interface IUnitOfWork : IDisposable { IPermitRepository Permit_Repository{ get; } IRebateRepository Rebate_Repository { get; } IBuildingTypeRepository BuildingType_Repository { get; } IEEProjectRepository EEProject_Repository { get; } IRebateLookupRepository RebateLookup_Repository { get; } IEEProjectTypeRepository EEProjectType_Repository { get; } void Save(); } UnitOfWork.cs public class UnitOfWork : IUnitOfWork { #region Private Members private readonly CEEPMSEntities context = new CEEPMSEntities(); private IPermitRepository permit_Repository; private IRebateRepository rebate_Repository; private IBuildingTypeRepository buildingType_Repository; private IEEProjectRepository eeProject_Repository; private IRebateLookupRepository rebateLookup_Repository; private IEEProjectTypeRepository eeProjectType_Repository; #endregion #region IUnitOfWork Implemenation public IPermitRepository Permit_Repository { get { if (this.permit_Repository == null) { this.permit_Repository = new PermitRepository(context); } return permit_Repository; } } public IRebateRepository Rebate_Repository { get { if (this.rebate_Repository == null) { this.rebate_Repository = new RebateRepository(context); } return rebate_Repository; } } } PermitRepository .cs public class PermitRepository : IPermitRepository { #region Private Members private CEEPMSEntities objectContext = null; private IObjectSet<Permit> objectSet = null; #endregion #region Constructors public PermitRepository() { } public PermitRepository(CEEPMSEntities _objectContext) { this.objectContext = _objectContext; this.objectSet = objectContext.CreateObjectSet<Permit>(); } #endregion public IEnumerable<RebateViewModel> GetRebatesByPermitId(int _permitId) { // need to implment } } PermitController .cs public class PermitController : Controller { #region Private Members IUnitOfWork CEEPMSContext = null; #endregion #region Constructors public PermitController(IUnitOfWork _CEEPMSContext) { if (_CEEPMSContext == null) { throw new ArgumentNullException("Object can not be null"); } CEEPMSContext = _CEEPMSContext; } #endregion } So here I am wondering how to generate a new Repository for example “TestRepository.cs” using same pattern where I can create more then one Repository object like RebateRepository rebateRepo = new RebateRepository () AddressRepository addressRepo = new AddressRepository() because , what ever Repository object I want to create I need an object of UnitOfWork first as implmented in the PermitController class. So if I would follow the same in each individual Repository class that would again break the priciple of Unit Of Work and create multiple instance of object context. So any idea or suggestion will be highly appreciated. Thank you

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  • Creating a mySQL query using PHP form dropdowns - If user ignores dropdown, do not filter by that pa

    - by user303043
    Hello, I am creating a simple MySQL query that will be built from the user selecting options from 2 dropdowns. The issue I am having is that I would like each drop down to default that if they do not actually choose an option, do not filter by that dropdown parameter. So, if they come in, and simply hit submit without choosing from a dropdown they should see everything. If they come in and pick from only one of the dropdowns, the query will basically ignore filtering by the other dropdown. I tried making <OPTION VALUE='any'>Choose but my query didn't know what to do with the 'any' and just shows no results. Here is my code. Thank you very much for whatever help you can provide. FORM <form method="POST" action="<?php echo $_SERVER['REQUEST_URI']; ?>"> <select name="GameType"> <OPTION VALUE='any'>Choose Game Type <option value="Game1">Option 1</option> <option value="Game2">Option 2</option> <option value="Game3">Option 3</option> </select> <select name="Instructor"> <OPTION VALUE='any'>Choose Instructor <option VALUE="InstructorA">Instructor A</option> <option value="InstructorB">Instructor B</option> <option value="InstructorC">Instructor C</option> </select> <input type='submit' value='Search Videos'> </form> MYSQL <?PHP connection stuff $db_handle = mysql_connect($server, $user_name, $password); $db_found = mysql_select_db($database, $db_handle); if ($db_found) { $SQL = "SELECT * FROM Videos WHERE GameType=\"{$_POST['GameType']}\" AND Instructor=\"{$_POST['Instructor']}\""; $result = mysql_query($SQL); while ($db_field = mysql_fetch_assoc($result)) { echo $db_field['ShortDescription'] . ", "; echo $db_field['LongDescription'] . ", "; echo $db_field['GameType'] . ", "; echo $db_field['NumberOfPlayers'] . ", "; echo $db_field['Instructor'] . ", "; echo $db_field['Stakes'] . ", "; echo $db_field['UserPermissionLevel'] . ", "; echo $db_field['DateCreated'] . "<BR>"; } mysql_close($db_handle); } else { print "Database NOT Found "; mysql_close($db_handle); } ?>

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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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  • how to invoke an activity of a library project from an android apps

    - by Austin
    I have an open source android code that I need to use in my android apps. It has all the source code as well as resource files, manifest files and class path. It can be compiled as a separate android apps. I have constraints for using the open source. 1. I can't change a single line of code. 2. I can't use it as a separate apps. These constraints are non negotiable. What I have done is I have compiled the open source as class library(in Eclipse: Project Properties-Android- Tick check box Is Library). This has resulted in generation of .class files(in bin) for the java files and resource files. This open source has an android activity that i want to open from my application. So I have linked the directory of these sets of class files in the source section of my java build path( in .classpath). I have declared the activity in my manifest file with proper action intent filters. Now when I am trying to call activity from my code, its not working. Cleaning and rebuilding doesn't help. However, if I build the open source project and my apps in the same workspace of eclipse and link the open source in my apps in exact same manner it works fine. I am not able to identify the difference. All settings seems to be same(all files are identical in both the cases). But only in the second case it works. I have tried it as jar file also. I have build the open source as project library and exported it into a jar file(excluding manifest file). But in that case I am getting the following error UNEXPECTED TOP-LEVEL EXCEPTION: java.lang.IllegalArgumentException: already added: .... Conversion to Dalvik format failed with error 1 This I guess is coming because the android library(2.2) has been included twice in my apps( one for building my apps & another for building the open source). I dont know how to avoid this. Cleaning the project doesn't help. What i require is to use the open source and invoking it's activities in my apps without violating the constraints. If i can use the open source as bunch of .class files then great, or else any other way will do fine. Please look into it and let me know. Thanks

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  • Undefined index: email in C:\wamp\www\emailvalidate.php on line 4

    - by klari
    I am new to Ajax. In my Ajax I get the following error message : Notice: Undefined index: address in C:\wamp\www\test\sample.php on line 11 I googled but I didn't get a solution for my specified issue. Here is what I did. HTML Form with Ajax (test1.php) <!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=utf-8" /> <title>Untitled Document</title> <script> function loadXMLDoc() { var xmlhttp; if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); } else {// code for IE6, IE5 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById("mydiv").innerHTML=xmlhttp.responseText; } } xmlhttp.open("POST","test2.php",true); xmlhttp.send(); } </script> </head> <body> <form id="form1" name="form1" method="post" action="sample.php"> <p> <label for="mail"></label> <input type="text" name="mail" id="mail" onblur="loadXMLDoc()" /> <input type="submit" name="submit" id="submit" value="Submit" /> </p> <p><div id = 'mydiv'></div></p> </form> </body> </html> test2.php <!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=utf-8" /> <title>Untitled Document</title> </head> <body> <?php echo "Your Address is ".$_POST['address']; ?> </body> </html> I am sure it is very simple issue but I don't know how to solve it. Any help ?

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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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  • sendmail.php needs some php work

    - by Chris
    I am having the hardest time to get a simple sendmail.php to work. My form html is <form action=sendmail.php id=contact-form method=post> <p> <label for=cf_name>Name *</label> <input id=cf_name name=cf_name placeholder='Enter your name...' required=required title=Name type=text /> </p> <p> <label for=cf_email>Email *</label> <input id=cf_email name=cf_email placeholder='Email address...' required=required title='Email address' type=email /> </p> <p> <label for=cf_subject>Subject *</label> <input id=cf_subject name=cf_subject placeholder='Specify subject...' required=required title=Subject type=text /> </p> <p> <label for=cf_message>Message *</label> <textarea id=cf_message name=cf_message placeholder='Message text...' required=required rows=10 title='Message text'></textarea> </p> <p> <input type=submit value='Send message'/> </p> </form> And my mailer script is: <? $cf_email = $_POST['cf_email'] ; $cf_message = $_POST['cf_message'] ; $cf_subject = $_POST['cf_subject'] ; $cf_name = $_POST['cf_name'] ; mail( "[email protected]", $cf_subject, $cf_message, $cf_name, $cf_email ); print "Congratulations your email has been sent"; ?> Just want an email to to go to my email. When it appears in the inbox, the subject they typed is the one that I will see as the subject in my inbox. The from will be their name. The email it came from will be their email And the message inside will be the message they wrote in the form. Please help.

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  • join 03 table in the database codeIgniter

    - by python
    with my table. person_id serial NOT NULL, firstname character varying(30) NOT NULL, lastname character varying(30), email character varying(50), username character varying(20) NOT NULL, "password" character varying(100) NOT NULL, gender character varying(10), dob date, accesslevel smallint NOT NULL, company_id integer NOT NULL,//Reference to table company position_id integer NOT NULL,//Reference to table position company_id serial NOT NULL, company_name character varying(80) NOT NULL, description character varying(255), address character varying(100) NOT NULL, In my controller ........................ // load data $persons = $this->person_model->get_paged_list(10,0); // generate table data $this->load->library('table'); $this->table->set_empty("&nbsp;"); $this->table->set_heading('No', 'FirstName', 'LastName','E-mail','Company''Gender', 'Date of Birth', 'Actions'); foreach ($persons as $person){ $this->table->add_row(++$i, $person->firstname, $person->lastname, $person->email, $person->company_name, //HOW CAN I GOT THE POSITION TITLE ?, strtoupper($person->gender)=='M'? 'Male':'Female', date('d-m-Y',strtotime($person->dob)), } My model <?php class Person_Model extends Model { private $person= 'person'; function Person(){ parent::Model(); } function list_all(){ $this->db->order_by('person_id','asc'); return $this->db->get($person); } function count_all(){ return $this->db->count_all($this->person); } function get_paged_list($limit = 0, $offset = 0) { $this->db->limit($limit, $offset); $this->db->select("person.*, company.company_name as company"); $this->db->from('person'); $this->db->join('company','person.company_id = company.company_id','left'); //MY QUESTION:? CAN I JOIN MORE WITH TABLE POSITION? $query = $this->db->get(); return $query->result(); } function get_by_id($id){ $this->db->where('person_id', $id); return $this->db->get($this->person); } function save($person){ $this->db->insert($this->person, $person); return $this->db->insert_id(); } function update($id, $person){ $this->db->where('person_id', $id); $this->db->update($this->person, $person); } function delete($id){ $this->db->where('person_id', $id); $this->db->delete($this->person); } } ?>

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  • Is there a way for a user to disable an AlertDialog completely?

    - by NewGuyChris
    In the app I'm making, I have an "if" statement where if two strings are saved to a certain string, an AlertDialog pops up. These strings will stay the same for some users, thus having this AlertDialog constantly pop up whenever they launch the activity where the ALertDialog is set to appear. Code (I have no setNegativeButton as of yet): private void SetWarning() { AlertDialog.Builder alert = new AlertDialog.Builder(this); alert.setTitle("Warning!"); alert.setMessage(R.string.Warning); alert.setPositiveButton("Ok", new DialogInterface.OnClickListener() { public void onClick(DialogInterface dialog, int whichButton) { //No action needed; just close the AlertDialog. } }); alert.show(); } Here is a segment of my code that makes this AlertDialog appear: SharedPreferences sharedPreferences = getSharedPreferences("MY_PREF", MODE_PRIVATE); String s = sharedPreferences2.getString("MEM1", ""); String s2 = sharedPreferences2.getString("MEM2", ""); if(s.equals("String1") && s2.equals("String2")) SetWarning(); Is there a way to make an "alert.setNegativeButton" method where if the user clicks it, the AlertDialog will NEVER appear again? I'm thinking of maybe somehow implementing another SavedPreferences somehow so it saves the users selection and will then prevent the AlertDialog from ever appearing again. So far, to no luck. I've searched to find nothing, other than people asking how to disable buttons in an AlertDialog. Thank you! New updated code: alert.setNegativeButton("Cancel", new DialogInterface.OnClickListener() { public void onClick(DialogInterface dialog, int whichButton) { //set sharedpreferences boolean called DONTSHOWAGAIN to true; SharedPreferences sharedPreferences2 = getSharedPreferences("MY_PREF", MODE_PRIVATE); Boolean dontShowAgain = sharedPreferences2.getBoolean("dontShowAgain ", false); SharedPreferences.Editor ed = sharedPreferences2.edit(); ed.putBoolean("dontShowAgain", true); ed.commit(); } }); alert.show(); } private void StringWarning() { SharedPreferences sharedPreferences2 = getSharedPreferences("MY_PREF", MODE_PRIVATE); String s = sharedPreferences2.getString("MEM1", ""); String s2 = sharedPreferences2.getString("MEM2", ""); if(s.equals("String1") && s2.equals("String2")){ if(!dontShowAgain){ SetWarningExamConflict(); } }

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  • Linking Post Title to Specific Page ID

    - by ThatMacLad
    I've created a form to update my websites homepage with content but I wanted to know how I could set it up so that a posts title links to a specific post ID. I'd also like to add a Read More link that directs anybody reading the blog to the correct post. Here is my PHP code: <html> <head> <title>Blog Name</title> </head> <body> <?php mysql_connect ('localhost', 'root', 'root') ; mysql_select_db ('tmlblog'); $sql = "SELECT * FROM php_blog ORDER BY timestamp DESC LIMIT 5"; $result = mysql_query($sql) or print ("Can't select entries from table php_blog.<br />" . $sql . "<br />" . mysql_error()); while($row = mysql_fetch_array($result)) { $date = date("l F d Y", $row['timestamp']); $title = stripslashes($row['title']); $entry = stripslashes($row['entry']); $password = $row['password']; $id = $row['id']; if ($password == 1) { echo "<p><strong>" . $title . "</strong></p>"; printf("<p>This is a password protected entry. If you have a password, log in below.</p>"); printf("<form method=\"post\" action=\"post.php?id=%s\"><p><strong><label for=\"username\">Username:</label></strong><br /><input type=\"text\" name=\"username\" id=\"username\" /></p><p><strong><label for=\"pass\">Password:</label></strong><br /><input type=\"password\" name=\"pass\" id=\"pass\" /></p><p><input type=\"submit\" name=\"submit\" id=\"submit\" value=\"submit\" /></p></form>",$id); print "<hr />"; } else { ?> <p><strong><?php echo $title; ?></strong><br /><br /> <?php echo $entry; ?><br /><br /> Posted on <?php echo $date; ?> <hr /></p> <?php } } ?> </body> </html> Thanks for any help. I really appreciate any input!

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  • Parse error in PHP login form

    - by user225269
    I'm trying to have a login form in php. But my current code doesnt work. Here is the form: <form name="form1" method="post" action="loginverify.php"> <td><font size="3">Username:</td> <td></td> <td><input type="text" name="uname" value="" maxlength="15"/><br/></td> <td><font size="3">Password:</td> <td></td> <td><input type="text" name="pword" value="" maxlength="15"/><br/></td> <tr> <td>&nbsp;</td> <td>&nbsp;</td> <td><input type="submit" name="Submit" value="Login" /></td> </form> And the verify.php <?php session_start(); ?> <?php $host="localhost"; $username="root"; $password="nitoryolai123$%^"; $db_name="login"; $tbl="users"; $connection=mysql_connect($host, $username, $password) or die("cannot connect"); mysql_select_db($db_name, $connection) or die("cannot select db"); $user=$_POST['uname'] $pass=$_POST['pword'] $sql="SELECT Username, Password from users where Username='$user' and Password='$pass'"; $result=mysql_query[$sql]; $count=mysql_num_rows($result); if($count==1){ $SESSION['Username']=$user; echo"<a href='searchmain.php'> CONTINUE</a>"; } else{ echo"wrong username or password"; echo"<a href='loginform.php'>Back</a>"; } ?> Is there something wrong with my code. I get this parse error at line 15, which is this: $pass=$_POST['pword'] But when I try to remove it.It goes to line 16 or line 17 again. What do I do

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  • Message passing chrome extension

    - by Mayur Kataria
    I wants to create an extension where content script will send message to background page and then on browser action means clicking on extension icon will access that background page and get some data.I am using chrome Version 23.0.1271.64 m on windows8. I am getting following error. Port error: Could not establish connection. Receiving end does not exist. I tried to solve the same. but people are using sendRequest which is not supported by chrome20+. i also found solution mentioned for chrome 20+. But not working. Please help. Below is the file contents. manifest.json { "name": "Test Extension", "version": "1.0", "manifest_version": 2, "description": "A test extension.", "background": "background.html", "content_scripts": [ { "matches": ["<all_urls>"], "js": ["jquery.js","content.js"] } ], "permissions": ["tabs", "http://*/", "https://*/"], "browser_action": { "default_icon": "icon.png", "default_popup": "popup.html" } } background.html <html> <head> <script src="background.js"></script> </head> <body> <h1>Wy</h1> </body> </html> background.js chrome.extension.onMessage.addListener(function(request, sender, sendResponse) { // Chrome 20+ alert(request); console.log('received in listener'); sendResponse({farewell: "goodbye"}); }); content.js $(function(){ console.log('start-sending message'); chrome.extension.sendMessage({greeting: "hello"},function(response){alert(response);}); console.log('end-sending message'); }); popup.html <!doctype html> <html> <head> <title>Getting Started Extension's Popup</title> </style> <!-- JavaScript and HTML must be in separate files for security. --> <script src="jquery.js"></script> <script src="popup.js"></script> </head> <body> </body> </html> popup.js $(function(){ var str_html = "<tr><td width='60%'>S</td><td width='40%'>15</td></tr><tr><td width='60%'>M</td><td width='40%'>25</td></tr>"; $('#sizes_container').html(str_html); var bkg = chrome.extension.getBackgroundPage(); console.log(bkg); });

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  • EJB / JSF java.lang.ClassNotFoundException: com.ericsantanna.jobFC.dao.DAOFactoryRemote from [Module "com.sun.jsf-impl:main" from local module loader

    - by Eric Sant'Anna
    I'm in my first time using EJB and JSF, and I can't resolve this: 20:23:12,457 Grave [javax.enterprise.resource.webcontainer.jsf.application] (http-localhost-127.0.0.1-8081-2) com.ericsantanna.jobFC.dao.DAOFactoryRemote from [Module "com.sun.jsf-impl:main" from local module loader @439db2b2 (roots: C:\jboss-as-7.1.1.Final\modules)]: java.lang.ClassNotFoundException: com.ericsantanna.jobFC.dao.DAOFactoryRemote from [Module "com.sun.jsf-impl:main" from local module loader @439db2b2 (roots: C:\jboss-as-7.1.1.Final\modules)] I'm getting this when I do an action like a selectOneMenu or a commandButton click. DAOFactory.class @Singleton @Remote(DAOFactoryRemote.class) public class DAOFactory implements DAOFactoryRemote { private static final long serialVersionUID = 6030538139815885895L; @PersistenceContext private EntityManager entityManager; @EJB private JobDAORemote jobDAORemote; /** * Default constructor. */ public DAOFactory() { // TODO Auto-generated constructor stub } @Override public JobDAORemote getJobDAO() { JobDAO jobDAO = (JobDAO) jobDAORemote; jobDAO.setEntityManager(entityManager); return jobDAO; } JobDAO.class @Stateless @Remote(JobDAORemote.class) public class JobDAO implements JobDAORemote { private static final long serialVersionUID = -5483992924812255349L; private EntityManager entityManager; /** * Default constructor. */ public JobDAO() { // TODO Auto-generated constructor stub } @Override public void insert(Job t) { entityManager.persist(t); } @Override public Job findById(Class<Job> classe, Long id) { return entityManager.getReference(classe, id); } @Override public Job findByName(Class<Job> clazz, String name) { return entityManager .createQuery("SELECT job FROM " + clazz.getName() + " job WHERE job.nome = :nome" , Job.class) .setParameter("name", name) .getSingleResult(); } ... TriggerFormBean.class @ManagedBean @ViewScoped @Stateless public class TriggerFormBean implements Serializable { private static final long serialVersionUID = -3293560384606586480L; @EJB private DAOFactoryRemote daoFactory; @EJB private TriggerManagerRemote triggerManagerRemote; ... triggerForm.xhtml (a portion with problem) </p:layoutUnit> <p:layoutUnit id="eastConditionPanel" position="center" size="50%"> <p:panel header="Conditions to Release" style="width:97%;height:97%;"> <h:panelGrid columns="2" cellpadding="3"> <h:outputLabel value="Condition Name:" for="conditionName" /> <p:inputText id="conditionName" value="#{triggerFormBean.newCondition.name}" /> </h:panelGrid> <p:commandButton value="Add Condition" update="conditionsToReleaseList" id="addConditionToRelease" actionListener="#{triggerFormBean.addNewCondition}" /> <p:orderList id="conditionsToReleaseList" value="#{triggerFormBean.trigger.conditionsToRelease}" var="condition" controlsLocation="none" itemLabel="#{condition.name}" itemValue="#{condition}" iconOnly="true" style="width:97%;heigth:97%;"/> </p:panel> </p:layoutUnit> In TriggerFormBean.class if comments daoFactory we get the same exception with triggerManagerRemote, both annotated with @EJB. I'm don't understand the relationship between my DAOFactory and the "Module com.sun.jsf-impl:main"... Thanks.

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  • I am unsure of how to access a persistence entity from a JSP page?

    - by pharma_joe
    Hi, I am just learning Java EE, I have created a Persistence entity for a User object, which is stored in the database. I am now trying to create a JSP page that will allow a client to enter a new User object into the System. I am unsure of how the JSP page interacts with the User facade, the tutorials are confusing me a little. This is the code for the facade: <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <title>Add User to System</title> </head> <body> <h2>Add User</h2> <h3>Please fill out the details to add a user to the system</h3> <form action=""> <label>Email:</label> <input type="text" name="email"><br /> <label>Password:</label> <input type="password" name="name"><br /> <label>Name:</label> <input type="text" name="name"><br /> <label>Address:</label> <input type="text" name="address"><br /> <label>Type:</label> <select name="type"> <option>Administrator</option> <option>Member</option> </select><br /> <input type="submit" value="Add" name="add"/> <input type="reset" value="clear" /> </form> </body> This is the code I have to add a new User object within the User facade class: @Stateless public class CinemaUserFacade { @PersistenceContext(unitName = "MonashCinema-warPU") private EntityManager em; public void create(CinemaUser cinemaUser) { em.persist(cinemaUser); } I am finding it a little difficult to get my head around the whole MVC thing, getting there but would appreciate it if someone could turn the light on for me!

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  • Paperclip not running tasks but not showing errors

    - by Trip
    This is strange. I just did a deploy to a cluster server, and since then, pictures have not been processing. Reading the logs, I usually do not get an error at all, but they never finish. However, on one particular image, I found this little bit at least, but this might not explain everything.. Any ideas? Processing PhotosController#edit (for 69.248.152.173 at 2010-05-27 04:25:12) [GET] Parameters: {"gallery_id"="2102", "action"="edit", "type"="photo", "id"="15453", "crop"="true", "controller"="photos", "organization_id"="470", "_"="1274959512393"} Rendering media/crop_photo ActionView::TemplateError (/data/HQ_Channel/releases/20100524111501/public/system/photos/15453/original/DSC05193.JPG is not recognized by the 'identify' command.) on line #4 of app/views/media/crop_photo.js.haml: 1: == $("#media_header").html('#{ escape_javascript(render :partial = 'media/crop_photo') }').slideDown("slow"); 2: 3: :plain 4: function updateForm(coords) 5: { 6: var rx = #{PHOTO_IMAGE_WIDTH} / coords.w; 7: var ry = #{PHOTO_IMAGE_HEIGHT} / coords.h; vendor/gems/thoughtbot-paperclip-2.3.1/lib/paperclip/geometry.rb:24:in `from_file' app/models/photo.rb:68:in `photo_geometry' app/views/media/crop_photo.js.haml:4:in `_run_haml_app47views47media47crop_photo46js46haml' haml (2.2.2) [v] lib/haml/helpers/action_view_mods.rb:13:in `render' app/controllers/photos_controller.rb:81:in `crop' app/controllers/photos_controller.rb:24:in `edit' haml (2.2.2) [v] rails/./lib/sass/plugin/rails.rb:19:in `process' lib/flash_session_cookie_middleware.rb:14:in `call' vendor/gems/hoptoad_notifier-2.2.2/lib/hoptoad_notifier/rack.rb:27:in `call' ** [Hoptoad] Failure: Net::HTTPClientError ** [Hoptoad] Environment Info: [Ruby: 1.8.6] [Rails: 2.3.3] [Env: production] ** [Hoptoad] Response from Hoptoad: No project exists with the given API key. Rendering /data/HQ_Channel/releases/20100524111501/public/500.html (500 Internal Server Error) And then a little later, I got this : ActionView::TemplateError (/data/HQ_Channel/releases/20100524111501/public/system/photos/15453/original/DSC05193.JPG is not recognized by the 'identify' command.) on line #4 of app/views/media/crop_photo.js.haml: 1: == $("#media_header").html('#{ escape_javascript(render :partial = 'media/crop_photo') }').slideDown("slow"); 2: 3: :plain 4: function updateForm(coords) 5: { 6: var rx = #{PHOTO_IMAGE_WIDTH} / coords.w; 7: var ry = #{PHOTO_IMAGE_HEIGHT} / coords.h; vendor/gems/thoughtbot-paperclip-2.3.1/lib/paperclip/geometry.rb:24:in `from_file' app/models/photo.rb:68:in `photo_geometry' app/views/media/crop_photo.js.haml:4:in `_run_haml_app47views47media47crop_photo46js46haml' haml (2.2.2) [v] lib/haml/helpers/action_view_mods.rb:13:in `render' app/controllers/photos_controller.rb:81:in `crop' app/controllers/photos_controller.rb:24:in `edit' haml (2.2.2) [v] rails/./lib/sass/plugin/rails.rb:19:in `process' lib/flash_session_cookie_middleware.rb:14:in `call' vendor/gems/hoptoad_notifier-2.2.2/lib/hoptoad_notifier/rack.rb:27:in `call' ** [Hoptoad] Failure: Net::HTTPClientError ** [Hoptoad] Environment Info: [Ruby: 1.8.6] [Rails: 2.3.3] [Env: production] ** [Hoptoad] Response from Hoptoad: No project exists with the given API key. Rendering /data/HQ_Channel/releases/20100524111501/public/500.html (500 Internal Server Error)

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  • jQuery Sales Tax

    - by CKallemeres
    Hello everyone! I have created a function (see below) that calculates a 7.5% sales tax. Now I need help doing the following: Have totalTax() take in 2 arguments one for the price and one for the tax. On submit (use the onSubmit event handler to call this function) have the function process the price and the tax by manipulating the arguments you passed in. Have the sales tax on the page update dynamically with what ever the sales tax is that you defined for the function 7.5 percent sales tax: Instead of using .innerHTML use jQuery to access these document elements and write to them: document.getElementById('requestedAmount' ).innerHTML = priceInput; document.getElementById('requestedTax' ).innerHTML = salesTax; document.getElementById('requestedTotal' ).innerHTML = totalAmount; Original Code: <script type="text/javascript"> $().ready(function() { // validate the comment form when it is submitted $("#inputForm").validate(); $("#priceInput").priceFormat({ prefix: '', limit: 5, centsLimit: 2 }); }); function totalTax(){ var priceInput = document.getElementById( 'priceInput' ).value; var salesTax = Math.round(((priceInput / 100) * 7.5)*100)/100; var totalAmount = (priceInput*1) + (salesTax * 1); document.getElementById( 'requestedAmount' ).innerHTML = priceInput; document.getElementById( 'requestedTax' ).innerHTML = salesTax; document.getElementById( 'requestedTotal' ).innerHTML = totalAmount; } </script> <body> <form class="cmxform" id="inputForm" method="get" action=""> <p> <label for="priceInput">Enter the price: </label> <input id="priceInput" name="name" class="required"/> </p> <p> <input class="submit" type="submit" value="Submit" onclick="totalTax();"/> </p> </form> <div>Entered price: <p id="requestedAmount"></p> </div> <div>7.5 percent sales tax: <p id="requestedTax"></p> </div> <div>Total: <p id="requestedTotal"> </p> </div>

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  • mysql_fetch_array() expects parameter 1 to be resource problem

    - by user225269
    I don't get it, I see no mistakes in this code but there is this error, please help: mysql_fetch_array() expects parameter 1 to be resource problem <?php $con = mysql_connect("localhost","root","nitoryolai123$%^"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("school", $con); $result = mysql_query("SELECT * FROM student WHERE IDNO=".$_GET['id']); ?> <?php while ($row = mysql_fetch_array($result)) { ?> <table class="a" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#D3D3D3"> <tr> <form name="formcheck" method="get" action="updateact.php" onsubmit="return formCheck(this);"> <td> <table border="0" cellpadding="3" cellspacing="1" bgcolor=""> <tr> <td colspan="16" height="25" style="background:#5C915C; color:white; border:white 1px solid; text-align: left"><strong><font size="2">Update Students</td> <tr> <td width="30" height="35"><font size="2">*I D Number:</td> <td width="30"><input name="idnum" onkeypress="return isNumberKey(event)" type="text" maxlength="5" id='numbers'/ value="<?php echo $_GET['id']; ?>"></td> </tr> <tr> <td width="30" height="35"><font size="2">*Year:</td> <td width="30"><input name="yr" onkeypress="return isNumberKey(event)" type="text" maxlength="5" id='numbers'/ value="<?php echo $row["YEAR"]; ?>"></td> <?php } ?> I'm just trying to load the data in the forms but I don't know why that error appears. What could possibly be the mistake in here?

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  • Google Maps and jQuery Tabs

    - by Dom
    Hello All, I have slight problem with Google maps included in simple jQuery Tabs. Below I pasted the code: jQuery: $(document).ready(function() { //Default Action $(".tab_content").hide(); $("ul.tabs li:first").addClass("active").show(); $(".tab_content:first").show(); //On Click Event $("ul.tabs li").click(function() { $("ul.tabs li").removeClass("active"); $(this).addClass("active"); $(".tab_content").hide(); var activeTab = $(this).find("a").attr("href"); $(activeTab).fadeIn(); return false; }); }); Here is the HTML for the tabs: <div class="bluecontainer"> <ul class="tabs"> <li><a href="#tab1">Tab1</a></li> <li><a href="#tab2">Tab2</a></li> <li><a href="#tab3">Tab3</a></li> <li><a href="#tab4">Tab4</a></li> </ul> <div class="tab_container"> <div id="tab1" class="tab_content"> <h2>Tab1</h2> </div> <div id="tab2" class="tab_content"> <h2>Tab2</h2> </div> <div id="tab3" class="tab_content"> <div>google Map</div> </div> <div id="tab4" class="tab_content"> <h2>Tab4</h2> </div> </div> </div> I really don't know what to do to. Is that a general problem with google maps or there is something with my tabs? But they are working just fine with everything else. Thank you for your help in advance

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  • form inside tabview doesn't work

    - by user3536737
    i am working with jsf and primefaces , and here is what 've tried well i want to creat a tabview that get data from an arraylist in my bean i get for exemple 4 tabs , and inside each one i've created a hidden panel where i have a form with 2 input text to update informations , do i display the panel when i click on the second button Update , after that my panel is not hidden anymore , and i set the new values and click on the second button to update the informations , the problem is that the updating and the execution is working only for the first tab , it means when i try to update the new informations it works for the first one and for the other tabs it doesn't here is the code <p:tab title="#{rr.nom_ressource}"> <h:panelGrid> <h:graphicImage value="Ressources/images/emp.jpg" style="vertical-align:middle" /> <span style="font-size:15px; width:170px; display:inline-block;"> Nom : #{rr.nom_ressource} Type: #{rr.type_ressource} Specification: #{rr.experience} </span> <h:commandButton image="Ressources/images/delete.jpg" actionListener="#{SelectBean.act}" update=":form" style="vertical-align:middle" > Update </h:commandButton> <h:commandButton update=":outPanel" actionListener="#{SelectBean.mod1()}" image="Ressources/images/update.png" style="vertical-align:middle" > Modifier </h:commandButton> <h:form id="form111"> <p:growl id="growl" showDetail="true" sticky="true" /> <p:panel rendered ="#{SelectBean.bol}" closable="true" toggleable="true" id="outPanel" styleClass="outPanel" widgetVar="outpanel"> <h:outputLabel value="Nom " /> <h:inputText value="#{SelectBean.nom}" /> <br/> <h:outputLabel value="Experience " /> <h:inputText value="#{SelectBean.exp}" /> <br/> <h:commandButton value="Update" action="#{SelectBean.done}"/> </p:panel> </h:form> </h:panelGrid> </p:tab> for my managedbean the code is correct i think the problem is here

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