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  • Error in inserting data into database

    - by Matthew
    When I try to run this code it will not insert the data into the database? <?php class Database { private $dsn; function __construct($dbname, $host, $user, $password, $enckey) { $this->dsn = "mysql:dbname=" . $dbname . ';host=' . $host; $this->user = $user; $this->password = $password; } private function createDSN() { return $this->dsn; } public function createConnection() { try { $dbh = new PDO(self::createDSN(), $this->user, $this->password); } catch (PDOException $e) { echo 'Connection failed: ' . $e->getMessage(); } return $dbh; } } $db = new Database('mytest', 'localhost', 'root', 'hashedpassword', null); $dbh = $db->createConnection(); $sql = $dbh->prepare("INSERT INTO contacts (firstname, lastname) VALUES (?,?)"); $sql->execute(array("abc", "xyz")); ?>

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  • Change post form data function into curl

    - by QLiu
    Hello Guys, In the old way in our website, when users clicks “logout” button. It runs a post form function; which will pass parameters (logout, sn) to external sites to execute “logout” function. Like: I do not want the users jump to the external site, therefore, i use curl to post data. (because we are in different domain, i guess Ajax request doesnot work ) Post the same data to execute logout function in external site. // create cURL resource $URL = "http://bswi.development.intra.local/"; //Initl curl $ch = curl_init(); curl_setopt($ch, CURLOPT_URL, $URL); // Load in the destination URL curl_setopt($ch, CURLOPT_HTTPAUTH, CURLAUTH_BASIC); //Normal HTTP request, not SSL curl_setopt($ch, CURLOPT_POSTFIELDS, "logout=1"); // receive server response ... curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); curl_exec ($ch); echo $content; curl_close ($ch); Do u think i am going in the right direction?

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  • how to get a text box with the value selected in select box

    - by udaya
    hi This is my view page here i have a select box <tr> <td>Chidren</td> <td>:</td> <td><select style="font-family: verdana; min-width: 52px;" id="ddlChildren" name="ddlChildren" class="required" onChange="return Check_Adult('dd1Age')" > <option value="">children</option> <option value="1">1</option> <option value="2">2</option> <option value="3">3</option> <option value="4">4</option></select> </td> </tr> with the value from select box if the value is one then i need to create a text box if the value is two then i need to create two ,,, and respectively In my Check_Adult javascript function ,,I did this function Check_Adult() { var Child= document.getElementById('ddlChildren').value; if( Child==1) { <? echo '<input type="text" name="child" id="child">' ?> } } But the text box is not created how to create it?

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  • Can I automate a loop query based on category with the page's title?

    - by jordaninternets
    To explain further, I need to get a page to display posts from a specific category. I want to automate this process so I don't have to make a template for each category. How would I do that? (keep in mind the person I'm building this is for has no coding experience.) The only way I could think off from the top of my head is to use the title or slug. So if the category is named the same thing as the slug, could I filter by category using the slug? Maybe this isn't the best way... what should I do? This is what I came up with, but it doesn't work, I'm sure due to improper use on may part, but I've been pouring over the WP codex and Google with no avail to tell me my problem. <?php $paged = (get_query_var('paged')) ? get_query_var('paged') : 1; $args= array( 'orderby' => 'date', 'order' => 'ASC', 'paged' => $paged, 'category_name' => echo the_title('\'','\'',) ); query_posts($args); ?>

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  • Zend Framework: How do I modify/format the form view generated with Zend_Dojo_Form elements

    - by pinardelrio
    I have created a form: <?php class Application_Form_Issue extends Zend_Dojo_Form { public function init() { $this->setName('issue'); $this->setMethod('post'); $id = new Zend_Form_Element_Hidden('id'); $id->addFilter('Int'); $date_recvd = new Zend_Dojo_Form_Element_DateTextBox('date_recvd'); $date_recvd->setLabel('Date Received') //->setRequired(true) /*->addValidator('NotEmpty'); */; More Form elements ... To view this form my view script is: <?php echo $this->form; ?> This all works just fine, with fully functional dojo form elements (datepicker, timepicker, etc) and successfully saving the data. However, now, I want to format the form that is generated with css. Such as grouping some elements and floating left or right, making some input text fields wider/narrower, etc. How? I realize I can modify the view script but it seems like that defeats the purpose of using Zend_Dojo_Form or Zend_Form. Is that a correct assumption?

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  • Query not being executed

    - by user2385241
    I'm trying to create a script that allows me to upload an image, grab the details sent through inputs (a description and chosen project number) and insert this information into a table. I currently have this function: public function NewEntry() { $connect = new dbconnect; $_SESSION['rnd'] = substr(number_format(time() * rand(),0,'',''),0,15); $allowedExts = array("gif", "jpeg", "jpg", "png"); $size = $_FILES["file"]["size"]; $path = $_FILES["file"]["name"]; $extension = pathinfo($path, PATHINFO_EXTENSION); $pr = $_POST['project']; $cl = $_POST['changelog']; $file = $_SESSION['rnd'] . "." . $extension; if (in_array($extension, $allowedExts) && $size < 200000000) { if ($_FILES["file"]["error"] == 0) { if (!file_exists("../uploads/" . $_SESSION['rnd'])) { move_uploaded_file($_FILES["file"]["tmp_name"], "../uploads/" . $_SESSION['rnd'] . "." . $extension); } } } else { echo "File validation failed."; } $row = $connect->queryExecute("INSERT INTO entries(project,file,changelog)VALUES($pr,$file,$cl)"); header('location:http://www.example.com/admin'); } When the form is posted the function runs, the image uploads but the query isn't executed. The dbconnect class isn't at fault as it's untampered and has been used in past projects. The error logs don't give any output and no MySQL errors show. Any ideas?

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  • Read all sub directories within a certain folder to display a random image.

    - by Andy
    I have this code i have been using....but i need a conditional where it will read all the sub directories of /bg to select an image as opposed to a specific folder if they were on a subpage. Heres my code so far which works perfectly for all subpages to display specific images: //This would tell us its on the homepage if it helps: $this->level() == 0 //This is the code so far $path = '/home/sites/mydomain.co.uk/public_html/public/images/bg/'.$this->slug; $homepagefile = URL_PUBLIC.'public/images/bg/'.$this->slug.'/main.jpg'; $bgimagearray = array(); $iterator = new DirectoryIterator($path); foreach ($iterator as $fileinfo) { if ($fileinfo->isFile() && !preg_match('\.jpg$/', $fileinfo->getFilename())) { $bgimagearray[] = "'" . $fileinfo->getFilename() . "'"; } } $bgimage = array_rand($bgimagearray); ?> <div id="bg"> <div> <table cellspacing="0" cellpadding="0"> <tr> <td><img src="<?php echo $file.trim($bgimagearray[$bgimage], "'"); ?>" alt=""/></td> </tr> </table> </div> </div> Any help would be appreciated, im sure its not rocket science but ive tried a few ways and cant get my head around it. Thanks in advance.

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  • Detect if Download is Complete

    - by user604138
    I have a very simple and standard PHP force download script. How do I check if/when the download has completed in order to notify the user on the clientside? I don't even need to show the progress in real time, I am only interested in the very specific event: "when the download completes". Based on my research, it seems like it would have to be determined from the serverside as there is noondownloadready event and I don't think it is possible to intercept browser events. So it seems that my best bet would be to compare bytes sent to total bytes with some sort of clientside/severside interaction. How would I go about checking the bytes sent from the server for a PHP forced download? is there some sort of global PHP variable that store these data that I can ping with AJAX? <?php header("Content-Type: video/x-msvideo"); header("Content-Disposition: attachment; filename=\"".basename($realpath)."\";"); ... $chunksize = 1 * (1024 * 1024); // how many bytes per chunk if ($size > $chunksize) { $handle = fopen($realpath, 'rb'); $buffer = ''; while (!feof($handle)) { $buffer = fread($handle, $chunksize); echo $buffer; ob_flush(); flush(); } fclose($handle); } else { readfile($realpath); } exit(); ?> The reason I need this: For the project I am working on, it is required that after the download starts, the page redirects to (or displays) a "please wait while the download completes" page. Then, once it is complete, it should redirect to (or display) a "Your download is complete, thank you" page. I am open to other ideas that would achieve the same result.

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  • How to remove background of a image and copy into another?

    - by Gustavo Pinent
    I'm researching about captchas. In this idea, the task is: create a image from a JPEG and remove the white background, than create another image from another JPEG, than create the final image by adding the second as a background, and copying the first one over this background preserving the transparent area created, of course. Here is the code: header("Content-Type: image/jpeg"); $nFundo = rand(0,4); $Dirs = array(rand(0,7), rand(0,7), rand(0,7), rand(0,7)); // Will be four times all $_SESSION["form_captcha"] = $Dirs; $image = ImageCreatetruecolor(320, 80); ImageAlphaBlending($image, FALSE); ImageSaveAlpha($image, TRUE); $image_seta = ImageCreateFromJPEG("_captcha-seta.jpg"); // Image do copy over $image_fundo = ImageCreateFromJPEG("_captcha-fundo-".$nFundo.".jpg"); // Image to make the background for($i=0; $i<4; $i++){ ImageCopy($image, $image_fundo, $i*80, 0, 0, 0, 80, 80); } // So far so good, a background with a pattern repeated four times $color_white = ImageColorAllocate($image_seta, 255, 255, 255); ImageColorTransparent($image_seta, $color_white); ImageSaveAlpha($image_seta, TRUE); for($i=0; $i<4; $i++){ $image_seta_rot = imageRotate($image_seta, $Dirs[$i]*45, $color_white); ImageCopyResampled($image, $image_seta_rot, $i*80, 0, 0, 0, 80, 80, 80, 80); // Try } echo(imagejpeg($image)); imagedestroy($image); I tried to replace $image_seta_rot by $image_seta ("Try" line) to see if the rotation is the problem, but even without rotation, the white wasn't removed and the image just "erase" the background created before. So the copy is failing or the white were never removed... I may create a PNG with transparent background, but will be interesting to learn how to make it dynamically, don't you think? Any ideas?

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  • Processing data from an AJAX request

    - by Josh K
    I have a PHP API I'm working with that outputs everything as JSON. I need to call one of the API methods and parse it out using an AJAX request. I am using jQuery (though it shouldn't matter). When I make the request it errors out with a "parsererror" as the textStatus and a "Syntax Error: invalid label" when I make the request. Simplified code: $.ajax ({ type: "POST", url: "http://mydomain.com/api/get/userlist/"+mid, dataType: "json", dataFilter: function(data, type) { /* Here we assume and pray */ users = eval(data); alert(users[1].id); }, success: function(data, textStatus, XMLHttpRequest) { alert(data.length); // Should be an array, yet is undefined. }, error: function(XMLHttpRequest, textStatus, errorThrown) { alert(textStatus); alert(errorThrown); }, complete: function(XMLHttpRequest, textStatus) { alert("Done"); } }); If I leave off the eval(data) then everything works fine. Well, except for data still being undefined in success. Note that I'm taking an array of objects in PHP and then passing them out through json_encode. Would that make any difference? There has been no progress made on this. I'm willing to throw more code up if someone believes they can help. Here is the PHP side of things private function _get_user_colors($id) { $u = new User(); $u->get_where(array('id' => $id)); $bar = array(); $bar['user'] = $u->stored; foreach($user->colors as $color) { $bar['colors'][] = $color; } echo(json_encode($bar)); } I have had zero issues using this with other PHP based scripts. I don't know why Javascript would take issue with it.

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  • Trying to join two independent forms

    - by user248959
    Hi, i'm trying to join two independent forms (login and register) in the same page. My idea is (just looking at the signin form): Create an action that shows both forms (partials): public function executeLoginAndRegister(sfWebRequest $request){ $this->form_signin = $this->getUser()->getAttribute('form_signin'); } Each partial calls to its action: form action="php? echo url_for('@sf_guard_signin') ?" method="post" In the actions i write this code public function executeSignin($request) { //... $this->form = new $MyFormclass(); if ($this->form->isValid()) { //... }else{ // save the form to show the error messages. $this-&gt;getUser()-&gt;setAttribute('form_signin', $this-&gt;form); return $this-&gt;forward('sfGuardAuth', 'loginAndRegister'); } } It works, but, for example, if i execute LoginAndRegister and submit incorrectly the signin form and I go to another page and then return to LoginAndRegister, i will find the submiting error messages... If i execute LoginAndRegister and submit incorrectly the signin form and open another browser tab, i will find the submiting error messages in the signin form of the second tab... Any idea? any better approach?

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  • Difficulty Inserting from HTML form into MySQL (created in Workbench)

    - by Chaya Cooper
    I created a MySQL database in Workbench 5.2.35 for purposes of storing and using information submitted in html forms but I'm having difficulty inserting records from the html form. The relevant SQL script was saved as Demo2.sql, the schema is C2F, and the table is customer_info. I wasn't sure if that was the problem, so I tried replacing the database name (Demo2) with the schema name, but that didn't work either. My html file includes: form action="insert.php" method="post" The insert.php file states: ?php $con = mysql_connect("localhost","root","****"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("Demo2", $con); $sql="INSERT INTO customer_info(fname, lname, user, password, reminder, answer) VALUES ('$_POST[fname]','$_POST[lname]','$_POST[user]','$_POST[password]','$_POST[reminder]','$_POST[answer]')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "1 record added"; mysql_close($con) ? I've also tried INSERT INTO c2f.customer_info(fname, lname, etc. and INSERT INTO 'c2f'.'customer_info'

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  • PHP Multi-Domain Sessions; ini_set Not Working?

    - by SumWon
    Hello, I'm trying to set it up so if you log in to my website the session carries over to all sub-domains of my website. For example, if you go to domain.com and log in, then go to sub.domain.com, you'll already be logged in at sub.domain.com. To my understanding, you would want to use ini_set('session.cookie_domain','.domain.com') and then session_start(), then set your session variables, but this isn't working. Example of what I'm doing: Code for domain.com: <?php ini_set('session.cookie_domain','.domain.com'); session_start(); $_SESSION['variable'] = 1; ?> Code for sub.domain.com: <?php session_start(); echo $_SESSION['variable']; ?> But $_SESSION['variable'] isn't set. I've also tried using ini_set() in the sub.domain.com code, but it made no difference. I've verified that setting session.cookie_domain is working by using ini_get(). What am I doing wrong? Thanks!

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  • Trigger alert when database entries are added, not when they are removed

    - by Jeremy
    I have a jQuery script running that makes a periodic AJAX call using the following code. var a = moment(); var dayOfMonth = a.format("MMM Do"); var timeSubmitted = a.format("h:mm a"); var count_cases = -1; var count_claimed = -1; setInterval(function(){ //check if new lead was added to the db $.ajax({ type : "POST", url : "inc/new_lead_alerts_process.php", dataType: 'json', cache: false, success : function(response){ $.getJSON("inc/new_lead_alerts_process.php", function(data) { if (count_cases != -1 && count_cases != data.count) { window.location = "new_lead_alerts.php?id="+data.id; } count_cases = data.count; }); } }); This is the PHP that runs with each call: $count = mysql_fetch_array(mysql_query("SELECT count(*) as count FROM leads")); $client_id = mysql_fetch_array(mysql_query("SELECT id, client_id FROM leads ORDER BY id DESC LIMIT 1")); echo json_encode(array("count" => $count['count'], "id" => $client_id['id'], "client_id" => $client_id['client_id'])); I need to change the code so that the alert only triggers when a new entry is added to the database, not when an existing entry is removed. As it stands, the alert fires on both events. Any help is greatly appreciated.

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  • possible to make codeigniter work with another framework?

    - by ajsie
    the situation is this. my client (who also is a programmer) asks me to develop an address book (with mysql database) with a lot of functions. then he can interact with some class methods i provide for him. kinda like an API. the situation is that the address book application is getting bigger and bigger, and i feel like its way better to use CodeIgniter to code it with MVC. i wonder if i can use codeigniter, then in some way give him the access to controller methods. eg. in a controller there are some functions u can call with the web browser. public function create_contact($information) {..} public function delete_contact($id) {..} public function get_contact($id) {..} however, these are just callable from web browser. how can i let my client have access to these functions like an API? then in his own application he can use: $result = $address_book-create_contact($information); if($result) { echo "Success"; } $contact = $address_book-get_contact($id); is this possible? cause i just know how to access the controller methods with the webbrowser. and i guess its not an option for him to use header(location) to access them. all suggestions to make this possible are welcomed! thanks

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  • entering datas of php page to database

    - by Rahima farzana.S
    I have a php form with two text boxes and i want to enter the text box values into the database. I have created the table (with two columns namely webmeasurementsuite id and webmeasurements id) I used the following syntax for creating table: CREATE TABLE `radio` ( `webmeasurementsuite id` INT NOT NULL, `webmeasurements id` INT NOT NULL ); Utilising the tutorial in the following link, I wrote the php coding but unfortunately the datas are not getting entered into the database. I am getting an error in the insert sql syntax. I checked it but i am not able to trace out the error.Can anyone correct me? I got the coding from http://www.webune.com/forums/php-how-to-enter-data-into-database-with-php-scripts.html $sql = "INSERT INTO $db_table(webmeasurementsuite id,webmeasurements id) values ('".mysql_real_escape_string(stripslashes($_REQUEST['webmeasurementsuite id']))."','".mysql_real_escape_string(stripslashes($_REQUEST['webmeasurements id']))."')"; echo$sql; My error is as follows: INSERT INTO radio(webmeasurementsuite id,webmeasurements id) values ('','')ERROR: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'id,webmeasurements id) values ('','')' at line 1

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  • PHP Error Form - Leave Contents of Form on Redirect

    - by user1371500
    I have a simple login form in which if an error occurs such as wrong password, I need it to be able to remember the username which was entered. Would I Go about doing this PHP or Javascript as I am not allowed to use JQuery. My current PHP - (Not Including the HTML Form) <?php //MySQl Connection mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db("clubresults") or die(mysql_error()); //Initiates New Session - Cookie session_start(); // Start a new session // Get the data passed from the form $username = $_POST['username']; $password = md5($_POST['pass']); // Do some basic sanitizing $username = mysql_real_escape_string($username); $password = mysql_real_escape_string($password); //Performs SQL Query to retrieve Login Details from DB $sql = "select * from admin_passwords where username = '$username' and password = '$password'"; $result = mysql_query($sql) or die ( mysql_error() ); //Assigns a Variable Count to 0 $count = 0; //Exectues a loop to increment on Successful Login while ($line = mysql_fetch_assoc($result)) { $count++; } //If count is equal to 1 Redirect user to the Members Page and Set Cookie if ($count == 1) { $_SESSION['loggedIn'] = "true"; header("Location: members.php"); // This is wherever you want to redirect the user to } else { //Else Echo that login was a failure. die('Login Failed. <a href=login.php>Click Here to Try Again</a>'); } ?> Any help would be appreciated. Cheers

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  • ajax form handling an array

    - by moata_u
    am trying to handle an array comes from php file after submitting the form data , the value of data after submitting the form is = ARRAY but i cant use this array in any way , any idea how can i handle this array !!!! Javascript : $('#file').live('change',function(){ $('#preview').html(''); $('#preview').html('<img src="loader.gif" />'); $('#data').ajaxForm(function(data){ $(data['toshow']).insertBefore('.pic_content').hide().fadeIn(1000); }).submit(); }); PHP : .... ....etc echo json_encode(array('toshow'=>somedata,'data'=>somedata)); data come from php file {"toshow":"\r\n\t\t\t\t\r\n\t\t<table class=\"out\">\r\n\t\t\t<tr ><td class=\"img\"><a title=\"2012-06-02 01-22-09\" rel=\"prettyPhoto\" href=\"img\/2012-06-02 01-22-09.284.jpg\"><img src=\"img\/thumb\/2012-06-02 01-22-09.284.jpg\"\/><\/a><\/td><\/tr>\r\n\t\t\t\r\n\t\t\t<td>\r\n\t\t\t\t<table cellSpacing=\"1\" cellPadding=\"0\">\r\n\t\t\t\t\t<tr><td class=\"data\"><span class=\"click\">2012-06-02 01-22-09<\/span><\/td><\/tr>\r\n\t\t\t\t\t<tr><td class=\"data\"><span class=\"click\">Download<\/span><\/td><\/tr>\r\n\t\t\t\t\t<tr><td class=\"data\"><a href=\"img\/2012-06-02 01-22-09.284.jpg\"><span class=\"click\">View<\/span><\/a><\/td><\/tr>\r\n\t\t\t\t<\/table>\r\n\t\t\t<\/td>\r\n\t\t\t<\/tr>\r\n\t\t<\/table>","span":"<span class='text'><img src='greencheck.png'\/>2012-06-02 01-22-09 Uploaded ,File Size =152Kb <\/span>"}

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  • codeigniter and JSON

    - by ole
    Hello all i having a problem that it only get 1 value in my database and its my title and i want to show content and username from the same table to. here is my JSON kode <script type="text/javascript"> $.getJSON( 'ajax/forumThreads', function(data) { alert(data[0].overskrift); alert(data[0].indhold); } ); </script> my controller <?php class ajax extends Controller { function forumThreads() { $this->load->model('ajax_model'); $data['forum_list'] = $this->ajax_model->forumList(); if ($data['forum_list'] !== false) { echo json_encode($data['forum_list']); } } } my model fle <?php class ajax_model extends Model { function forumList() { $this->db->select('overskrift', 'indhold', 'brugernavn', 'dato'); $this->db->order_by('id', 'desc'); $this->db->limit(5); $forum_list = $this->db->get('forum_traad'); if($forum_list->num_rows() > 0) { return $forum_list->result_array(); } else { return false; } } }

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  • Display attribute from XML using PHP

    - by user560411
    Hello. I need to display the id attribute of CD from the following XML file. I display correctly everything except the id. Any help would be appreciate. display code <?php $doc = new DOMDocument(); $doc->load( 'insert.xml' ); $CATEGORIES = $doc->getElementsByTagName( "CD" ); foreach( $CATEGORIES as $CD ) { $TITLES = $CD->getElementsByTagName( "TITLE" ); $TITLE = $TITLES->item(0)->nodeValue; $BANDS= $CD->getElementsByTagName( "BAND" ); $BAND= $BANDS->item(0)->nodeValue; $YEARS = $CD->getElementsByTagName( "YEAR" ); $YEAR = $YEARS->item(0)->nodeValue; echo "<b>$TITLE - $BAND - $YEAR\n</b><br>"; } ?> XML <?xml version="1.0" encoding="utf-8"?> <MY_CD> <CATEGORIES> <CD id="3231"> <TITLE>NEVER MIND THE BOLLOCKS</TITLE> <BAND>SEX PISTOLS</BAND> <YEAR>1977</YEAR> </CD> <CD id="2453"> <TITLE>NEVERMIND</TITLE> <BAND>NIRVANA</BAND> <YEAR>1991</YEAR> </CD> </CATEGORIES> </MY_CD>

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  • How to change a recursive function for count files and catalogues?

    - by user661999
    <?php function scan_dir($dirname) { $file_count = 0 ; $dir_count = 0 ; $dir = opendir($dirname); while (($file = readdir($dir)) !== false) { if($file != "." && $file != "..") { if(is_file($dirname."/".$file)) ++$file_count; if(is_dir($dirname."/".$file)) { ++ $dir_count; scan_dir($dirname."/".$file); } } } closedir($dir); echo "There are $dir_count catalogues and $file_count files.<br>"; } $dirname = "/home/user/path"; scan_dir($dirname); ?> Hello, I have a recursive function for count files and catalogues. It returns result for each catalogue. But I need a common result. How to change the script? It returns : There are 0 catalogues and 3 files. There are 0 catalogues and 1 files. There are 2 catalogues and 14 files. I want: There are 2 catalogues and 18 files.

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  • How can I achieve this kind of relationship (inheritance, composition, something else)?

    - by Tim
    I would like to set up a foundation of classes for an application, two of which are person and student. A person may or may not be a student and a student is always a person. The fact that a student “is a” person led me to try inheritance, but I can't see how to make it work in the case where I have a DAO that returns an instance of person and I then want to determine if that person is a student and call student related methods for it. class Person { private $_firstName; public function isStudent() { // figure out if this person is a student return true; // (or false) } } class Student extends Person { private $_gpa; public function getGpa() { // do something to retrieve this student's gpa return 4.0; // (or whatever it is) } } class SomeDaoThatReturnsPersonInstances { public function find() { return new Person(); } } $myPerson = SomeDaoThatReturnsPersonInstances::find(); if($myPerson->isStudent()) { echo 'My person\'s GPA is: ', $myPerson->getGpa(); } This obviously doesn't work, but what is the best way to achieve this effect? Composition doesn't sond right in my mind because a person does not “have a” student. I'm not looking for a solution necessarily but maybe just a term or phrase to search for. Since I'm not really sure what to call what I'm trying to do, I haven't had much luck. Thank you!

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  • PHP & HTML Purifier Error: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given

    - by TaG
    I'm trying to Integrate HTML Purifier http://htmlpurifier.org/ to filter my user submitted data but I get the following error below. And I was wondering how can I fix this problem? I get the following error. on line 22: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given line 22 is. if (mysqli_num_rows($dbc) == 0) { Here is the php code. if (isset($_POST['submitted'])) { // Handle the form. require_once '../../htmlpurifier/library/HTMLPurifier.auto.php'; $config = HTMLPurifier_Config::createDefault(); $config->set('Core.Encoding', 'UTF-8'); // replace with your encoding $config->set('HTML.Doctype', 'XHTML 1.0 Strict'); // replace with your doctype $purifier = new HTMLPurifier($config); $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT users.*, profile.* FROM users INNER JOIN contact_info ON contact_info.user_id = users.user_id WHERE users.user_id=3"); $about_me = mysqli_real_escape_string($mysqli, $purifier->purify($_POST['about_me'])); $interests = mysqli_real_escape_string($mysqli, $purifier->purify($_POST['interests'])); if (mysqli_num_rows($dbc) == 0) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"INSERT INTO profile (user_id, about_me, interests) VALUES ('$user_id', '$about_me', '$interests')"); } if ($dbc == TRUE) { $dbc = mysqli_query($mysqli,"UPDATE profile SET about_me = '$about_me', interests = '$interests' WHERE user_id = '$user_id'"); echo '<p class="changes-saved">Your changes have been saved!</p>'; } if (!$dbc) { // There was an error...do something about it here... print mysqli_error($mysqli); return; } }

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  • pathogen#infect not updating the runtimepath

    - by Taylor Price
    I have started working with pathogen.vim with gvim on Windows, following Tim Pope's setup guide at his github repository here. However, I'm running into the problem that pathogen#infect() does not seem to be modifying the runtimepath (as seen by running :echo &runtimepath in gvim). The simple test case _vimrc that I came up with is as follows. Please note that pathogen gets loaded just fine. "Set a base directory. let $BASE_DIR='H:\development\github\vimrc' "Source pathogen since it's not in the normal autoload directory. source $BASE_DIR\autoload\pathogen.vim "Start up pathogen call pathogen#infect() "call pathogen#infect('$BASE_DIR\functions') Neither running pathogen#infect() without an argument (which should add the bundles directory under the vimfiles directory) nor specifying a directory to contain files works. Substituting the pathogen#infect() call with pathogen#runtime_prepend_subdirectories('$BASE_DIR\functions'), which is what pathogen#infect() does fails to change the runtimepath as well. Any ideas that I've missed? Any more information that would be helpful? My repository with the non-trivial example is here.

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  • Zend_Dojo_Form not rendering in layout

    - by Grant Collins
    Hi, I have a quick question about adding Zend_Dojo_Form into Zend_layouts. I have a Zend_Dojo_Form that I want to display in the layout that is used for a particular controller. I can add the form to the layout without any issue however the dojo elements fail to render, as they would do if I added the form to a standard view. Is there any reason why this would be the case? Do I need to do something to the layout so that it will enable the components for this embedded form in the layout. Any other dojo enabled forms that are added in the view using this layout work fine. My form is created in the usual way: class QuickAddJobForm extends Zend_Dojo_Form{ public function init(){ $this->setName('quickaddjobfrm') ->setMethod('post') ->setAction('/addjob/start/); /*We now create the elements*/ $jobTitle = new Zend_Dojo_Form_Element_TextBox('jobtitle', array( 'trim' => true ) ); $jobTitle->setAttrib('style', 'width:200px;') ->addFilter('StripTags') ->removeDecorator('DtDdWrapper') ->removeDecorator('HtmlTag') ->removeDecorator('Label'); .... $this->addElements(array($jobTitle, ....)); In the controller I declare the layout and the form in the init function: public function init(){ $this->_helper->layout->setLayout('add-layout'); $form = new QuickAddJobForm(); $form->setDecorators(array(array('ViewScript', array('viewScript' => 'quickAddJobFormDecorator.phtml')))); $this->_helper->layout()->quickaddjob = $form; In my layout Where I want the form I have: echo $this->layout()->quickaddjob; Why would adding this form in the layout fail to render/add the Dojo elements? All that is currently being displayed are text boxes, rather than some of the other components such as ComboBoxes/FilteringSelects etc... Thanks in advance.

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