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  • PHP Sessions and Passing Session ID

    - by Jason McCreary
    I have an API where I am passing the session id back and forth between calls. I set up the session like so: // start API session session_name('apikey'); session_id($data['apikey']); // required to link session session_start(); Although I named my session and am passing the session id via GET and POST using the name, PHP does not automatically resume that session. It always creates a new one unless I set the explicitly set the session id. I found some old user comments on www.php.net that said unless the session id is the first parameter PHP won't set it automatically. This seems odd, but even when I call tried it still didn't work: rest_services.php?apikey=sdr6d3subaofcav53cpf71j4v3&q=testing I have used PHP for years, but am a little confused on why I needed to explicitly set the session with session_id() when I am naming the session and passing it's key accordingly. UPDATE It seems I wasn't clear. My question is why is setting the session ID with session_id() required when I am passing the id, using the session name apikey, via $_GET or $_POST. Theoretically this is no different than PHP's SID when cookies are disabled. But for me it doesn't work unless I explicitly set the session ID. Why?

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  • Displaying data from a linked table and displaying it as a list with HTML/PHP/MySQL

    - by user1672694
    I have three tables. students studentID | FirstName | LastName | Email | Form course CCode | Title courseenrolement courseenrolementid | studentID | ccode | complete | scode With the website, I have a page where I can view all the current enrolements and I wish to be able to view the list displaying the first name, surname and course title. I know I could do it with the following SQL (for the names): SELECT FirstName, LastName FROM student, courseenrolement WHERE courseenrolement.studentID = student.studentID But I am unsure how to get this to work using HTML/PHP. At present I only know how to display the studentID and CCode from the courseenrolement table using the following code: <ul> <?php foreach ($courseenrolements as $ce): ?> <li> <form action="" method="post"> <div> <?php htmlout($ce['studentID']); ?> <?php htmlout($ce['CCode']); ?> <input type="hidden" name="courseenrolementid" value="<?php echo $ce['courseenrolementid']; ?>"> <input type="submit" name="action" value="Edit"> <input type="submit" name="action" value="Delete"> </div> </form> </li> <?php endforeach; ?> </ul> which displays this: But I would like the names and course title. I managed to get it to show the names etc in the dropdown on the 'Add new' form, so I would assume it will be similar, but just unsure on how exactly. Thanks in advance

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  • php: autoload exception handling.

    - by YuriKolovsky
    Hello again, I'm extending my previous question (Handling exceptions within exception handle) to address my bad coding practice. I'm trying to delegate autoload errors to a exception handler. <?php function __autoload($class_name) { $file = $class_name.'.php'; try { if (file_exists($file)) { include $file; }else{ throw new loadException("File $file is missing"); } if(!class_exists($class_name,false)){ throw new loadException("Class $class_name missing in $file"); } }catch(loadException $e){ header("HTTP/1.0 500 Internal Server Error"); $e->loadErrorPage('500'); exit; } return true; } class loadException extends Exception { public function __toString() { return get_class($this) . " in {$this->file}({$this->line})".PHP_EOL ."'{$this->message}'".PHP_EOL . "{$this->getTraceAsString()}"; } public function loadErrorPage($code){ try { $page = new pageClass(); echo $page->showPage($code); }catch(Exception $e){ echo 'fatal error: ', $code; } } } $test = new testClass(); ?> the above script is supposed to load a 404 page if the testClass.php file is missing, and it works fine, UNLESS the pageClass.php file is missing as well, in which case I see a "Fatal error: Class 'pageClass' not found in D:\xampp\htdocs\Test\PHP\errorhandle\index.php on line 29" instead of the "fatal error: 500" message I do not want to add a try/catch block to each and every class autoload (object creation), so i tried this. What is the proper way of handling this?

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  • Forcing user to new page in php. (PHP newbie)

    - by JohnC
    Hello I'm a newbie web programmer. My background is writing Windows applications with sql. I'm putting together my 1st data entry screens in Php. I have a search form that links to a form that displays records in a grid. On each row of the grid I have a delete url to allow the user to remove a record. This links to a form delete.php (which calls the sql to remove the record). Ideally I would like to automatically take the user back to the search form rather than forcing the user to click on a link to do so. I have used ob_start with the header to do this elsewhere but cannot get it to work on this page. Is there another way to do it? (Using php 5 as part of LAMP) file delete.php <?php $id = $_GET['recordID']; //ob_start(); require_once('connections/local.php'); mysql_select_db($database_local, $local); mysql_query("DELETE FROM user_access WHERE id = {$id}") or die(mysql_error()); echo("Record ".$id." deleted"); echo("<br>"); //header("location:http://localhost/search7.htm); //ob_flush(); echo("<a href=\"http://localhost/search7.htm\">Search for Members</a>"); ?>

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  • Trouble decoding JSON string with PHP

    - by Anthony
    I'm trying to send an array of objects from JS to PHP using JSON. I have an array of players as follows: var player; var players = new Array(); //loop for number of players player = new Object(); player.id = theID; players[i] = player; Then my AJAX call looks like this: JSONplayers = JSON.stringify(players); $.ajax({ type: "POST", url: "php/ajax_send_players.php", data: { "players" : JSONplayers } On the PHP side the decode function looks like this $players = $_REQUEST['players']; echo var_dump($players); $players = json_decode($players); echo 'players: ' .$players. '--'. $players[0] . '--'. $players[0]->id; Debugging in chrome, the JSON players var looks like this before it is sent: JSONplayers: "[{"id":"Percipient"},{"id":"4"}]" And when I vardump in PHP it looks OK, giving this: string(40) "[{\"id\":\"Percipient\"},{\"id\":\"4\"}]" But I can't access the PHP array, and the echo statement about starting with players: outputs this: players: ---- Nothing across the board...maybe it has something to do with the \'s in the array, I am new to this and might be missing something very simple. Any help would be greatly appreciated. note I've also tried json_decode($players, true) to get it as an assoc array but get similar results.

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  • PHP JQuery: Where to specify uploadify destination folder

    - by Eamonn
    I have an uploadify script running, basic setup. Works fine when I hard code the destination folder for the images into uploadify.php - now I want to make that folder dynamic. How do I do this? I have a PHP variable $uploadify_path which contains the path to the folder I want. I have switched out my hard coded $targetPath = path/to/directory for $targetPath = $uploadify_path in both uploadify.php and check_exists.php, but it does not work. The file upload animation runs, says it is complete, yet the directory remains empty. The file is not hiding out somewhere else either. I see there is an option in the Javascript to specify a folder. I tried this also, but to no avail. If anyone could educate me on how to pass this variable destination to uploadify, I'd be very grateful. I include my current code for checking (basically default): The Javascript <script type="text/javascript"> $(function() { $('#file_upload').uploadify({ 'swf' : 'uploadify/uploadify.swf', 'uploader' : 'uploadify/uploadify.php', // Put your options here }); }); </script> uploadify.php $targetPath = $_SERVER['DOCUMENT_ROOT'] . $uploadify_path; // Relative to the root if (!empty($_FILES)) { $tempFile = $_FILES['Filedata']['tmp_name']; $targetFile = $targetPath . $_FILES['Filedata']['name']; // Validate the file type $fileTypes = array('jpg','jpeg','gif','png'); // File extensions $fileParts = pathinfo($_FILES['Filedata']['name']); if (in_array($fileParts['extension'],$fileTypes)) { move_uploaded_file($tempFile,$targetFile); echo '1'; } else { echo 'Invalid file type.'; } }

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  • Somewhat lost with jquery + php + json

    - by Luis Armando
    I am starting to use the jquery $.ajax() but I can't get back what I want to...I send this: $(function(){ $.ajax({ url: "graph_data.php", type: "POST", data: "casi=56&nada=48&nuevo=98&perfecto=100&vales=50&apenas=70&yeah=60", dataType: "json", error: function (xhr, desc, exceptionobj) { document.writeln("El error de XMLHTTPRequest dice: " + xhr.responseText); }, success: function (json) { if (json.error) { alert(json.error); return; } var output = ""; for (p in json) { output += p + " : " + json[p] + "\n"; } document.writeln("Results: \n\n" + output); } }); }); and my php is: <?php $data = $_POST['data']; function array2json($data){ $json = $data; return json_encode($json); } ?> and when I execute this I come out with: Results: just like that I used to have in the php a echo array2json statement but it just gave back gibberish...I really don't know what am I doing wrong and I've googled for about 3 hours just getting basically the same stuff. Also I don't know how to pass parameters to the "data:" in the $.ajax function in another way like getting info from the web page, can anyone please help me? Edit I did what you suggested and it prints the data now thank you very much =) however, I was wondering, how can I send the data to the "data:" part in jQuery so it takes it from let's say user input, also I was checking the php documentation and it says I'm allowed to write something like: json_encode($a,JSON_HEX_TAG|JSON_HEX_APOS|JSON_HEX_QUOT|JSON_HEX_AMP) however, if I do that I get an error saying that json_encode accepts 1 parameter and I'm giving 2...any idea why? I'm using php 5.2

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  • how to use a PHP Constant that gets pulled from a database

    - by Ronedog
    Can you read out the name of a PHP constant from a database and use it inside of a php variable, to display the value of the constant for use in a menu? For example here's what I'm trying to accomplish In SQL: select menu_name AS php_CONSTANT where menu_id=1 the value returned would be L_HOME which is the name of a CONSTANT in a php config page. The php config page looks like this define('L_HOME','Home'); and gets loaded before the database call. The php usage would be $db_returned_constant which has a value of L_HOME that came from the db call, then I would place this into a string such as $string = '<ul><li>' . $db_returned_constant . '</li></ul>' and thus return a string that looks like $string = '<ul><li><a href="#" onclick="path_from_db">Home</a></li></ul>'. To sum up what I'm trying to do Load a config file based on the language preference query the db to return the menu name, which is the name of a CONSTANT in the config file loaded in step one, and also retrieve the menu_link which is used in the "onclick" event. Use a php variable to hold the name of the CONSTANT Place the variable into a string that gets echo'd out to create the menu displaying the value of the CONSTANT. I hope this makes enough sense...is it even possible to use a constant like this? Thanks.

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  • Flex + PHP + ValueObjects

    - by Tempname
    I have a php/flex value object that I am using to transmit data to/from in my application. Everything works great php-flex, but I am having an issue with flex-php. In my MergeTemplateService.php service I have the following code. This is the method that flex hits directly: function updateTemplate($valueObject){ $object = DAOFactory::getMergeTemplateDAO()->update($valueObject); return $object; } I am passing a value object that from flex looks like this: (com.rottmanj.vo::MergeTemplateVO)#0 communityID = 0 creationDate = (null) enterpriseID = 0 lastModifyDate = (null) templateID = 2 templateName = "My New Test Template" userID = 0 The issue I am having is that my updateTemplate method sees the value object as an array and not an object. In my amfphp globals.php I have set my voPath as: $voPath = "services/class/dto/"; Any help with this is greatly appreciated Here are my two value objects: AS3 VO: package com.rottmanj.vo { [RemoteClass(alias="MergeTemplate")] public class MergeTemplateVO { public var templateID:int; public var templateName:String; public var communityID:int; public var enterpriseID:int; public var userID:int; public var creationDate:String; public var lastModifyDate:String public function MergeTemplateVO(data:Object = null):void { if(data != null) { templateID = data.templateID; templateName = data.templateName; communityID = data.communityID; enterpriseID = data.enterpriseID; userID = data.userID; creationDate = data.creationDate; lastModifyDate = data.lastModifyDate; } } } } PHPVO: <?php class MergeTemplate{ var $templateID; var $templateName; var $communityID; var $enterpriseID; var $userID; var $creationDate; var $lastModifyDate; var $_explictType = 'MergeTemplate'; } ?>

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  • .NET HttpModule does not send form variables to PHP file on RewritePath()

    - by jammus
    Hello friends. We have an application running on IIS 6 which uses a custom HttpModule to rewrite urls. This works great (well done us) except in the case where the Context.RewritePath destination is a .php file. The php file is executed as expected, however the $_POST collection is empty meaning it cannot access any forms which are submitted to rewritten urls. The problem does not exist when rewriting to .aspx files as the Request.Form collection is fine. My question therefore has two parts: Why is the $_POST collection not being populated? Is there a way to ensure that the .php $_POST collection is correctly populated after a rewrite? I don't have much to show in the way of code. There's just a simple: context.RewritePath(newPath); once the HttpModule has figured out where to send the request. Edit: Interestingly, if I do var_dump(file_get_contents('php://input')); in the PHP file (method described here) the contents of the form is displayed. So the data is reaching the PHP script but not the $_POST array.

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  • PHP IDE with Integrated Web Server

    - by seth
    Note: This is not another "What is the best PHP IDE?" question. I'm looking for a PHP IDE with a specific feature, namely an integrated / embedded (php enabled) web server; ideally with xdebug pre-bundled. I already know that Aptana 1.5 has this functionality (and some older versions of Zend Studio as well), but Aptana 1.5 hasn't been supported for quite some time and as we make the transition to PHP 5.3 and beyond, it's usefulness will diminish significantly. I've looked at some options including Eclipse PDT and NetBeans, but it seems every PHP IDE relies on a separate local/remote web server to actually interpret the code. I know installing a web server locally is fairly trivial, but this is for a classroom solution, where installing, configuring, and maintaining a web server on 1000 machines is simply not feasible. A remote server solution will also not work due to the need to use debugging functionality (xdebug currently requires a hardcoded IP for the debug client). This seems like such an obvious feature/plugin for a PHP IDE, but my research thus far has turned up no results.

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  • URL routing in an MVC framework - PHP

    - by Walderman
    I'm developing an MVC framework in PHP from scratch; mostly for the learning experience but this could easily end up in a live project. I went through this tutorial as a base and I've expanded from there. Requests are made like this: examplesite.com/controller/action/param1/param2/ and so on... And this is my .htaccess file: RewriteEngine on RewriteCond %{REQUEST_FILENAME} !-f RewriteCond %{REQUEST_FILENAME} !-d RewriteRule ^(.*)$ index.php?rt=$1 [L,QSA] So all requests go to index.php and they are routed to the correct controller and action from there. If no controller or action is given, then the default 'index' is assumed for both. I have an index controller with an index action, which is supposed to be the home page of my site. I can access it by going to examplesite.com (since the index part is assumed). It has some images, a link to a stylesheet, and some scripts. They are linked with paths relative to index.php. I thought this would be fine since all request go to index.php and all content is simply included in this page using php. This works if I go to examplesite.com. I will see all of the images and styles, and scripts will run. However, if I go to examplesite.com/index, I am routed to the correct part of the site, but all of the links don't work. Does the browser think I am in a different folder? I would like to be able to use relative paths for all of the content in my site, because otherwise I need to use absolute paths everywhere to make sure things will show up. Is this possible?

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  • trouble connecting to MySql DB (PHP)

    - by user332817
    Hi I have the following PHP code to connect to my db. <?php ob_start(); $host="localhost"; // Host name $username="root"; // Mysql username $password=""; // Mysql password $db_name="test"; // Database name $tbl_name="members"; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); ?> however I get the following error: Warning: mysql_connect() [function.mysql-connect]: [2002] A connection attempt failed because the connected party did not (trying to connect via tcp://localhost:3306) in C:\Program Files (x86)\EasyPHP-5.3.2i\www\checklogin.php on line 11 Warning: mysql_connect() [function.mysql-connect]: A connection attempt failed because the connected party did not properly respond after a period of time, or established connection failed because connected host has failed to respond. in C:\Program Files (x86)\EasyPHP-5.3.2i\www\checklogin.php on line 11 Fatal error: Maximum execution time of 30 seconds exceeded in C:\Program Files (x86)\EasyPHP-5.3.2i\www\checklogin.php on line 11 I am able to add a db/tables via phpmyadmin but I cant connect using php. here is a screenshot of my phpmyadmin page: http://img294.imageshack.us/img294/1589/sqls.jpg any help would be appreciated, thanks in advance.

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  • Mimic Coldfusion's debug output in PHP?

    - by TekiusFanatikus
    I'm trying to mimic Coldfusion's debug output in PHP. Here's an example of what it looks like (ie. Execution Time section): I've turned to XDebug. Ideally, the exception stack error output would be what I'd be looking for. However, it only shows up when an exception occurs. I also tried something like (in our CMS-ish app) this (original question here): $content.= "<?php xdebug_start_trace('e:/xdebug/trace');?>"; $content.= "<?php require('".$page['file_'.LG]."'); ?>"; $content.= "<?php xdebug_stop_trace();?>"; ... $content.= "<?php echo readfile('e:/xdebug/trace.xt');?>"; However, I get an insane, browser crashing HTML table dropped at the bottom of page. Not very efficient. My php.ini config: xdebug.trace_format = 2 xdebug.collect_vars = 1 xdebug.collect_params = 4 xdebug.dump_globals = 1 xdebug.dump.SERVER = 'REQUEST_URI' xdebug.show_local_vars = 1 xdebug.show_mem_delta = 1 I'm just wondering if someone has already done something similar?

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  • PHP active page code - I can't figure out parse error

    - by dmschenk
    I'm trying to build an active page menu with PHP and MySQL and am having a difficult time fixing the error. In the while statement I have an if statement that is giving me fits. Basically I think I'm saying that "thispage" is equal to the "title" based on pageID and as the menu is looped through if "thispage" is equal to "title" then echo id="active". Thanks <?php mysql_select_db($database_db_connection, $db_connection); $query_rsDaTa = "SELECT * FROM pages WHERE pagesID = 4"; $rsDaTa = mysql_query($query_rsDaTa, $db_connection) or die(mysql_error()); $row_rsDaTa = mysql_fetch_assoc($rsDaTa); $totalRows_rsDaTa = mysql_num_rows($rsDaTa); $query_rsMenu = "SELECT * FROM menu WHERE online = 1 ORDER BY menuPos ASC"; $rsMenu = mysql_query($query_rsMenu, $db_connection) or die(mysql_error()); $thisPage = ($row_rsDaTa['title']); ?> <link href="../css/MainStyle.css" rel="stylesheet" type="text/css" /> <h2><?php echo $thisPage; ?></h2> <div id="footcontainer"> <ul id="footlist"> <?php while($row_rsMenu = mysql_fetch_assoc($rsMenu)) { echo (" <li" . <?php if ($thisPage==$row_rsDaTa['title']) echo id="active"; ?> . "<a href=\"../" . $row_rsMenu['menuURL'] . "\">" . $row_rsMenu['menuName'] . "</a></li>\n"); } echo "</ul>\n"; ?> </div> <?php mysql_free_result($rsMenu); mysql_free_result($rsDaTa); ?>

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  • PHP While loop seperating unique categories from multiple 'Joined' tables

    - by Hob
    I'm pretty new to Joins so hope this all makes sense. I'm joining 4 tables and want to create a while loop that spits out results nested under different categories. My Tables categories id | category_name pages id | page_name | category *page_content* id | page_id | image_id images id | thumb_path My current SQL join <?php $all_photos = mysql_query(" SELECT * FROM categories JOIN pages ON pages.category = categories.id JOIN image_pages ON image_pages.page_id = pages.id JOIN images ON images.id = image_pages.image_id ");?> The result I want from a while loop I would like to get something like this.... Category 1 page 1 Image 1, image 2, image 3 page 2 Image 2, image 4 Category 2 page 3 image 1 page 4 image 1, image 2, image 3 I hope that makes sense. Each image can fall under multiple pages and each page can fall under multiple categories. at the moment I have 2 solutions, one which lists each category several times according to the the amount of pages inside them: eg. category 1, page 1, image 1 - category 1, page 1, image 2 etc One that uses a while loop inside another while loop inside another while loop, resulting in 3 sql queries. <?php while($all_page = mysql_fetch_array($all_pages)) { ?> <p><?=$all_page['page_name']?></p> <?php $all_images = mysql_query("SELECT * FROM images JOIN image_pages ON image_pages.page_id = " . $all_page['id'] . " AND image_pages.image_id = images.id"); ?> <div class="admin-images-block clearfix"> <?php while($all_image = mysql_fetch_array($all_images)) { ?> <img src="<?=$all_image['thumb_path']?>" alt="<?=$all_image['title']?>"/> <?php } ?> </div> <?php } } ?

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  • PHP unlink OR rewrite own/current file by itself

    - by Email
    Hi Task: Cut or erase a file after first walk-through. i have an install file called "index.php" which creates another php file. <? /* here some code*/ $fh = fopen($myFile, 'w') or die("can't open file"); $stringData = "<?php \n echo 'hallo, *very very long text*'; \n ?>"; fwrite($fh, $stringData); /*herecut"/ /*here some code */ after the creation of the new file this file is called and i intent to erase the filecreation call since it is very long and only needed on first install. i therefor add to the above code echo 'hallo, *very very long text*'; \n ***$new= file_get_contents('index.php'); \n $findme = 'habanot'; $pos = strpos($new, $findme); if ($pos === false) { $marker='herecut';\n $new=strstr($new,$marker);\n $new='<?php \n /*habanot*/\n'.$new;\n $fh = fopen('index.php', 'w') or die 'cant open file'); $stringData = $new; fwrite($fh, $stringData); fclose($fh);*** ?>"; fwrite($fh, $stringData);]} Isnt there an easier way or a function to modify the current file or even "self destroy" a file after first call? Regards

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  • PHP, MySQL: Security concern; Page loads in a weird way

    - by Devner
    Hi all, I am testing the security of my website. I am using the following URL to load a PHP page in my website, on localhost: http://localhost/domain/user/index.php/apple.php When I do this, the page is not loading normally; Instead the images, icons used in the page simply vanish/disappear from the page. Only text appears. And also on any link I click on this page, it brings me to this same page again without navigating to the required page. So if I have hyperlinks to other pages, such as "SEARCH", which points to search.php, instead of navigating to the search.php page, it refreshes the index.php page and just appends the page name of the destination page to the end of the URL. For example, say I used the link above. It then loads the index.php page minus the images at it's will. When I click on the "Search" link to navigate to the search page, I see the following in the URL: http://localhost/domain/user/index.php/search.php I have a redirection configured to a 404 error page in my .htaccess file, but the page does not redirect to the 404 error page. Notice the search.php towards the end of the URL above. Any other link that I click, reloads the index.php page and just appends the destination page name to the end of the URL like I have shown above. I was expecting to see a 404 Error but that does not happen. The URL should not even be able to load the page because I do NOT have a "index.php" folder in my website. What can I do to solve this? All help is appreciated. Thank you.

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  • PHP class_exists always returns true

    - by Ali
    I have a PHP class that needs some pre-defined globals before the file is included: File: includes/Product.inc.php if (class_exists('Product')) { return; } // This class requires some predefined globals if ( !isset($gLogger) || !isset($db) || !isset($glob) ) { return; } class Product { ... } The above is included in other PHP files that need to use Product using require_once. Anyone who wants to use Product must however ensure those globals are available, at least that's the idea. I recently debugged an issue in a function within the Product class which was caused because $gLogger was null. The code requiring the above Product.inc.php had not bothered to create the $gLogger. So The question is how was this class ever included if $gLogger was null? I tried to debug the code (xdebug in NetBeans), put a breakpoint at the start of Product.inc.php to find out and every time it came to the if (class_exists('Product')) clause it would simply step in and return thus never getting to the global checks. So how was it ever included the first time? This is PHP 5.1+ running under MAMP (Apache/MySQL). I don't have any auto loaders defined.

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  • Pass a variable from the source file to an included file in PHP

    - by Alpha1
    For my website I want to store the general format of the site in a single PHP file in a single location, and each of the different pages content in the local location of the page. I then want to pass the title and content address to the included file via a variable. However I can't get the included format file to read the variables storing the title and content data. AKA, the called file for the individual page would be: <?php $title = 'Some Title'; $source_file = 'content.php'; readfile('http:...../format.php'); ?> The format file would be: <html> ... <title> <?php echo $title; ?> </title> ... <?php include($source_file); ?> ... I recall reading somewhere I need to include something to get the variables at the start of the format file, however I can't remember what it is or find where I found that information.

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  • How to avoid notice in php when one of the conditions is not true

    - by user225269
    I've notice that when one of the two conditions in a php if statement is not true. You get an undefined index notice for the statement that is not true. And the result in my case is a distorted web page. For example, this code: <?php session_start(); if (!isset($_SESSION['loginAdmin']) && ($_SESSION['loginAdmin'] != '')) { header ("Location: loginam.php"); } else { include('head2.php'); } if (!isset($_SESSION['login']) && ($_SESSION['login'] != '')) { header ("Location: login.php"); } else { include('head3.php'); } ?> If one of the if statements is not true. The one that is not true will give you a notice that it is undefined. In my case it says that the session 'login' is not defined. If session 'LoginAdmin' is used. What can you recommend that I would do in order to avoid these undefined index notice.

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  • Unable to Get values from Web Form to a PHP Class to Display

    - by kentrenholm
    I am having troubles getting the value from my variables submitted via a web form using a PHP class file. Here is my structure of the web page: Order Form Page Process.php Page Book.php Page I can easily get the user data entered (on Order Form Page), process, and display it on the Process.php page. The issue is that I must create a Book class and print the details of the data using the Book class. I have an empty constructor printing out "created" so I know my constructor is being called. I also am able to print the word "title" so I know I can print to the screen by using the Book class. My issue is that I can't get values in my variables in the Book class. Here is my variable declaration: private $title; Here is my printDetails function: public function printDetails () { echo "Title: " . $this->title . "<br />"; } Here is my new instance of the book class: $bookNow = new book; Here are my get and set functions: function __getTitle($title) { return $this->$title; } function __setTitle($title,$value) { $this->$title = $value; } I do have four other variables that I'm looking to display as well. Each of those have their own variable declaration, a line in printDetails, and their own setter and getter. Lastly, I also have a call to the Book class in my process PHP. It looks like this: function __autoload($book) { include $book . '.php'; } $bookNow = new book(); Any help, much appreciated. It must be something so very small (I'm hoping).

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  • JQuery to PHP function and back Ajaxed

    - by Xaris
    Hi all, i have a set of php function that i want to call on different events mostly onclick with jquery async (ajax). The first function is called on load $(document).ready(function() { $("#div2").hide('slow'); $("#div1").empty().html('<img src="ajax-loader.gif" />'); $.ajax( { type: "POST", url: "WebFunctions.php", data: {'func':'1'}, success: function(html) { $("#div1").show('slow').html(html) } }); The Data: {'func':'1'} -- is a switch statement on the php side switch($_POST['func']) { case '1': getParents(); break; case '2': getChilds(params); break; case '3': getChildObjects(params); break; default: } "This functions are calls to a soap server" <-- irrelevant. So when that function finishes i get an array which contains IDs and Names. I echo the names but i want the ID for reference so when i click on the echoed name i can call an other php function with parameter the ID of the name... How do i get rid of the switch statement?? How do i call properly php functions and pass params to it??? How can i save this IDs so when i click on an item with that id an other php function is called?? Plz feel free to ask any question, any answer is welcome :)

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  • Include php code within echo from a random text

    - by lisa
    I want to display a php code at random and so for I have <?php // load the file that contain thecode $adfile = "code.txt"; $ads = array(); // one line per code $fh = fopen($adfile, "r"); while(!feof($fh)) { $line = fgets($fh, 10240); $line = trim($line); if($line != "") { $ads[] = $line; } } // randomly pick an code $num = count($ads); $idx = rand(0, $num-1); echo $ads[$idx]; ?> The code.txt has lines like <?php print insert_proplayer( array( "width" => "600", "height" => "400" ), "http://www.youtube.com/watch?v=xnPCpCVepCg"); ?> Proplayer is a wordpress plugin that displays a video. The codes in code.txt work well, but not when I use the pick line from code.txt. Instead of the full php line I get: "width" => "600", "height" => "400" ), "http://www.youtube.com/watch?v=xnPCpCVepCg"); ?> How can I make the echo show the php code, rather than a txt version of the php code?

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  • How to load data on option box in php

    - by user225269
    I'm having trouble loading the mysql data on the html form option box using php. Here's my code: <?php $con = mysql_connect("localhost","myuname","mypassword"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("school", $con); $idnum= mysql_real_escape_string($_POST['idnum']); $result = mysql_query("SELECT * FROM student WHERE IDNO='$idnum'"); ?> <?php while ( $row = mysql_fetch_array($result) ) { ?> <tr> <td width="30" height="35"><font size="2">*I D Number:</td> <td width="30"><input name="idnum" type="text" maxlength="5" value="<?php echo $row["IDNO"]; ?>" readonly="readonly"></td> </tr> My problem is loading it here: <td><font size="2">Gender</td> <td> <select name="gender" id="gender"> <font size="2"> <option value="<?php echo $line['IDNO']; ?> "><?php $line['GENDER'] ; ?></option> </select></td></td> The table looks like this: IDNO | GENDER 123 | M 321 | F What am I supposed to do?To load the exact gender corresponding to the IDNO?

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