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  • Fastcgi 500 error on preg_match_all in PHP

    - by Bertvan
    Hi, I'm trying to set up some exotic PHP code (I'm no expert), and I get a FastCGI Error 500 on a PHP line containing 'preg_match_all'. When I comment out the line, the page is returned with a 200 (but not how it was meant to be). The code is parsing php, html and javascript content loaded from the database and is composing them to return the finished page. Now, by placing around some error_log entries I could determine that the line with the preg_match_all is the cause of the 500. However the line is hit multiple times during the loading of the page and on other occasions, the line does not cause an error. Here's how it looks like exactly: preg_match_all ("/(<([\w]+)[^>]*>)((?:.|\n)*)(<\/\\2>)/", $part['data'], $tags, PREG_PATTERN_ORDER|PREG_OFFSET_CAPTURE); The subject string is a piece of text that looks like: <script> ... some javascript functions ... </script> [Edit:] This is code that is up and running correctly elsewhere, so this very well could be a PHP setting or environment difference. I'm using PHP 5.2.13 on IIS6 with FastCGI. [Edit:] Nothing is mentioned in the log files. At least not in the ones I checked: IIS Logs Event Logs PHP Log Any thoughts or direction would be welcome.

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  • search form in php via ajax

    - by fusion
    i've a search form wherein the database query has been coded in php and the html file calls this php file via ajax to display the results in the search form. the problem is, i would like the result to be displayed in the same form as search.html; yet while the ajax works, it goes to search.php to display the results. search.html: <!DOCTYPE html> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8"> <script src="scripts/search_ajax.js" type="text/javascript"></script> </head> <body> <form id="submitForm" method="post"> <div class="wrapper"> <div class="field"> <input name="search" id="search" /> </div><br /> <input id="button1" type="submit" value="Submit" class="submit" onclick="run_query();" /><br /> </div> <div id="searchContainer"> </div> </form> </body> </html> if i add action="search.php" to the form tag, it displays the result but on search.php. i'd like it to display on the same form [i.e search.html, and not search.php] if i just add the javascript function [as done above], it displays nothing on search.html

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  • Netbeans PHP Validation sees endif as a syntax error

    - by Asaf
    I have this part of code <?php for ($j=0; $j < $count; $j++): ?> <?php if(isset(votes[$j])): ?> <dt>something something</dt> <dd> <span><?php echo $result; ?>%</span> <div class="bar"> </div> </dd> <?php else: ?> <dt>info</dt> <dd> <span>0</span> <div class="bar"> <div style="width: 0px"></div> </div> </dd> <?php endif; ?> <?php endfor; ?> now Netbeans insists that on the endif line (near the end) there's a syntax error: Error Syntax error: expected: exit, identifier, variable, function... Is there some sort of known problem with the validation of endif on Netbeans ?

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  • PHP header location redirect causing 500 Internal Server error

    - by Globalz
    Hi, I I keep getting a 500 Internal Server Error when the script below reaches the header('location:php_email_thankyou.php'). Im not sure what is causing this, as I can place the header expression before or after the if statements and it works fine. In firebug it mentions a GET request for the php_email_thankyou.php page not sure if that means anything... <?php ini_set('display_errors', 'On'); error_reporting(E_ALL | E_STRICT); include('php/cl/cl_val.php'); $val = new Validate; $print_errors = false; if (isset($_POST['email(email)'])){ if(isset($_SERVER['HTTP_X_REQUESTED_WITH'])) { $validation = $val->clean($_POST); if (isset($validation['send'])) { header('location:php_email_thankyou.php'); exit(); } else { print json_encode($validation); exit(); } } else { $validation = $val->clean($_POST); } } ?> <!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"> Thanks heaps!

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  • PHP: Making my code simpler/shorter welcome message

    - by Karem
    Any suggestion to make this welcome message shorter: <?php if(isset($_SESSION['user_id'])) { if(isSet($_SESSION['1stTime'])){ ?> <strong id="welcome" style="font-size: 10px;"> <a href="logout.php"> Logga ut </a> </strong> <?php }else{ $_SESSION['1stTime'] = time(); ?> <script> $(document).ready(function() { $("#welcome").fadeIn("slow"); setTimeout(function(){ $("#welcome").fadeOut("slow"); setTimeout(function(){ $("#welcome").html("<a href='logout.php'>Logga ut</a>"); $("#welcome").fadeIn(); }, 800); }, 5000); }); </script> <strong id="welcome" style="display: none; color: #FFF; font-size: 10px;">Hej, <?php echo $FULL; ?>!</strong> <?php } } ?> First it checks if you are signed in. Next if 1stTime is set, if it is then show "Log out" in swedish, if it isnt, then introduce with "Hi, NAME", and then change to "Log out" after 5 seconds(jquery) + set the session How can i make this simpler?

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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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  • 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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  • 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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  • 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: documentElemnt->childNodes warning

    - by jun
    $xml = file_get_contents(example.com); $dom = new DomDocument(); $dom->loadXML($xml); $items = $dom->documentElement; foreach($items->childNodes as $item) { $childs = $item->childNodes; foreach($childs as $i) { echo $i->nodeValue . "<br />"; } } Now I get this warning in every 2nd foreach: Warning: Invalid argument supplied for foreach() in file_example.php on line 14 Please help guys. Thanks!

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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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