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• swfupload session problem destroy session

  - by saquib

  Hello Friends, I have a problem with swfupload. I am passing session_id() like this /upload-file.php?s=189477fcfa1ec7f630e70a09e1e84cae but its not maintaining session and destroying my current session (logging me out) here is code in file upload. <?php if(isset($_GET['s'])) { session_id($_GET['s']); session_start(); require_once 'admin/class/user.php'; $u = new User(); //Check for user logged in if($u->islogged() == FALSE) { header("location: index.php"); exit(); code continue ..... } because am not logged in server redirect me to the index.php this is swfupload debugger window output SWF DEBUG: ----- END SWF DEBUG OUTPUT ---- SWF DEBUG: SWF DEBUG: Event: fileDialogStart : Browsing files. Multi Select. Allowed file types: *.jpg SWF DEBUG: Select Handler: Received the files selected from the dialog. Processing the file list... SWF DEBUG: Event: fileQueued : File ID: SWFUpload_0_0 SWF DEBUG: Event: fileDialogComplete : Finished processing selected files. Files selected: 1. Files Queued: 1 SWF DEBUG: StartUpload: First file in queue SWF DEBUG: Event: uploadStart : File ID: SWFUpload_0_0 SWF DEBUG: ReturnUploadStart(): File accepted by startUpload event and readied for upload. Starting upload to /upload-file.php?s='189477fcfa1ec7f630e70a09e1e84cae' for File ID: SWFUpload_0_0 SWF DEBUG: Event: uploadProgress (OPEN): File ID: SWFUpload_0_0 SWF DEBUG: Event: uploadProgress: File ID: SWFUpload_0_0. Bytes: 317793. Total: 317793 SWF DEBUG: Event: uploadError: HTTP ERROR : File ID: SWFUpload_0_0. HTTP Status: 302. SWF DEBUG: Event: uploadComplete : Upload cycle complete. SWF DEBUG: StartUpload: First file in queue SWF DEBUG: StartUpload(): No files found in the queue.

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• Conditional operator in Mako using Pylons

  - by Antoine Leclair

  In PHP, I often use the conditional operator to add an attribute to an html element if it applies to the element in question. For example: <select name="blah"> <option value="1"<?= $blah == 1 ? ' selected="selected"' : '' ?>> One </option> <option value="2"<?= $blah == 2 ? ' selected="selected"' : '' ?>> Two </option> </select> I'm starting a project with Pylons using Mako for the templating. How can I achieve something similar? Right now, I see two possibilities that are not ideal. Solution 1: <select name="blah"> % if blah == 1: <option value="1" selected="selected">One</option> % else: <option value="1">One</option> % endif % if blah == 2: <option value="2" selected="selected">Two</option> % else: <option value="2">Two</option> % endif </select> Solution 2: <select name="blah"> <option value="1" % if blah == 1: selected="selected" % endif >One</option> <option value="2" % if blah == 2: selected="selected" % endif >Two</option> </select> In this particular case, the value is equal to the variable tested (value="1" = blah == 1), but I use the same pattern in other situations, like <?= isset($variable) ? ' value="$variable" : '' ?>. I am looking for a clean way to achieve this using Mako.

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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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• Ternary operator

  - by Antoine Leclair

  In PHP, I often use the ternary operator to add an attribute to an html element if it applies to the element in question. For example: <select name="blah"> <option value="1"<?= $blah == 1 ? ' selected="selected"' : '' ?>> One </option> <option value="2"<?= $blah == 2 ? ' selected="selected"' : '' ?>> Two </option> </select> I'm starting a project with Pylons using Mako for the templating. How can I achieve something similar? Right now, I see two possibilities that are not ideal. Solution 1: <select name="blah"> % if blah == 1: <option value="1" selected="selected">One</option> % else: <option value="1">One</option> % endif % if blah == 2: <option value="2" selected="selected">Two</option> % else: <option value="2">Two</option> % endif </select> Solution 2: <select name="blah"> <option value="1" % if blah == 1: selected="selected" % endif >One</option> <option value="2" % if blah == 2: selected="selected" % endif >Two</option> </select> In this particular case, the value is equal to the variable tested (value="1" = blah == 1), but I use the same pattern in other situations, like <?= isset($variable) ? ' value="$variable" : '' ?>. I am looking for a clean way to achieve this using Mako.

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• zend session exception on zend_session::start with forms

  - by Grant Collins

  Hi I'm having issues with trying to use Zend_Form_SubForm and sessions. My controller is in essance acting a wizard showing different subforms depending on the stage of the wizard. Using the example I am planning on storing the forms in a session namespace. My controller looks like this. include 'mylib/Form/addTaskWizardForm.php'; class AddtaskController extends Zend_Controller_Action{ private $config = null; private $log = null; private $subFormSession = null; /** * This function is called and initialises the global variables to this object * which is the configuration details and the logger to write to the log file. */ public function init(){ $this->config = Zend_Registry::getInstance()->get('config'); $this->log = Zend_Registry::getInstance()->get('log'); //set layout $this->_helper->layout->setLayout('no-sidemenus'); //we need to get the subforms and $wizardForms = new addTaskWizardForm(); $this->subFormSession = new Zend_Session_Namespace('addTaskWizardForms'); if(!isset($this->subFormSession->subforms)){ $this->subFormSession->subforms = $wizardForms; } } /** * The Landing page controller for the site. */ public function indexAction(){ $form = $this->subFormSession->subforms->getSubForm('start'); $this->view->form = $form; } However this is causing the application session to crash out with Uncaught exception 'Zend_Session_Exception' with message 'Zend_Session::start() Any idea why this is having issues with the Zend Session?? thanks.

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• Wrong colors when merging images with PHP

  - by OfficeJet

  Hi, I want to get images ID's and creat from files a merged image according to the given ID's. This code is called by ajax and return the image file name (which is the server time to prevent browser caching). code: if (isset($_REQUEST['items'])){ $req_items = $_REQUEST['items']; } else { $req_items = 'a'; } $items = explode(',',$req_items); $bg_img = imagecreatefrompng('bg.png'); for ($i=0; $i<count($items); $i++){ $main_img = $items[$i].'-large.png'; $image = imagecreatefrompng($main_img); $image_tc = imagecreatetruecolor(300, 200); imagecopy($image_tc,$image,0,0,0,0,300,200); $black = imagecolorallocate($image_tc, 0, 0, 0); imagecolortransparent($image_tc, $black); $opacity = 100; $bg_width = 300; $bg_height = 200; $dest_x = 0;//$image_size[0] - $bg_width - $padding; $dest_y = 0;//$image_size[1] - $bg_height - $padding; imagecopymerge($bg_img, $image_tc, $dest_x, $dest_y, 0, 0, $bg_width, $bg_height, $opacity) ; } $file = $_SERVER['REQUEST_TIME'].'.jpg'; imagejpeg($bg_img, $file, 100); echo $file; imagedestroy($bg_img); imagedestroy($image); die(); The images are shown exactly as I want but with wrong colors. I lately added the part with imagecreatetruecolor and imagecolortransparent, and still got wrong results. I also saved the PNG itself on a 24 bit format and also later as 8 bit - not helping. every ideas is very welcomed ! Thanks

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• javascript, php, cookies

  - by kennedy

  When i declare mac = 123, my internet explorer and firefox will keep refresh non-stop. And if i declare mac = getMacAddress it returns a value 1... I'm able to do a document.write(getMacAddress()) and it would able to display the mac address nicely. 1) Why my explorer will keep refreshing non-stop when i code it manually with "123" 2) why is the document.write able to display out, and when i store it to the cookie, somehow it didnt mange to capture into the cookie and it return a value of "1". Anyone help? create_users.php <script language="JavaScript"> function getMacAddress(){ document.macaddressapplet.setSep( "-" ); return (document.macaddressapplet.getMacAddress()); } function setCookie(c_name,value) { document.cookie = c_name + "=" +escape(value); } //error checking //var mac = getMacAddress(); var mac = "123"; setCookie('cookie_name',mac); window.location = "checkAvailability.php"; </script> checkAvailability.php $javascript_cookie_value = isset($_COOKIE["cookie_name"]) ? $_COOKIE["cookie_name"] : 1; mysql_query("INSERT INTO test (mac) VALUES ('$javascript_cookie_value')");

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• PHP Overloading, singleton instance

  - by jamalali81

  I've sort of created my own MVC framework and am curious as to how other frameworks can send properties from the "controller" to the "view". Zend does something along the lines of $this->view->name = 'value'; My code is: file: services_hosting.php class services_hosting extends controller { function __construct($sMvcName) { parent::__construct($sMvcName); $this->setViewSettings(); } public function setViewSettings() { $p = new property; $p->banner = '/path/to/banners/home.jpg'; } } file: controller.php class controller { public $sMvcName = "home"; function __construct($sMvcName) { if ($sMvcName) { $this->sMvcName = $sMvcName; } include('path/to/views/view.phtml'); } public function renderContent() { include('path/to/views/'.$this->sMvcName.'.phtml'); } } file: property.php class property { private $data = array(); protected static $_instance = null; public static function getInstance() { if (null === self::$_instance) { self::$_instance = new self(); } return self::$_instance; } public function __set($name, $value) { $this->data[$name] = $value; } public function __get($name) { if (array_key_exists($name, $this->data)) { return $this->data[$name]; } } public function __isset($name) { return isset($this->data[$name]); } public function __unset($name) { unset($this->data[$name]); } } In my services_hosting.phtml "view" file I have: <img src="<?php echo $this->p->banner ?>" /> This just does not work. Am I doing something fundamentally wrong or is my logic incorrect? I seem to be going round in circles at the moment. Any help would be very much appreciated.

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• Get the last checked checkboxes...

  - by Sara

  Hi everyone, I'm not sure how to accomplish this issue which has been confusing me for a few days. I have a form that updates a user record in MySQL when a checkbox is checked. Now, this is how my form does this: if (isset($_POST['Update'])) { $paymentr = $_POST['paymentr']; //put checkboxes array into variable $paymentr2 = implode(', ', $paymentr); //implode array for mysql $query = "UPDATE transactions SET paymentreceived=NULL"; $result = mysql_query($query); $query = "UPDATE transactions SET paymentdate='0000-00-00'"; $result = mysql_query($query); $query = "UPDATE transactions SET paymentreceived='Yes' WHERE id IN ($paymentr2)"; $result = mysql_query($query); $query = "UPDATE transactions SET paymentdate=NOW() WHERE id IN ($paymentr2)"; $result = mysql_query($query); foreach ($paymentr as $v) { //should collect last updated records and put them into variable for emailing. $query = "SELECT id, refid, affid FROM transactions WHERE id = '$v'"; $result = mysql_query($query) or die("Query Failed: ".mysql_errno()." - ".mysql_error()."<BR>\n$query<BR>\n"); $trans = mysql_fetch_array($result, MYSQL_ASSOC); $transactions .= '<br>User ID:'.$trans['id'].' -- '.$trans['refid'].' -- '.$trans['affid'].'<br>'; } } Unfortunately, it then updates ALL the user records with the latest date which is not what I want it to do. The alternative I thought of was, via Javascript, giving the checkbox a value that would be dynamically updated when the user selected it. Then, only THOSE checkboxes would be put into the array. Is this possible? Is there a better solution? I'm not even sure I could wrap my brain around how to do that WITH Javascript. Does the answer perhaps lie in how my mysql code is written? Thanks - I sincerely appreciate it!!!

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• Php 5 and mysql connecting gives error ...code and error in there check out and plz..plz help....

  - by user309381

  **mysql_query() [function.mysql-query]: Access denied for user 'SYSTEM'@'localhost' (using password: NO) in C:\wamp\www\photo_gallery\includes\database.php on line 56 Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in C:\wamp\www\photo_gallery\includes\database.php on line 56 The Query has problemAccess denied for user 'SYSTEM'@'localhost' (using password: NO) i have set the password for root and grant privilege all for root.Why it soes show like SYSTEM@localhost i dont have SYSTEM .** class MySQLDatabase { public $connection; function _construct() { $this-open_connection(); } public function open_connection() { /* $DB_SERVER = "localhost"; $DB_USER = "root"; $DB_PASS = ""; $DB_NAME = "photo_gallery";*/ $this-connection = mysql_connect($DBSERVER,$DBUSER,$DBPASS); if(!$this-connection) { die("Database Connection Failed" . mysql_error()); } else { $db_select = mysql_select_db($DBNAME,$this-connection); if(!$db_select) { die("Database Selection Failed" . mysql_error()); } } } function mysql_prep($value) { if (get_magic_quotes_gpc()) { $value = stripslashes($value); } // Quote if not a number if (!is_numeric($value)) { $value = "'" . mysql_real_escape_string($value) . "'"; } return $value; } public function close_connection() { if(isset($this->connection)) { mysql_close($this->connection); unset($this->connection); } } public function query($sql) { //$sql = "SELECT*FROM users where id = 1"; $result = mysql_query($sql); $this->confirm_query($result); //$found_user = mysql_fetch_assoc($result); //echo $found_user; return $found_user; } private function confirm_query($result) { if(!$result) { die("The Query has problem" . mysql_error()); } } } $database = new MySQLDatabase(); ?

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• Codeigniter: Simple ajax submit not working

  - by Kevin Brown

  jQuery: $('#welcome-message').submit(function() { // inside event callbacks 'this' is the DOM element so we first // wrap it in a jQuery object and then invoke ajaxSubmit $(this).ajaxSubmit({}); $(this).fadeTo(400, 0, function () { // Links with the class "close" will close parent $(this).slideUp(400); }); return false; }); Controller: function welcome_message() { if(isset($_POST['welcome_message'])) { $this-_my_private_function(); } } private function _my_private_function() { $id = $this->session->userdata('id'); $profile['welcome_message'] = '0'; $this->db->update('be_user_profiles',$profile, array('user_id' => $id)); redirect('home', 'location'); } html: <?php print form_open('home/welcome_message',array('class'=>'horizontal','id'=>'welcome-message'))?> <p> Before you can complete the assessment, you need to complete your profile. Once that's done you'll be ready! After you have completed the assessment, you will be able to view the results from your profile. </p> <input type="checkbox" value="0" name="welcome_message" checked="false"> Don't show me this again </input> <p> <input class="button submit" type="submit" class="close-box" value="Close" /> </p> <?php print form_close()?> This should be working. My hypothesis is that the data isn't being passed to the function...but I really have no idea! It works when I visit the function in the url.

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• Drupal administration theme doesn't apply to Blocks pages (admin/build/block)

  - by hfidgen

  A site I'm creating for a customer in D6 has various images overlaying parts of the main content area. It looks very pretty and they have to be there for the general effect. The problem is, if you use this theme in the administration pages, the images get in the way of everything. My solution was to create a custom admin theme, based on the default one, which has these image areas disabled in the output template files - page.tpl.php The problem is that when you try and edit the blocks page, it uses the default theme and half the blocks admin settings are unclickable behind the images. I KNOW this is by design in Drupal, but it's annoying the hell out of me and is edging towards "bug" rather than "feature" in my mind. It also appears that there is no way of getting around it. You can edit /modules/blocks/block.admin.inc to force Drupal to show the blocks page in the chosen admin theme. BUT whichever changes you then make will not be transferred to the default theme, as Drupal treats each theme separately and each theme can have different block layouts. :x function block_admin_display($theme = NULL) { global $custom_theme; // If non-default theme configuration has been selected, set the custom theme. // $custom_theme = isset($theme) ? $theme : variable_get('theme_default', 'garland'); // Display admin theme $custom_theme = variable_get('admin_theme', '0'); // Fetch and sort blocks $blocks = _block_rehash(); usort($blocks, '_block_compare'); return drupal_get_form('block_admin_display_form', $blocks, $theme); } Can anyone help? the only thing I can think of is to push the $content area well below the areas where the image appear and use blocks only for content display. Thanks!

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• Apache ReWriteEngine throwing 500 Internal Server Error for too many internal redirects... why?!?!?!

  - by Stephen G

  I'm trying to implement a new ReWrite rule on my local dev machine. I have 13 rules set up already, and all work fine (even as of this writing). However, for some reason the newest one is throwing me 500 Internal Server Errors. The ReWrite rule is: RewriteRule stuff/public_html/vault/mystuff/view/(.*) /stuff/public_html/vault/mystuff/view/index.php?stuff=$1 RewriteRule stuff/public_html/vault/mystuff/view/(.*)/ /stuff/public_html/vault/mystuff/view/index.php?stuff=$1 Checked my apache logs and got this: [Thu Jan 13 22:07:43 2011] [error] [client ::1] mod_rewrite: maximum number of internal redirects reached. Assuming configuration error. Use 'RewriteOptions MaxRedirects' to increase the limit if neccessary., referer: http://localhost:8888/stuff/public_html/vault/mystuff/all/index.php?curr=7 On the script I am trying to redirect to view/index.php?stuff=$1, there is nothing that even remotely resembles a redirect of any kind. I do have a very, very basic session verifier being called at the top of the landing script, which is as follows: //Start session session_start(); //Check whether the session variable SESS_MEMBER_ID is present or not if(!isset($_SESSION['SESS_MEMBER_ID']) || (trim($_SESSION['SESS_MEMBER_ID']) == '')) { header("location: ".$root_http.""); exit(); } However, when I access the page directly, it acts as it should, and there is no redirect. All of my other ReWrite rules and their corresponding landing pages are set up the exact same way. This is blowing my mind. Any help, PLEASE!?

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• What is the best way to identify which form has been submitted?

  - by Rupert

  Currently, when I design my forms, I like to keep the name of the submit button equal to the id of the form. Then, in my php, I just do if(isset($_POST['submitName'])) in order to check if a form has been submitted and which form has been submitted. Firstly, are there any security problems or design flaws with this method? One problem I have encountered is when I wish to overlay my forms with javascript in order to provide faster validation to the user. For example, whilst I obviously need to retain server side validation, it is more convenient for the user if an error message is displayed inline, upon blurring an input. Additionally, it would be good to provide entire form validation, upon clicking the submit button. Therefore, when the user clicks on the form's submit button, I am stopping the default action, doing my validation, and then attempting to renable the traditional submit functionality, if the validation passes. In order to do this, I am using the form.submit() method but, unfortunately, this doesn't send the submit button variable (as it should be as form.submit() can be called without any button being clicked). This means my PHP script fails to detect that the form has been submitted. What is the correct way to work around this? It seems like the standard solution is to add a hidden field into the form, upon passing validation, which has the name of form's id. Then when form.submit() is called, this is passed along in place of the submit button. However, this solution seems very ungraceful to me and so I am wondering whether I should: a) Use an alternative method to detect which form has been submitted which doesn't rely rely on passing the submit button. If so what alternative is there? Obviously, just having an extra hidden field from the start isn't any better. b) Use an alternative Javascript solution which allows me to retain my non-Javascript design. For example, is there an alternative to form.submit() which allows me to pass in extra data? c) Suck it up and just insert a hidden field using Javascript.

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• ORDER BY column_name help (via link in HTML table view) (PHP MySQL

  - by Derek

  My output for my table in HTML has several columns such as userid, name, age, dob. The table heading is simply the title of the column name, I want this to be a link, and when clicked, the selected column is sorted in order, ASC, and then DESC (on next click). I thought this was pretty straight forward but I'm having some difficulty. So far, I have produced this, and no output is taken, apart from the URL works by displaying 'users.php?orderby=userid' <?php if(isset($_GET['orderby'])){ $orderby = $_GET['orderby']; $query_sv = "SELECT * FROM users BY ".mysql_real_escape_string($orderby)." ASC"; } //default query else{ $query_sv = "SELECT * FROM users BY user_id DESC"; } ?> <tr> <th><a href="<?php echo $_SERVER['php_SELF']."?orderby=userid";?>">User ID</a></th> Hoefully if I get this working, I can sort the users by D.O.B. next also using the same principles. Does anyone have any ideas?

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• What are some methods to prevent double posting in a form? (PHP)

  - by jpjp

  I want to prevent users from accidentally posting a comment twice. I use the PRG (post redirect get) method, so that I insert the data on another page then redirect the user back to the page which shows the comment. This allows users to refresh as many times as they want. However this doesn't work when the user goes back and clicks submit again or when they click submit 100 times really fast. I don't want 100 of the same comments. I looked at related questions on SO and found that a token is best. But I am having trouble using it. //makerandomtoken(20) returns a random 20 length char. <form method="post" ... > <input type="text" id="comments" name="comments" class="commentbox" /><br/> <input type="hidden" name="_token" value="<?php echo $token=makerandomtoken(20); ?>" /> <input type="submit" value="submit" name="submit" /> </form> if (isset($_POST['submit']) && !empty($comments)) { $comments= mysqli_real_escape_string($dbc,trim($_POST['comments'])); //how do I make the if-statment to check if the token has been already set once? if ( ____________){ //don't insert comment because already clicked submit } else{ //insert the comment into the database } } So I have the token as a hidden value, but how do I use that to prevent multiple clicking of submit. METHODS: someone suggested using sessions. I would set the random token to $_SESSION['_token'] and check if that session token is equal to the $_POST['_token'], but how do I do that? When I tried, it still doesn't check

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• Parsing the Youtube API with DOM

  - by Kirk

  I'm using the Youtube API and I can retrieve the date information without a problem, but don't know how to retrieve the description information. My Code: <?php $v = "dQw4w9WgXcQ"; $url = "http://gdata.youtube.com/feeds/api/videos/". $v; $doc = new DOMDocument; $doc->load($url); $pub = $doc->getElementsByTagName("published")->item(0)->nodeValue; $desc = $doc->getElementsByTagName("media:description")->item(0)->nodeValue; echo "<b>Video Uploaded:</b> "; echo date( "F jS, Y", strtotime( $pub ) ); echo '<br>'; if (isset ($desc)) { echo "<b>Description:</b> "; echo $desc; echo '<br>'; } ?> Here's a link to the feed: http://gdata.youtube.com/feeds/api/videos/dQw4w9WgXcQ?prettyprint=true And the excerpt of code I don't know how to retrieve data from: <media:group> <media:description type='plain'>Music video by Rick Astley performing Never Gonna Give You Up. (C) 1987 PWL</media:description> </media:group> Thanks in advance.

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• i dont understand error while connecting php and mysql? user denied ? plz help me out to solve. ?

  - by user309381

  class MySQLDatabase { public $connection; function _construct() { $this->open_connection();} public function open_connection() {$this->connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS); if(!$this->connection){die("Database Connection Failed" . mysql_error());} else{$db_select = mysql_select_db(DB_NAME,$this->connection); if(!$db_select){die("Database Selection Failed" . mysql_error()); } }} public function close_connection({ if(isset($this->connection)){ mysql_close($this->connection); unset($this->connection);}} public function query(/*$sql*/){ $sql = "SELECT*FROM users where id = 1"; $result = mysql_query($sql); $this->confirm_query($result); //return $result;while( $found_user = mysql_fetch_assoc($result)) { echo $found_user ['username']; } } private function confirm_query($result) { if(!$result) { die("The Query has problem" . mysql_error()); } } } $database = new MySQLDatabase(); $database->open_connection(); $database->query(); $database->close_connection(); I am getting error like denied for user system@locahost(using password no).i have also other database but it runs fine and i dont also i have set the password after encountered the error what else can do to solve plz help ?

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• PHP static function self:: in joomla JFactory class explanation?

  - by Carbon6

  Hi I'm looking at the code of Joomla and trying to figure out what exactly happends in this function. index.php makes a call to function $app = JFactory::getApplication('site'); jfactory.php code public static function getApplication($id = null, $config = array(), $prefix='J') { if (!self::$application) { jimport('joomla.application.application'); self::$application = JApplication::getInstance($id, $config, $prefix); } return self::$application; } application.php code.. public static function getInstance($client, $config = array(), $prefix = 'J') { static $instances; if (!isset($instances)) { $instances = array(); } ....... more code ........ return $instances[$client]; } Now I cannot figure out in function getApplication why is self:$application used. self::$application = JApplication::getInstance($id, $config, $prefix); $application is always null, what is the purpose of using this approach. I tryied modifying it to $var = JApplication::getInstance($id, $config, $prefix); and returnig it but it doesn't work. I would be very glad if someone with more knowledge could explain what is happening here detailed as possible. Many thanks.

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• PHP & MySQL Image deletion problem?

  - by IMAGE

  I have this script that deletes a users avatar image that is stored in a folder name thumbs and another named images and the image name is stored in a mysql database. But for some reason the script deletes all the users info for example if the users id is 3 all of the users info like first name last name age and so are deleted as well, basically everything is deleted including the user how do I fix this so only the images and image name is deleted? Here is the code. $user_id = '3'; if (isset($_POST['delete_image'])) { $a = "SELECT * FROM users WHERE avatar = '". $avatar ."' AND user_id = '". $user_id ."'"; $r = mysqli_query ($mysqli, $a) or trigger_error("Query: $a\n<br />MySQL Error: " . mysqli_error($mysqli)); if ($r == TRUE) { unlink("../members/" . $user_id . "/images/" . $avatar); unlink("../members/" . $user_id . "/images/thumbs/" . $avatar); $a = "DELETE FROM users WHERE avatar = '". $avatar ."' AND user_id = '". $user_id ."'"; $r = mysqli_query ($mysqli, $a) or trigger_error("Query: $a\n<br />MySQL Error: " . mysqli_error($mysqli)); } }

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• AJAX Post Not Sending Data?

  - by Jascha

  I can't for the life of me figure out why this is happening. This is kind of a repost, so forgive me, but I have new data. I am running a javascript log out function called logOut() that has make a jQuery ajax call to a php script... function logOut(){ var data = new Object; data.log_out = true; $.ajax({ type: 'POST', url: 'http://www.mydomain.com/functions.php', data: data, success: function() { alert('done'); } }); } the php function it calls is here: if(isset($_POST['log_out'])){ $query = "INSERT INTO `token_manager` (`ip_address`) VALUES('logOutSuccess')"; $connection->runQuery($query); // <-- my own database class... // omitted code that clears session etc... die(); } Now, 18 hours out of the day this works, but for some reason, every once in a while, the POST data will not trigger my query. (this will last about an hour or so). I figured out the post data is not being set by adding this at the end of my script... $query = "INSERT INTO `token_manager` (`ip_address`) VALUES('POST FAIL')"; $connection->runQuery($query); So, now I know for certain my log out function is being skipped because in my database is the following data: if it were NOT being skipped, my data would show up like this: I know it is being skipped for two reasons, one the die() at the end of my first function, and two, if it were a success a "logOutSuccess" would be registered in the table. Any thoughts? One friend says it's a janky hosting company (hostgator.com). I personally like them because they are cheap and I'm a fan of cpanel. But, if that's the case??? Thanks in advance. -J

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• mysql_connect() acces denied for system@localhost(using password NO)

  - by user309381

  class MySQLDatabase { public $connection; function _construct() { $this-open_connection(); } public function open_connection() { $this-connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS); if(!$this-connection) { die("Database Connection Failed" . mysql_error()); } else { $db_select = mysql_select_db(DB_NAME,$this-connection); if(!$db_select) { die("Database Selection Failed" . mysql_error()); } } } function mysql_prep($value) { if (get_magic_quotes_gpc()) { $value = stripslashes($value); } // Quote if not a number if (!is_numeric($value)) { $value = "'" . mysql_real_escape_string($value) . "'"; } return $value; } public function close_connection() { if(isset($this->connection)) { mysql_close($this->connection); unset($this->connection); } } public function query(/*$sql*/) { $sql = "SELECT*FROM users where id = 1"; $result = mysql_query($sql); $this->confirm_query($result); //return $result; while( $found_user = mysql_fetch_assoc($result)) { echo $found_user ['username']; } } private function confirm_query($result) { if(!$result) { die("The Query has problem" . mysql_error()); } } } $database = new MySQLDatabase(); $database-open_connection(); $database-query(); $database-close_connection(); ?

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• How to print contents from a session variable by looping in a foreach statement

  - by itsover9000

  im trying to write a code where can print and loop through the contents of my session variable by using a foreach statement here is my code <form class="form form-inline" method = "post" action="reportmaker.php"> <select name="rfield"> <option value="">--Select Field--</option> <?php $sc2=mysql_query("SELECT * from searchcolumn s left join report_fields r on s.scol_id=r.field_id where s.category != 'wh'"); foreach($sc2 as $sc){ ?> <option value="<?php echo $sc[advsearch_col]; ?>"><?php echo $sc[advsearch_name]; ?></option> <?php } ?> </select> <button type="submit" value = "submit" id="add" name="add" class="btn pull-right">Add More</button> </form> <?php if(isset($_POST['add'])) { $_SESSION['temp'][]=$_POST['rfield']; } if($_SESSION[temp][]!=""){ foreach($_SESSION[temp][] as $temp) { echo $temp; } } ?> the error that appears with this code is Fatal error: Cannot use [] for reading the line where the error is is this if($_SESSION[temp][]!=""){ i need to print the contents of the session array and this is the only way i know how is there a way to fix this? thanks =========EDIT thanks for the answers guys i finally got it

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• How to emulate mod_rewrite in PHP

  - by Tyler Crompton

  I have a few URLs that I want to map to certain files via PHP. Currently, I am just using mod_rewrite in Apache. However, my application is getting too large for the rewriting to be done with regular expressions. So I created a file router.php that does the rewriting. I understand to do a redirect I could just send the Location: header. However, I don't always want to do a redirect. For example, I may want /api/item/ to map to the file /herp/derp.php relative to the document root. I need to preserve the HTTP method as well. "No problem," I thought. I made my .htaccess have the following snippet. RewriteEngine On RewriteRule ^api/item/$ /cgi-bin/router.php [L] And my router.php file looks as follows: <?php $uri = parse_url($_SERVER['REQUEST_URI']); $query = isset($uri['query']) ? $uri['query'] ? array(); // some code that modifies the query require_once "{$_SERVER['DOCUMENT_ROOT']}/herp/derp.php?" . http_build_query($query); ?> However, this doesn't work, because the OS is looking for a file named derp.php?some=query. How can I simulate a rewrite rule such as RewriteRule ^api/item/$ /herp/derp/ [L] in PHP. In other words, how do I tell the server to process a different URL than requested and preserve the query and HTTP method without causing a redirect? Note: Using variables set in router.php is less than desirable and is bad structure since it's only supposed to be responsible for handling URLs. I am open to using a light-weight third party solution.

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• Draw a comparison between an integer in a specific row and a count

  - by XCoderX

  This is a follow-up question to this one: Check for specific integer in a row WHERE user = $name I want a user to be able to comment on my site for exactly five times a day. After this five times, the user has to wait 24 hours. In order to accomplish that I raise a counter in my MYSQL database, right next to the user. So where the name of the user is, there is where the counter gets raised. When it reaches 5 it should stop counting and reset after 24 hours. In order to check the time I use a timestamp. I check if the timestamp is older than 24 hours. If that is the case, the counter gets reseted (-5) and the user can comment again. In order to do that, I use the following code, however it never stops at five, my guess is that my comparison is wrong somehow: $counter = "SELECT FROM table VALUES CommentCounterReset WHERE Name = '$name'"; if(!isset($_SESSION['ts'])); { $_SESSION['ts'] = time(); } if ($counter >= 5) { if (time() - $_SESSION['ts'] <= 60*60*24){ echo "You already wrote five comments."; } else { $sql = "UPDATE table SET CommentCounterReset = CommentCounterReset-5 WHERE Name = '$name'"; } } else { $sql = "UPDATE table SET CommentCounterReset = CommentCounterReset+1 WHERE Name = '$name'"; echo "Your comment has been added."; }

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