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  • where are wrong in my php code ????

    - by user318068
    hi all, <td align="center" bgcolor="#FFFFFF"><?php echo '<label onclick="window.open('profilephp.php?member=$row['MemberID']','mywindow')">'.{$row['MemberName']}.'</label>';?><br /> <?php echo "<p align='center'><img width='100' height='100' src={$row['MemberImg']} alt='' /></p>";?></td></tr> Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in C:\xampp\htdocs\home - Copy\membercopy.php on line 141 I really don't know where it went wrong. Please help,

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  • Can't serve HTML5 video through PHP on Safari/Mac (5.0)

    - by JKS
    I'm encountering a strange bug in Safari where, when I serve MP4 video through PHP (to obfuscate the file beneath the document root with a token-based authentication system), Safari for some reason fires the <video>'s onerror event, and the video never loads (I can't get any useful information out of the event object sent to onerror — everything is undefined). When I access the PHP script directly (i.e., the video is not embedded in a page), the video controls appear momentarily before flashing to a QuickTime question mark. When I access the MP4 file directly, it works as expected. What's bizarre is that the embedded video works perfectly in the latest version of Chrome for Mac. Here are the headers when accessed through PHP: Connection:Keep-Alive Content-Disposition:inline; filename="test.mp4" Content-Length:5558749 Content-Type:video/mp4 Date:Tue, 22 Jun 2010 01:24:25 GMT Keep-Alive:timeout=10, max=29 Server:Apache/2.2.15 (CentOS) mod_ssl/2.2.15 0.9.8l DAV/2 mod_auth_passthrough/2.1 FrontPage/5.0.2.2635 X-Powered-By:PHP/5.2.13 And here are the headers when test.mp4 is accessed directly: Accept-Ranges:bytes Connection:Keep-Alive Content-Length:5558749 Content-Type:video/mp4 Date:Tue, 22 Jun 2010 01:26:45 GMT Etag:"1c04757-54d1dd-489944c5a6400" Keep-Alive:timeout=10, max=30 Last-Modified:Tue, 22 Jun 2010 01:25:36 GMT Server:Apache/2.2.15 (CentOS) mod_ssl/2.2.15 0.9.8l DAV/2 mod_auth_passthrough/2.1 FrontPage/5.0.2.2635 The only differing headers are: Accept-Ranges (which I don't think is necessary), Etag, Last-Modified, Content-Disposition, and X-Powered-By. Not only can Chrome handle the PHP-served video fine, but when I use the same script to load the MP4 through a Flash player, it also works fine. I just can't figure out what Safari is choking on. EDIT: Also, when I change the content disposition to "attachment", Safari will download the MP4 file just fine.

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  • PHP: Over-writing session variables

    - by Tom
    Hi, Question related to PHP memory-handling from someone not yet very experienced in PHP: If I set a PHP session variable of a particular name, and then set a session variable of the exact same name elsewhere (during the same session), is the original variable over-written, or does junk accumulate in the session? In other words, should I be destroying a previous session variable before creating a new one of the same name? Thank you.

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  • will php/apache ever support multi threading?

    - by fayer
    i mainly focus on the web, i think i will never create desktop applications. so i think it's better for me to focus on typical web languages like php. i know an advantage java has over php is multi threading though. will php ever support this feature in the future? thanks

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  • php Dollar amount Regular Expression

    - by Thildemar
    I am have completed javascript validation of a form using Regular Expressions and am now working on redundant verification server-side using PHP. I have copied this regular expression from my jscript code that finds dollar values, and reformed it to a PHP friendly format: /\$?((\d{1,3}(,\d{3})*)|(\d+))(\.\d{2})?$/ Specifically: if (preg_match("/\$?((\d{1,3}(,\d{3})*)|(\d+))(\.\d{2})?$/", $_POST["cost"])){} While the expression works great in javascript I get : Warning: preg_match() [function.preg-match]: Compilation failed: nothing to repeat at offset 1 when I run it in PHP. Anyone have a clue why this error is coming up?

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  • How should I architect JasperReports with a PHP front+backend system

    - by Itay Moav
    Our system is written completely in PHP. For various business reasons (which are a given) I need to build the reports of the system using JasperReports. What architecture should I use? Should I put the Jasper as a stand alone server (if possible) and let the php query against it, should I have it generate the reports with a cron, and then let the PHP scoop up the files and send them to the web client/browser...

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  • PHP Include and sort by variable within file

    - by Jason Hoax
    I have written this PHP include-script but now I'm trying to sort the included files out by variables WITHIN the included php's. In other words, in each included PHP file there is a rating, now I want the ratings to be read so that when they are included they will be sorted out from highest to lowest. (scores are like 6.0 to 9.0) Kind Regards! $location = 'experiments/visualizations'; foreach (glob("$location/*.php") as $filename) { include $filename; } The included files are named randomly like: File1: $filename = "AAAA"; $projecttitle = "Project Name"; $description = "This totally explains the product"; $score = "7.6"; File 2: $filename = "BBBB"; $projecttitle = "Project Name2" $description = "This totally explains the product"; $score = "9.6"; As you can see 9.6 is higher than 7.6 but PHP sorts the includes out by name instead of variables within the file. I tried sorting, but I can't get it fixed. Help!

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  • jquery with php loading file

    - by Marcus Solv
    I'm trying to use jquery with a simple php code: $('#some').click(function() { <?php require_once('some1.php?name="some' + index + '"'); ?> }); It shows no error, so I don't know what is wrong. In some1 I have: <?php //Start session session_start(); //Include database connection details require_once('../sql/config.php'); //Connect to mysql server $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } //Select database $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } //Function to sanitize values received from the form. Prevents SQL injection function clean($str) { $str = @trim($str); if(get_magic_quotes_gpc()) { $str = stripslashes($str); } return mysql_real_escape_string($str); } //Sanitize the POST values $name = clean($_GET['name']); ?> It's not complete because I want to make a sql command (insert). I want when I click in #some to execute that file (create a entry in the table that isn't define yet).

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  • Hide URL of PHP page

    - by manoj singhal
    I want to hide the URL of my PHP page; that is, I don't want to write /register.php directly in the href tag, I want to write /register/ and have it open the register.php page directly. I want to do that for all the webpages.

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  • Nginx: Loopback connection via PHP's getimage size crashes server (Magento's CMS)

    - by Alex
    We were able to trace down a problem that is crashing our NGINX server running Magento until the following point: Background info: Magento Backend has a CMS function with a WYSIWYG editor. This editor loads some pictures via a controller in magento (cms/directive). When we set the NGINX error_log level to info, we get the following lines (line break inserted for better readability): 2012/10/22 18:05:40 [info] 14105#0: *1 client closed prematurely connection, so upstream connection is closed too while sending request to upstream, client: XXXXXXXXX, server: test.local, request: "GET index.php/admin/cms_wysiwyg/directive/___directive/BASEENCODEDIMAGEURL,,/ HTTP/1.1", upstream: "fastcgi://127.0.0.1:9024", host: "test.local" When checking the code in the debugger, the following call does never return (in ´Varien_Image_Adapter_Abstract::getMimeType()` # $this->_fileName is http://test.local/skin/adminhtml/base/default/images/demo-image-not-existing.gif` # $_SERVER['REQUEST_URI'] = http://test.local/admin/cms_wysiwyg/directive/___directive/BASEENCODEDIMAGEURL list($this->_imageSrcWidth, $this->_imageSrcHeight, $this->_fileType, ) = getimagesize($this->_fileName); The filename requests is an URL to the same server which is requesting the script a link to a static .gif that is not existing. Sample URL: http://test.local/skin/adminhtml/base/default/images/demo-image-not-existing.gif When the above line executed, any subsequent request to the NGNIX server does not respond any more. After waiting for around 10 minutes, the NGINX server starts answering requests again. I tried to reproduce the error with a simple test script that only calls getimagesize() with the given URL - but this not crash. It simple leads to an exception saying that the URL could not be loaded (which is fine as the URL is wrong)

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  • PHP echo query result in Class??

    - by Jerry
    Hi all I have a question about PHP Class. I am trying to get the result from Mysql via PHP. I would like to know if the best practice is to display the result inside the Class or store the result and handle it in html. For example, display result inside the Class class Schedule { public $currentWeek; function teamQuery($currentWeek){ $this->currentWeek=$currentWeek; } function getSchedule(){ $connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS); if (!$connection) { die("Database connection failed: " . mysql_error()); } $db_select = mysql_select_db(DB_NAME,$connection); if (!$db_select) { die("Database selection failed: " . mysql_error()); } $scheduleQuery=mysql_query("SELECT guest, home, time, winner, pickEnable FROM $this->currentWeek ORDER BY time", $connection); if (!$scheduleQuery){ die("database has errors: ".mysql_error()); } while($row=mysql_fetch_array($scheduleQuery, MYSQL_NUMS)){ //display the result..ex: echo $row['winner']; } mysql_close($scheduleQuery); //no returns } } Or return the query result as a variable and handle in php class Schedule { public $currentWeek; function teamQuery($currentWeek){ $this->currentWeek=$currentWeek; } function getSchedule(){ $connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS); if (!$connection) { die("Database connection failed: " . mysql_error()); } $db_select = mysql_select_db(DB_NAME,$connection); if (!$db_select) { die("Database selection failed: " . mysql_error()); } $scheduleQuery=mysql_query("SELECT guest, home, time, winner, pickEnable FROM $this->currentWeek ORDER BY time", $connection); if (!$scheduleQuery){ die("database has errors: ".mysql_error()); // create an array } $ret = array(); while($row=mysql_fetch_array($scheduleQuery, MYSQL_NUMS)){ $ret[]=$row; } mysql_close($scheduleQuery); return $ret; // and handle the return value in php } } Two things here: I found that returned variable in php is a little bit complex to play with since it is two dimension array. I am not sure what the best practice is and would like to ask you experts opinions. Every time I create a new method, I have to recreate the $connection variable: see below $connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS); if (!$connection) { die("Database connection failed: " . mysql_error()); } $db_select = mysql_select_db(DB_NAME,$connection); if (!$db_select) { die("Database selection failed: " . mysql_error()); } It seems like redundant to me. Can I only do it once instead of calling it anytime I need a query? I am new to php class. hope you guys can help me. thanks.

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  • INSERT DELAYED on locked tables blocks PHP processes to continue

    - by sw0x2A
    Our webservers write some tracking information into a MySQL database (using INSERT DELAYED into MyISAM table). When a huge SELECT query is executed on this table or when it is locked for another reason, the webserver processes (with INSERT DELAYED) are waiting for the database and in some cases the MaxServer limit is reached in Apaches, so they will stop serving requests. We use INSERT DELAYED because The DELAYED option for the INSERT statement is a MySQL extension to standard SQL that is very useful if you have clients that cannot or need not wait for the INSERT to complete. This is a common situation when you use MySQL for logging and you also periodically run SELECT and UPDATE statements that take a long time to complete. Quote from MySQL documentation. I am wondering why the Apache processes are waiting for the INSERT DELAYED to finish. And what can I do to just send the data and forget about it. (Since this is logging data, I do not care if we lose some entries.) Even when the table is locked the PHP script should just go on and should not wait for an answer of MySQL. (We do not want to setup Master-slave for this table but we are thinking about move this data to some NoSQL database. But for now I would like to know why INSERT DELAYED is not working as expected.)

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  • PHP if statement - select two different get variables?

    - by arsoneffect
    Below is my example script: <li><a <?php if ($_GET['page']=='photos' && $_GET['view']!=="projects"||!=="forsale") { echo ("href=\"#\" class=\"active\""); } else { echo ("href=\"/?page=photos\""); } ?>>Photos</a></li> <li><a <?php if ($_GET['view']=='projects') { echo ("href=\"#\" class=\"active\""); } else { echo ("href=\"/?page=photos&view=projects\""); } ?>>Projects</a></li> <li><a <?php if ($_GET['view']=='forsale') { echo ("href=\"#\" class=\"active\""); } else { echo ("href=\"/?page=photos&view=forsale\""); } ?>>For Sale</a></li> I want the PHP to echo the "href="#" class="active" only when it is not on the two pages: ?page=photos&view=forsale or ?page=photos&view=projects

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  • jQuery ajax request to php, how to return plain text only

    - by jyoseph
    I am making an ajax request to a php page and I cannot for the life of me get it to just return plain text. $.ajax({ type: 'GET', url: '/shipping/test.php', cache: false, dataType: 'text', data: myData, success: function(data){ console.log(data) } }); in test.php I am including this script to get UPS rates. I am then calling the function with $rate = ups($dest_zip,$service,$weight,$length,$width,$height); I am doing echo $rate; at the bottom of test.php. When viewed in a browser shows the rate, that's great. But when I request the page via ajax I get a bunch of XML. Pastie here: http://pastie.org/1416142 My question is, how do I get it so I can just return the plain text string from the ajax call, where the result data will be a number? Edit, here's what I see in Firebug- Response tab: HTML tab:

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  • PHP language specification ?

    - by Rolf
    Hi, as I know there is an official document for Java (JLS), I'd like to know if it's also the case of PHP language. I found the "Language Reference" section on the PHP manual, but it doesn't look as detailed as the JLS. The thing is I have a good practical knowledge of PHP but I'm miserably clueless about what REALLY happens under the hood. If there isn't any official document, could you recommend me some good books to read ? Thanks in advance ! Rolf

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  • PHP & WP: Render Certain Markup Based on True False Condition

    - by rob
    So, I'm working on a site where on the top of certain pages I'd like to display a static graphic and on some pages I would like to display an scrolling banner. So far I set up the condition as follows: <?php $regBanner = true; $regBannerURL = get_bloginfo('stylesheet_directory'); //grabbing WP site URL ?> and in my markup: <div id="banner"> <?php if ($regBanner) { echo "<img src='" . $regBannerURL . "/style/images/main_site/home_page/mock_banner.jpg' />"; } else { echo 'Slider!'; } ?> </div><!-- end banner --> In my else statement, where I'm echoing 'Slider!' I would like to output the markup for my slider: <div id="slider"> <img src="<?php bloginfo('stylesheet_directory') ?>/style/images/main_site/banners/services_banners/1.jpg" alt="" /> <img src="<?php bloginfo('stylesheet_directory') ?>/style/images/main_site/banners/services_banners/2.jpg" alt="" /> <img src="<?php bloginfo('stylesheet_directory') ?>/style/images/main_site/banners/services_banners/3.jpg" alt="" /> ............. </div> My question is how can I throw the div and all those images into my else echo statement? I'm having trouble escaping the quotes and my slider markup isn't rendering.

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  • The youtube API sometimes throws error: Call to a member function children() on a non-object

    - by Anna Lica
    When i launch the php script, sometime works fine, but many other times it retrieve me this errror Fatal error: Call to a member function children() on a non-object in /membri/americanhorizon/ytvideo/rilevametadatadaurlyoutube.php on line 21 This is the first part of the code // set feed URL $feedURL = 'http://gdata.youtube.com/feeds/api/videos/dZec2Lbr_r8'; // read feed into SimpleXML object $entry = simplexml_load_file($feedURL); $video = parseVideoEntry($entry); function parseVideoEntry($entry) { $obj= new stdClass; // get nodes in media: namespace for media information $media = $entry->children('http://search.yahoo.com/mrss/'); //<----this is the doomed line 21 UPDATE: solution adopted for ($i=0 ; $i< count($fileArray); $i++) { // set feed URL $feedURL = 'http://gdata.youtube.com/feeds/api/videos/'.$fileArray[$i]; // read feed into SimpleXML object $entry = simplexml_load_file($feedURL); if ( is_object($entry)) { $video = parseVideoEntry($entry); echo ($video->description."|".$video->length); echo "<br>"; } else { $i--; } } In this mode i force the script to re-check the file that caused the error

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  • PHP post programmatically

    - by Tural Teyyuboglu
    After user registration, website send activation code to email. something like that. www.domain.com/?activate=<code> I'm creating 2 variants of activation: 1.manual 2.auto Lets say we have index.php. 1.Manual method. When someone wants to activate user manually all things are obvious: User opens page www.domain.com/?activate Index.php checks with following script and includes div file (which contains activation form) if (isset($_GET['activate'])) { $page='activate'; $divfile = 'path to div.php'; } include $divfile; Then page sends form data via ajax to activation.php file. 2.Auto method. Lets say user clicked directly to www.domain.com/?activate=<code>. What I wanna do is, to check if(!empty($_GET['activate'])), if all right ... I can't figure out how to act?! Programmatically send something like POST to activation.php or what? Please help. Thx in advance

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  • Will php code exit after echo for ajax?

    - by Steve
    I am running a typical php-engined ajax webpage. I use echo to return a html string from the php code. My question is, if I have some other code after the echo, will those code get executed? Or echo behaves similar to exit, which immediately return and stop running the php code? Thanks.

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