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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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  • Validate Form with PHP AND Javascriipt?

    - by J M 4
    Is it possible to validate a form with PHP AND Javascript? I am currently able to do both using my existing form but only on an individual basis. My overall goal is this: Validate form using javascript client side and present any errors to the user immediately If javascript validation passes, a flag is created and then the PHP script can begin. When doing my javascript validation, i use the following code within the form tag: <form id="Enroll_Form" action="review.php" method="post" name="Enroll_Form" onsubmit="return Enroll_Form_Validator(this)" language="javascript"> if I want to process the PHP validation, I am forced to rename the action to the PHP_SELF file (or simply the same file name i'm using) and remove the 'onsubmit' function. Any ideas?

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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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  • PHP Array Key & value Question

    - by Nano HE
    Hi I writed a test code as below. <?php $url = 'http://localhost/events/result/cn_index.php?login'; print_r(parse_url($url)); echo parse_url($url, PHP_URL_PATH); ?> Output Array ( [scheme] => http [host] => localhost [path] => /events/result/cn_index.php [query] => login ) /events/result/cn_index.php Now I inserted the line below echo array[query]; // I want to echo 'login', but failed. How to the the value of 'login'?

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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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  • 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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  • 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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  • handle json request in PHP

    - by wo_shi_ni_ba_ba
    When making an ajax call, when contentType is set to application/json instead of the default x-www-form-urlencoded, server side (in PHP) can't get the post parameters. in the following working example, if I set the contentType to "application/json" in the ajax request, PHP $_POST would be empty. why does this happen? How can I handle a request where contentType is application/json properly in PHP? $.ajax({ cache: false, type: "POST", url: "xxx.php", //contentType: "application/json", processData: true, data: {my_params:123}, success: function(res){ }, complete: function(XMLHttpRequest, text_status) { } });

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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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  • 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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  • PHP: Object Oriented Programming -> Operator

    - by oman9589
    So I've been reading through the book PHP Solutions, Dynamic Web Design Made Easy by David Powers. I read through the short section on Object Oriented PHP, and I am having a hard time grasping the idea of the - operator. Can anyone try to give me a solid explanation on the - operator in OOP PHP? Example: $westcost = new DateTimeZone('America/Los_Angeles'); $now->setTimezone($westcoast); Also,a more general example: $someObject->propertyName Thanks

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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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  • 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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  • 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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  • 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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  • 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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  • 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 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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  • Socket php server not showing messages sent from android client

    - by Mj1992
    Hi I am a newbie in these kind of stuff but here's what i want to do. I am trying to implement a chat application in which users will send their queries from the website and as soon as the messages are sent by the website users.It will appear in the android mobile application of the site owner who will answer their queries .In short I wanna implement a live chat. Now right now I am just simply trying to send messages from android app to php server. But when I run my php script from dreamweaver in chrome the browser keeps on loading and doesn't shows any output when I send message from the client. Sometimes it happened that the php script showed some outputs which I have sent from the android(client).But i don't know when it works and when it does not. So I want to show those messages in the php script as soon as I send those messages from client and vice versa(did not implemented the vice versa for client but help will be appreciated). Here's what I've done till now. php script: <?php set_time_limit (0); $address = '127.0.0.1'; $port = 1234; $sock = socket_create(AF_INET, SOCK_STREAM, 0); socket_bind($sock, $address, $port) or die('Could not bind to address'); socket_listen($sock); $client = socket_accept($sock); $welcome = "Roll up, roll up, to the greatest show on earth!\n? "; socket_write($client, $welcome,strlen($welcome)) or die("Could not send connect string\n"); do{ $input=socket_read($client,1024,1) or die("Could not read input\n"); echo "User Says: \n\t\t\t".$input; if (trim($input) != "") { echo "Received input: $input\n"; if(trim($input)=="END") { socket_close($spawn); break; } } else{ $output = strrev($input) . "\n"; socket_write($spawn, $output . "? ", strlen (($output)+2)) or die("Could not write output\n"); echo "Sent output: " . trim($output) . "\n"; } } while(true); socket_close($sock); echo "Socket Terminated"; ?> Android Code: public class ServerClientActivity extends Activity { private Button bt; private TextView tv; private Socket socket; private String serverIpAddress = "127.0.0.1"; private static final int REDIRECTED_SERVERPORT = 1234; @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.main); bt = (Button) findViewById(R.id.myButton); tv = (TextView) findViewById(R.id.myTextView); try { InetAddress serverAddr = InetAddress.getByName(serverIpAddress); socket = new Socket(serverAddr, REDIRECTED_SERVERPORT); } catch (UnknownHostException e1) { e1.printStackTrace(); } catch (IOException e1) { e1.printStackTrace(); } bt.setOnClickListener(new OnClickListener() { public void onClick(View v) { try { EditText et = (EditText) findViewById(R.id.EditText01); String str = et.getText().toString(); PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(socket.getOutputStream())),true); out.println(str); Log.d("Client", "Client sent message"); } catch (UnknownHostException e) { tv.setText(e.getMessage()); e.printStackTrace(); } catch (IOException e) { tv.setText(e.getMessage()); e.printStackTrace(); } catch (Exception e) { tv.setText(e.getMessage()); e.printStackTrace(); } } }); } } I've just pasted the onclick button event code for Android.Edit text is the textbox where I am going to enter my text. The ip address and port are same as in php script.

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