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  • Trouble decoding JSON string with PHP

    - by Anthony
    I'm trying to send an array of objects from JS to PHP using JSON. I have an array of players as follows: var player; var players = new Array(); //loop for number of players player = new Object(); player.id = theID; players[i] = player; Then my AJAX call looks like this: JSONplayers = JSON.stringify(players); $.ajax({ type: "POST", url: "php/ajax_send_players.php", data: { "players" : JSONplayers } On the PHP side the decode function looks like this $players = $_REQUEST['players']; echo var_dump($players); $players = json_decode($players); echo 'players: ' .$players. '--'. $players[0] . '--'. $players[0]->id; Debugging in chrome, the JSON players var looks like this before it is sent: JSONplayers: "[{"id":"Percipient"},{"id":"4"}]" And when I vardump in PHP it looks OK, giving this: string(40) "[{\"id\":\"Percipient\"},{\"id\":\"4\"}]" But I can't access the PHP array, and the echo statement about starting with players: outputs this: players: ---- Nothing across the board...maybe it has something to do with the \'s in the array, I am new to this and might be missing something very simple. Any help would be greatly appreciated. note I've also tried json_decode($players, true) to get it as an assoc array but get similar results.

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  • sorting a list of objects by value php

    - by Mike
    I have a list of objects: class Beer { var $name; var $id; var $style; var $brewery; var $rate; function getName() { return $this->name; } function getID() { return $this->id; } function getStyle() { return $this->style; } function getBrewery() { return $this->brewery; } function getRate() { return $this->rate; } } After doing some research online on how to accomplish this, this is my code: usort($localBeersList, "cmp"); function cmp($a, $b) { if ($a->getRate() == $b->getRate()) { return 0; } return ($a->getRate() < $b->getRate()) ? -1 : 1; } If I try and output my list after this I do not get anything.

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  • PHP JQuery: Where to specify uploadify destination folder

    - by Eamonn
    I have an uploadify script running, basic setup. Works fine when I hard code the destination folder for the images into uploadify.php - now I want to make that folder dynamic. How do I do this? I have a PHP variable $uploadify_path which contains the path to the folder I want. I have switched out my hard coded $targetPath = path/to/directory for $targetPath = $uploadify_path in both uploadify.php and check_exists.php, but it does not work. The file upload animation runs, says it is complete, yet the directory remains empty. The file is not hiding out somewhere else either. I see there is an option in the Javascript to specify a folder. I tried this also, but to no avail. If anyone could educate me on how to pass this variable destination to uploadify, I'd be very grateful. I include my current code for checking (basically default): The Javascript <script type="text/javascript"> $(function() { $('#file_upload').uploadify({ 'swf' : 'uploadify/uploadify.swf', 'uploader' : 'uploadify/uploadify.php', // Put your options here }); }); </script> uploadify.php $targetPath = $_SERVER['DOCUMENT_ROOT'] . $uploadify_path; // Relative to the root if (!empty($_FILES)) { $tempFile = $_FILES['Filedata']['tmp_name']; $targetFile = $targetPath . $_FILES['Filedata']['name']; // Validate the file type $fileTypes = array('jpg','jpeg','gif','png'); // File extensions $fileParts = pathinfo($_FILES['Filedata']['name']); if (in_array($fileParts['extension'],$fileTypes)) { move_uploaded_file($tempFile,$targetFile); echo '1'; } else { echo 'Invalid file type.'; } }

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  • Somewhat lost with jquery + php + json

    - by Luis Armando
    I am starting to use the jquery $.ajax() but I can't get back what I want to...I send this: $(function(){ $.ajax({ url: "graph_data.php", type: "POST", data: "casi=56&nada=48&nuevo=98&perfecto=100&vales=50&apenas=70&yeah=60", dataType: "json", error: function (xhr, desc, exceptionobj) { document.writeln("El error de XMLHTTPRequest dice: " + xhr.responseText); }, success: function (json) { if (json.error) { alert(json.error); return; } var output = ""; for (p in json) { output += p + " : " + json[p] + "\n"; } document.writeln("Results: \n\n" + output); } }); }); and my php is: <?php $data = $_POST['data']; function array2json($data){ $json = $data; return json_encode($json); } ?> and when I execute this I come out with: Results: just like that I used to have in the php a echo array2json statement but it just gave back gibberish...I really don't know what am I doing wrong and I've googled for about 3 hours just getting basically the same stuff. Also I don't know how to pass parameters to the "data:" in the $.ajax function in another way like getting info from the web page, can anyone please help me? Edit I did what you suggested and it prints the data now thank you very much =) however, I was wondering, how can I send the data to the "data:" part in jQuery so it takes it from let's say user input, also I was checking the php documentation and it says I'm allowed to write something like: json_encode($a,JSON_HEX_TAG|JSON_HEX_APOS|JSON_HEX_QUOT|JSON_HEX_AMP) however, if I do that I get an error saying that json_encode accepts 1 parameter and I'm giving 2...any idea why? I'm using php 5.2

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  • php: autoload exception handling.

    - by YuriKolovsky
    Hello again, I'm extending my previous question (Handling exceptions within exception handle) to address my bad coding practice. I'm trying to delegate autoload errors to a exception handler. <?php function __autoload($class_name) { $file = $class_name.'.php'; try { if (file_exists($file)) { include $file; }else{ throw new loadException("File $file is missing"); } if(!class_exists($class_name,false)){ throw new loadException("Class $class_name missing in $file"); } }catch(loadException $e){ header("HTTP/1.0 500 Internal Server Error"); $e->loadErrorPage('500'); exit; } return true; } class loadException extends Exception { public function __toString() { return get_class($this) . " in {$this->file}({$this->line})".PHP_EOL ."'{$this->message}'".PHP_EOL . "{$this->getTraceAsString()}"; } public function loadErrorPage($code){ try { $page = new pageClass(); echo $page->showPage($code); }catch(Exception $e){ echo 'fatal error: ', $code; } } } $test = new testClass(); ?> the above script is supposed to load a 404 page if the testClass.php file is missing, and it works fine, UNLESS the pageClass.php file is missing as well, in which case I see a "Fatal error: Class 'pageClass' not found in D:\xampp\htdocs\Test\PHP\errorhandle\index.php on line 29" instead of the "fatal error: 500" message I do not want to add a try/catch block to each and every class autoload (object creation), so i tried this. What is the proper way of handling this?

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  • I need help understanding how this jQuery filter function works, line-by-line, if possible

    - by user717236
    Here is the HTML: <div> <h3>text</h3> </div> <div> <h3>moretext</h3> </div> <div> <h3>123</h3> </div>?????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????? Here is the JS: var rv1_wlength = $("div").filter(function() { return $(this).find("h3").filter(function () { return $(this).text() != "123"; }).length; }); var rv1_wolength = $("div").filter(function() { return $(this).find("h3").filter(function () { return $(this).text() != "123"; }); }); var rv2 = $("div").find("h3").filter(function() { return $(this).text() != "123"; }); alert(rv1_wlength.text()); // text // moretext alert(rv1_wolength.text()); // text // moretext // 123 alert(rv2.text());? // textmoretext I don't understand why the first two methods print the elements on each line, whereas the second method concatenates them. "rv2" is a jQuery object. Then, what are the first two (rv1_wlength and rv1_wolength)? Furthermore, I don't understand why the inclusion of the length property makes all the difference in filtering the elements. The second method does nothing, since it returns all the elements. The first method, with the only change being the addition of the length property, correctly filters the elements. I would very much like a line-by-line explanation. I would sincerely appreciate any feedback. Thank you.

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  • PHP: Making my code simpler/shorter welcome message

    - by Karem
    Any suggestion to make this welcome message shorter: <?php if(isset($_SESSION['user_id'])) { if(isSet($_SESSION['1stTime'])){ ?> <strong id="welcome" style="font-size: 10px;"> <a href="logout.php"> Logga ut </a> </strong> <?php }else{ $_SESSION['1stTime'] = time(); ?> <script> $(document).ready(function() { $("#welcome").fadeIn("slow"); setTimeout(function(){ $("#welcome").fadeOut("slow"); setTimeout(function(){ $("#welcome").html("<a href='logout.php'>Logga ut</a>"); $("#welcome").fadeIn(); }, 800); }, 5000); }); </script> <strong id="welcome" style="display: none; color: #FFF; font-size: 10px;">Hej, <?php echo $FULL; ?>!</strong> <?php } } ?> First it checks if you are signed in. Next if 1stTime is set, if it is then show "Log out" in swedish, if it isnt, then introduce with "Hi, NAME", and then change to "Log out" after 5 seconds(jquery) + set the session How can i make this simpler?

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  • PHP Sessions and Passing Session ID

    - by Jason McCreary
    I have an API where I am passing the session id back and forth between calls. I set up the session like so: // start API session session_name('apikey'); session_id($data['apikey']); // required to link session session_start(); Although I named my session and am passing the session id via GET and POST using the name, PHP does not automatically resume that session. It always creates a new one unless I set the explicitly set the session id. I found some old user comments on www.php.net that said unless the session id is the first parameter PHP won't set it automatically. This seems odd, but even when I call tried it still didn't work: rest_services.php?apikey=sdr6d3subaofcav53cpf71j4v3&q=testing I have used PHP for years, but am a little confused on why I needed to explicitly set the session with session_id() when I am naming the session and passing it's key accordingly. UPDATE It seems I wasn't clear. My question is why is setting the session ID with session_id() required when I am passing the id, using the session name apikey, via $_GET or $_POST. Theoretically this is no different than PHP's SID when cookies are disabled. But for me it doesn't work unless I explicitly set the session ID. Why?

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  • how to callback a lua function from a c function

    - by pierr
    Hi, I have a c function test_callback accepting a point to a function as the parameter and It will "callback" that function. //typedef int(*data_callback_t)(int i); int test_callback(data_callback_t f) { f(3); } int datacallback(int a ) { printf("called back %d\n",a); return 0; } //example test_callback(datacallback); // print : called back 3 Now, I want to wrap test_callback so that they can be called from lua, suppose the name is lua_test_callback ;and also the input parameter to it would be a lua function. How should I achieve this goal? function lua_datacallback (a ) print "hey , this is callback in lua" ..a end lua_test_callback(lua_datacallback) //expect to get "hey this is callback in lua 3 "

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  • PHP getimagesize is not working when is called from a function in function.php (Wordpress)?

    - by janoChen
    PHP getimagesize is not working when is called from a function in function.php. function.php: add_action('wp_head', 'get_image_size'); function get_image_size() { global $width; // I thought this would solve the problem but it didn't global $height; // I thought this would solve the problem but it didn't list($width, $height, $type, $attr) = getimagesize($options['logo']); echo "Image width " .$width; echo "<BR>"; echo "Image height " .$height; echo "<BR>"; } $options['logo'] is returning http://localhost/wordpress/wp-content/uploads/2010/12/logo4.png so the image is being displayed. I also did var_dump to $width and $height but they didn't show up. Any suggestions?

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  • accessing private variable from member function in PHP

    - by Carson Myers
    I have derived a class from Exception, basically like so: class MyException extends Exception { private $_type; public function type() { return $this->_type; //line 74 } public function __toString() { include "sometemplate.php"; return ""; } } Then, I derived from MyException like so: class SpecialException extends MyException { private $_type = "superspecial"; } If I throw new SpecialException("bla") from a function, catch it, and go echo $e, then the __toString function should load a template, display that, and then not actually return anything to echo. This is basically what's in the template file <div class="<?php echo $this->type(); ?>class"> <p> <?php echo $this->message; ?> </p> </div> in my mind, this should definitely work. However, I get the following error when an exception is thrown and I try to display it: Fatal error: Cannot access private property SpecialException::$_type in C:\path\to\exceptions.php on line 74 Can anyone explain why I am breaking the rules here? Am I doing something horribly witty with this code? Is there a much more idiomatic way to handle this situation? The point of the $_type variable is (as shown) that I want a different div class to be used depending on the type of exception caught.

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  • Mimic Coldfusion's debug output in PHP?

    - by TekiusFanatikus
    I'm trying to mimic Coldfusion's debug output in PHP. Here's an example of what it looks like (ie. Execution Time section): I've turned to XDebug. Ideally, the exception stack error output would be what I'd be looking for. However, it only shows up when an exception occurs. I also tried something like (in our CMS-ish app) this (original question here): $content.= "<?php xdebug_start_trace('e:/xdebug/trace');?>"; $content.= "<?php require('".$page['file_'.LG]."'); ?>"; $content.= "<?php xdebug_stop_trace();?>"; ... $content.= "<?php echo readfile('e:/xdebug/trace.xt');?>"; However, I get an insane, browser crashing HTML table dropped at the bottom of page. Not very efficient. My php.ini config: xdebug.trace_format = 2 xdebug.collect_vars = 1 xdebug.collect_params = 4 xdebug.dump_globals = 1 xdebug.dump.SERVER = 'REQUEST_URI' xdebug.show_local_vars = 1 xdebug.show_mem_delta = 1 I'm just wondering if someone has already done something similar?

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  • What are function pointers good for ?

    - by gramm
    Hi, I have trouble seing the utility of the function pointers. I guess it may be useful in some case (it exists, after all), but I can't think of a case where it's better or unavoidable to use a function pointer. Could you give some example of good use of function pointers (in C or C++)? Many thanks :)

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  • Forcing user to new page in php. (PHP newbie)

    - by JohnC
    Hello I'm a newbie web programmer. My background is writing Windows applications with sql. I'm putting together my 1st data entry screens in Php. I have a search form that links to a form that displays records in a grid. On each row of the grid I have a delete url to allow the user to remove a record. This links to a form delete.php (which calls the sql to remove the record). Ideally I would like to automatically take the user back to the search form rather than forcing the user to click on a link to do so. I have used ob_start with the header to do this elsewhere but cannot get it to work on this page. Is there another way to do it? (Using php 5 as part of LAMP) file delete.php <?php $id = $_GET['recordID']; //ob_start(); require_once('connections/local.php'); mysql_select_db($database_local, $local); mysql_query("DELETE FROM user_access WHERE id = {$id}") or die(mysql_error()); echo("Record ".$id." deleted"); echo("<br>"); //header("location:http://localhost/search7.htm); //ob_flush(); echo("<a href=\"http://localhost/search7.htm\">Search for Members</a>"); ?>

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  • C++ inheritance and member function pointers

    - by smh
    In C++, can member function pointers be used to point to derived (or even base) class members? EDIT: Perhaps an example will help. Suppose we have a hierarchy of three classes X, Y, Z in order of inheritance. Y therefore has a base class X and a derived class Z. Now we can define a member function pointer p for class Y. This is written as: void (Y::*p)(); (For simplicity, I'll assume we're only interested in functions with the signature void f() ) This pointer p can now be used to point to member functions of class Y. This question (two questions, really) is then: Can p be used to point to a function in the derived class Z? Can p be used to point to a function in the base class X?

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  • PHP unlink OR rewrite own/current file by itself

    - by Email
    Hi Task: Cut or erase a file after first walk-through. i have an install file called "index.php" which creates another php file. <? /* here some code*/ $fh = fopen($myFile, 'w') or die("can't open file"); $stringData = "<?php \n echo 'hallo, *very very long text*'; \n ?>"; fwrite($fh, $stringData); /*herecut"/ /*here some code */ after the creation of the new file this file is called and i intent to erase the filecreation call since it is very long and only needed on first install. i therefor add to the above code echo 'hallo, *very very long text*'; \n ***$new= file_get_contents('index.php'); \n $findme = 'habanot'; $pos = strpos($new, $findme); if ($pos === false) { $marker='herecut';\n $new=strstr($new,$marker);\n $new='<?php \n /*habanot*/\n'.$new;\n $fh = fopen('index.php', 'w') or die 'cant open file'); $stringData = $new; fwrite($fh, $stringData); fclose($fh);*** ?>"; fwrite($fh, $stringData);]} Isnt there an easier way or a function to modify the current file or even "self destroy" a file after first call? Regards

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  • php lampp permissions of fopen function

    - by marmoushismail
    hi i'm programming php using: netbeans 6.8 lampp for ubuntu (xampp) apache which came with xampp $fh = fopen("testfile2.txt", 'w') or die("Failed to create file"); $text ="hello man cool good"; fwrite($fh, $text) or die("Could not write to file"); fclose($fh); echo "File 'testfile.txt' written successfully"; //i get the next error: Warning: fopen(testfile2.txt) [function.fopen]: failed to open stream: Permission denied in /home/marmoush/allprojects/phpprojects/myindex.php on line 91 Failed to create file anyway i know what this error is; it's about folder and files permissions; i looked into the folder permission tab made access available for "others" group ( to read and write) the program worked result was a file (test.txt) so i looked at the created file permission it appears to be that (php , xampp or whoever) creates file with (nobody permission) I have 2 QUESTIONS: 1- what if i need the file created by (php code and xampp ) to have the "root or user or myname" permissions ?? where to set this setting 2-also my concern (what if i send this files to actual web server will it make nobody permissions also nobody ? when they create files

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  • how to use a PHP Constant that gets pulled from a database

    - by Ronedog
    Can you read out the name of a PHP constant from a database and use it inside of a php variable, to display the value of the constant for use in a menu? For example here's what I'm trying to accomplish In SQL: select menu_name AS php_CONSTANT where menu_id=1 the value returned would be L_HOME which is the name of a CONSTANT in a php config page. The php config page looks like this define('L_HOME','Home'); and gets loaded before the database call. The php usage would be $db_returned_constant which has a value of L_HOME that came from the db call, then I would place this into a string such as $string = '<ul><li>' . $db_returned_constant . '</li></ul>' and thus return a string that looks like $string = '<ul><li><a href="#" onclick="path_from_db">Home</a></li></ul>'. To sum up what I'm trying to do Load a config file based on the language preference query the db to return the menu name, which is the name of a CONSTANT in the config file loaded in step one, and also retrieve the menu_link which is used in the "onclick" event. Use a php variable to hold the name of the CONSTANT Place the variable into a string that gets echo'd out to create the menu displaying the value of the CONSTANT. I hope this makes enough sense...is it even possible to use a constant like this? Thanks.

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  • creating new instance fails PHP

    - by as3isolib
    I am relatively new to PHP and having some decent success however I am running into this issue: If I try to create a new instance of the class GenericEntryVO, I get a 500 error with little to no helpful error information. However, if I use a generic object as the result, I get no errors. I'd like to be able to cast this object as a GenericEntryVO as I am using AMFPHP to communicate serialize data with a Flex client. I've read a few different ways to create constructors in PHP but the typical 'public function Foo()' for a class Foo was recommended for PHP 5.4.4 //in my EntryService.php class public function getEntryByID($id) { $link = mysqli_connect("localhost", "root", "root", "BabyTrackingAppDB"); if (mysqli_connect_errno()) { printf("Connect failed: %s\n", mysqli_connect_error()); exit(); } $query = "SELECT * FROM Entries WHERE id = '$id' LIMIT 1"; if ($result = mysqli_query($link, $query)) { // $entry = new GenericEntryVO(); this is where the problem lies! while ($row = mysqli_fetch_row($result)) { $entry->id = $row[0]; $entry->entryType = $row[1]; $entry->title = $row[2]; $entry->description = $row[3]; $entry->value = $row[4]; $entry->created = $row[5]; $entry->updated = $row[6]; } } mysqli_free_result($result); mysqli_close($link); return $entry; } //my GenericEntryVO.php class <?php class GenericEntryVO { public function __construct() { } public $id; public $title; public $entryType; public $description; public $value; public $created; public $updated; // public $properties; } ?>

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  • PHP IDE with Integrated Web Server

    - by seth
    Note: This is not another "What is the best PHP IDE?" question. I'm looking for a PHP IDE with a specific feature, namely an integrated / embedded (php enabled) web server; ideally with xdebug pre-bundled. I already know that Aptana 1.5 has this functionality (and some older versions of Zend Studio as well), but Aptana 1.5 hasn't been supported for quite some time and as we make the transition to PHP 5.3 and beyond, it's usefulness will diminish significantly. I've looked at some options including Eclipse PDT and NetBeans, but it seems every PHP IDE relies on a separate local/remote web server to actually interpret the code. I know installing a web server locally is fairly trivial, but this is for a classroom solution, where installing, configuring, and maintaining a web server on 1000 machines is simply not feasible. A remote server solution will also not work due to the need to use debugging functionality (xdebug currently requires a hardcoded IP for the debug client). This seems like such an obvious feature/plugin for a PHP IDE, but my research thus far has turned up no results.

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  • PHP class_exists always returns true

    - by Ali
    I have a PHP class that needs some pre-defined globals before the file is included: File: includes/Product.inc.php if (class_exists('Product')) { return; } // This class requires some predefined globals if ( !isset($gLogger) || !isset($db) || !isset($glob) ) { return; } class Product { ... } The above is included in other PHP files that need to use Product using require_once. Anyone who wants to use Product must however ensure those globals are available, at least that's the idea. I recently debugged an issue in a function within the Product class which was caused because $gLogger was null. The code requiring the above Product.inc.php had not bothered to create the $gLogger. So The question is how was this class ever included if $gLogger was null? I tried to debug the code (xdebug in NetBeans), put a breakpoint at the start of Product.inc.php to find out and every time it came to the if (class_exists('Product')) clause it would simply step in and return thus never getting to the global checks. So how was it ever included the first time? This is PHP 5.1+ running under MAMP (Apache/MySQL). I don't have any auto loaders defined.

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  • PHP, MySQL: Security concern; Page loads in a weird way

    - by Devner
    Hi all, I am testing the security of my website. I am using the following URL to load a PHP page in my website, on localhost: http://localhost/domain/user/index.php/apple.php When I do this, the page is not loading normally; Instead the images, icons used in the page simply vanish/disappear from the page. Only text appears. And also on any link I click on this page, it brings me to this same page again without navigating to the required page. So if I have hyperlinks to other pages, such as "SEARCH", which points to search.php, instead of navigating to the search.php page, it refreshes the index.php page and just appends the page name of the destination page to the end of the URL. For example, say I used the link above. It then loads the index.php page minus the images at it's will. When I click on the "Search" link to navigate to the search page, I see the following in the URL: http://localhost/domain/user/index.php/search.php I have a redirection configured to a 404 error page in my .htaccess file, but the page does not redirect to the 404 error page. Notice the search.php towards the end of the URL above. Any other link that I click, reloads the index.php page and just appends the destination page name to the end of the URL like I have shown above. I was expecting to see a 404 Error but that does not happen. The URL should not even be able to load the page because I do NOT have a "index.php" folder in my website. What can I do to solve this? All help is appreciated. Thank you.

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