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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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  • Learning PHP - start out using a framework or no?

    - by Kevin Torrent
    I've noticed a lot of jobs in my area for PHP. I've never used PHP before, and figure if I can get more opportunities if I pick it up then it might be a good idea. The problem is that PHP without any framework is ugly and 99% of the time really bad code. All the tutorials and books I've seen are really lousy - it never shows any kind of good programming practice but always the quick and dirty kind of way of doing things. I'm afraid that trying to learn PHP this way will just imprint these bad practices in my head and make me waste time later trying to unlearn them. I've used C# in the past so I'm familiar with OOP and software design patterns and similar. Should I be trying to learn PHP by using one of the better known frameworks for it? I've looked at CakePHP, Symfony and the Zend Framework so far; Zend seems to be the most flexible without being too constraining like Cake and Symfony (although Symfony seemed less constraining than CakePHP which is trying too hard to be Ruby on Rails), but many tutorials for Zend I've seen assume you already know PHP and want to learn to use the framework. What would be my best opportunity for learning PHP, but learning GOOD PHP that uses real software engineering techniques instead of spaghetti code? It seems all the PHP books and resources either assume you are just using raw PHP and therefore showcase bade practices, or that you already know PHP and therefore don't even touch on parts of the language.

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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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  • 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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  • Multiple XML/XSLT files in PHP, transform one with XSLT and add others but process it first with PHP

    - by ipalaus
    I am processing XML files transformations with XSLT in PHP correctly. Actually I use this code: $xml = new DOMDocument; $xml->LoadXML($xml_contents); $xsl = new DOMDocument; $xsl->load($xsl_file); $proc = new XSLTProcesoor; $proc->importStyleSheet($xsl); echo $proc->transformToXml($xml); $xml_contents is the XML processed with PHP, this is done by including the XML file first and then assigning $xml_contents = ob_get_contents(); ob_end_clean();. This forces to process the PHP code on the XML, and it works perfectly. My problem is that I use more than one XML file and this XML files has PHP code on it that need to be processed AND have a XSLT file associated to process the data. Actually I'm including this files in XSLT with the next code: <!-- First I add the XML file --> <xsl:param name="menu" select="document('menu.xml')" /> <!-- Next I add the transformations for menu.xml file --> <xsl:include href="menu.xsl" /> <!-- Finally, I process it on the actual ("parent") XML --> <xsl:apply-templates select="$menu/menu" /> My questiion is how I can handle this. I need to add mutiple XML(+XSLT) files to my first XML file that will containt PHP so it needs to be processed. Thank you in advance!

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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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  • 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 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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  • PHP app.yaml, new to GAE, not sure of how to setup

    - by chetweewax
    I have a single page website built on bootstrap 3, that I am trying to move to Google Apps Engine. I Scaffold my sites using php, and all the content is showing but not the styles and javascript. My site is basically set up as follows _/js/bootstrap.js _/js/custom.js _/fonts/glypicon ...etc _/css/bootstrap.css _/css/custom.css _/php/ .. all my php files go here ... index.php can someone help me setup my app.yaml for this? I am new to GAE, and am a little confused by this.

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  • PHP socket UDP communication

    - by Ghedeon
    Server works fine, but the problem is the client doesn't receive anything. server.php <?php $buf_size = 1024; $socket = stream_socket_server("udp://127.0.0.1:3127", $errno, $errstr, STREAM_SERVER_BIND); do { $str = stream_socket_recvfrom($socket, $buf_size, 0, $peer); $str = "abc"; stream_socket_sendto($socket, $str, strlen($str), 0, $peer); } while (true); ?> client.php <?php $fp = stream_socket_client("udp://127.0.0.1:3127", $errno, $errstr); if (!$fp) { echo "$errno - $errstr<br />\n"; } else { fwrite($fp, "1 2 3"); echo fread($fp, 15); fclose($fp); } ?>

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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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  • Never getting a JSON response when running server-side PHP proxy script but I do with others

    - by Dohk
    I'm on PHP 5.3.4 and Apache 2.2 btw So I'm using (or trying to use) Simple PHP Proxy (Simple PHP Proxy) I enter a URL at his example page at SPP Example Page and it works fine, I see the JSON response and all the headers. However, when I copy the exact URL, only changing the URL to now have localhost, I get both empty headers and no JSON. Assuming that the script on his site is the same I downloaded, could this be due to a multitude of things or a setting in Apache and/or the PHP ini? So for example: benalman.com/code/projects/php-simple-proxy/ba-simple-proxy.php?url=http://github.com/&full_headers=1&full_status=1 That will get me a ton of info back Now changing to localhost http://localhost/ba-simple-proxy.php?url=http://github.com/&full_headers=1&full_status=1 {"headers":[],"status":{"url":"https:\/\/github.com\/","content_type":"text\/html","http_code":301,"header_size":194,"request_size":182,"filetime":-1,"ssl_verify_result":0,"redirect_count":1,"total_time":0.094,"namelookup_time":0,"connect_time":0.047,"pretransfer_time":0,"size_upload":0,"size_download":185,"speed_download":1968,"speed_upload":0,"download_content_length":185,"upload_content_length":0,"starttransfer_time":0,"redirect_time":0.047,"certinfo":[]},"contents":null} I even went basic and just used some curl and of course, empty objects being returned other than false for my content and the url I set in my JSON. Any help is deeply appreciated or any ideas.

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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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  • 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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  • PHP Ajax not working

    - by Kostis
    I have 3 buttons on my page and depending on which one the user is clickingi want to run through ajax call a delete query in my database. When the user clicks on a button the javascript function seems to work but it doesn't run the query in php script. The html page: <?php session_start(); ?> <!DOCTYPE HTML> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-7"> <script> function myFunction(name) { var r=confirm("Are you sure? This action cannot be undone!"); if (r==true) { alert(name); // check if is getting in if statement and confirm the parameter's value var xmlhttp; if (str.length==0) { document.getElementById("clearMessage").innerHTML=""; return; } if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); } else {// code for IE6, IE5 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById("clearMessage").innerHTML= responseText; } } xmlhttp.open("GET","clearDatabase.php?q="+name,true); xmlhttp.send(); } else alert('pff'); } </script> </head> <body> <div id="wrapper"> <div id="header"></div> <div id="main"> <?php if (session_is_registered("username")){ ?> <!--<a href="#">???a????s? pa?a??? µ???µ?t??</a><br /> <a href="#">???a????s? pa?a??? s??ed????</a><br /> <a href="#">???a????s? push notifications</a><br />--> <input type="button" value="???a????s? pa?a??? µ???µ?t??" onclick="myFunction('messages')" /> <input type="button" value="???a????s? pa?a??? s??ed????" onclick="myFunction('conferences')" /> <input type="button" value="???a????s? push notifications" onclick="myFunction('notifications')" /> <div id="clearMessage"></div> <?php } else echo "Login first."; ?> </div> <div id="footer"></div> </div> </body> </html> and the php script: <?php if (isset($_GET["q"])) $q=$_GET["q"]; $host = "localhost"; $database = "dbname"; $user = "dbuser"; $pass = "dbpass"; $con = mysql_connect($host,$user,$pass) or die(mysql_error()); mysql_select_db($database,$con) or die(mysql_error()); if ($q=="messages") $query = "DELETE FROM push_message WHERE time_sent IS NOT NULL"; else if ($q=="conferences") $query = "DELETE FROM push_message WHERE time_sent IS NOT NULL"; else if ($q=="notifications") { $query = "DELETE FROM push_friend WHERE time_sent IS NOT NULL"; } $res = mysql_query($query,$con) or die(mysql_error()); if ($res) echo "success"; else echo "failed"; mysql_close($con); ?>

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  • php database image show problem

    - by Termedi
    here is the code <?php session_start(); if(!isset($_SESSION['user_name'])) { header('Location: login.php'); } $conn = mysql_connect("localhost", "root", "") or die("Can no connect to Database Server"); ?> <html> <head> </head> <body> <center> <div id="ser"> <form action="" method="post"> <label for="file">Card No:</label> <input type="text" name="card_no" id="card_no" class="fil" onKeyUp="CardNoLength()" onKeyDown="CardNoLength()" onKeyPress="CardNoLength()"/> <input type="submit" name="search" value="Search" class="btn" onClick="return CardNoLengthMIN()"/> </form> </div> </center> <br/><hr style="border: 1px solid #606060 ;" /> <center><a href="index.php">Home</a></center> <br/> <center> <?php if(isset($_POST['card_no'])) { if($conn) { if(mysql_select_db("img_mgmt", $conn)) { $sql = "select * from temp_images where card_no='".trim($_POST['card_no'])."'"; $result = mysql_query($sql); $image = mysql_fetch_array($result); if(isset($image['card_no'])) { //echo "<img src=\"".$image['file_path']."\" alt=\"".$image['card_no']."\" width=\"250\" height=\"280\"/>"; header("Content-type: image/jpeg"); echo $image['img_content']; } else { echo "<p style=\"color:red;\">Sorry, Your search came with no results ! <br/> Try with different card number"; } } else { echo "Database selection error: ".mysql_error(); } } else { echo "Could not connect: ".mysql_error(); } } ?> </center> </body> </html> But it after executing the script it shows: Cannot modify header information - headers already sent by (output started at C:\xampp\htdocs\img\search.php:61) in C:\xampp\htdocs\img\search.php on line 77

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  • Creating a session user login php

    - by user2419393
    I'm stuck on how to create a session for a user who logs in. I got the part of checking to make sure the log in information corresponds with the database information, but is stuck on how to take the email address and store into a session. Here is my php code below. <?php include '../View/header.php'; session_start(); require('../model/database.php'); $email = $_POST['username']; $password = $_POST['password']; $sql = "SELECT emailAddress FROM customers WHERE emailAddress ='$email' AND password = '$password'"; $result = mysql_query($sql, $db); if (!$result) { echo "DB Error, could not query the database\n"; echo 'MySQL Error: ' . mysql_error(); exit; } while ($row = mysql_fetch_assoc($result)) { echo $row['emailAddress']; } mysql_free_result($result); ?>

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  • showing error on uploading a big file using php

    - by user1489969
    I have created a php code to display the upload option to upload multiple files as below: <?php $f_id= $_GET["id"]; ?> <title>Upload File</title> <form enctype="multipart/form-data" method="post" action="upload_hal_mult.php?id=<?php echo $f_id;?>" > <input type="hidden" name="MAX_FILE_SIZE" value="10000000"> <input id="infile" type="file" name="infile[]" multiple="true" /> <input type="submit" value="upload" name="file_uploaded" / > <br> <br> </form> So this will call "upload_hal_mult.php" when "upload" button is clicked. And the code for that is as follows: <title>Upload Results</title> <?php define("MAX_SIZE",10000000); $f_id= $_GET["id"]; $dir_name="dir_hal_".$f_id; $u=0; if (!is_dir($dir_name)) mkdir($dir_name); $dir=$dir_name."/"; $file_realname = $_FILES['infile']['name']; for ($i = 0; $i < count($_FILES['infile']['name']); $i++) { $ext = substr(strrchr($_FILES['infile']['name'][$i], "."), 1); $fname = substr($_FILES['infile']['name'][$i],0,strpos($_FILES['infile']['name'][$i], ".")); $fPath = $fname."_(".substr(md5(rand() * time()),0,4).")".".$ext"; echo "files size=".$_FILES["infile"]["size"][$i]."\n"; if($_FILES["infile"]["size"][$i]>MAX_SIZE) echo('File uploaded exceeds maximum upload size.'); if(($_FILES['infile']['error'][i]==0) && move_uploaded_file($_FILES['infile']['tmp_name'][$i], $dir . $fPath)) { $u=$u+1; ?> <!--<script type="text/javascript">setTimeout("window.close();", 1300);</script>--> <?php echo "Upload is successful\n"; } else echo "if stmt failed so error \n"; } if($u!=count($_FILES['infile']['name'])) echo "Error"; else echo "count is correct"; ?> This upload works correctly for files of size<10MB. But for files of size10MB, it's not echoing 'File uploaded exceeds maximum upload size.' as expected. Its also not uploading the file of size10MB. But the $u gets incremented. But none of the statements like "Upload is successful" or "if stmt failed so error" are being echoed as well. However the statement "count is correct" is being displayed, and this shows the $u got incremented somehow even though the echo statements didnt work! Can someone please point out the error I am doing here? I thought its simply a matter of 'if/else' statements but it seems more than that to me. Please help me out if you have any clue. Thanks

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  • Write to the second line of a PHP file

    - by Woz
    I have a php file that I want to add an include path to on the second line. I need to open the file and inset a line of code on line 2. I have tried a few techniques none of which are working but I think it has something to do with the text I am trying to write and possibly not escaping character correctly as I am not too familiar with file writing. So here is the file I want to write to: $file = $_SERVER['DOCUMENT_ROOT'].'/'.$domaindir.'/test.php'; Here is the piece of text I want to insert into the file: $dbfile = "include('".$_SERVER['DOCUMENT_ROOT']."/".$domaindir."/web_".$dbname.".inc.php');"; Then what I was doing was a string replace but all it did was bump the "session_start();" bit to a newline! Can anyone point me in the direction of a tutorial that might tell me how to insert this into the second line of my php file or indeed if anyone has any ideas? I can say for sure that the path to the PHP file is fully tested so i know its not that the file is not being open or written to. Any ideas would be much appreciated. Thanks in advance.

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  • how to run imagemagik commands in PHP ?

    - by user345804
    Hi, i am trying to bend text using imagemagik in PHP. but the commands shown in the website are not working. http://www.fmwconcepts.com/imagemagick/texteffect/index.php how can i run these scripts in PHP ? somebody please help me.. NB :-t \'SOME ARCHBOTTOM TEXT\' -s outline -e arch-bottom -d 1.0 -f Arial -p 48 -c skyblue -b white -o black -l 1 -u lightpink

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  • Getting search results from Twitter in php

    - by Mark Mayo
    I'm attempting to put together a little mashup with some twitter APIs. However, the whole area is new to me (I'm more of an embedded developer dabbling). And frustratingly, every tutorial I am trying in Php is either out of date, not doing what it claims to do, it or is broken. Essentially, I just want a nice bit of example code - say, an HTML file, a connection.js for the JQuery magic, and a php file - 'getsearch' which contains the relevant Curl calls to the API to just return the results for a given search term. Followed the tutorial to the letter at http://www.reynoldsftw.com/2009/02/using-jquery-php-ajax-with-the-twitter-api/ and even downloaded the guy's code and chucked it on my webserver, but it just seems to sit there. I'm relatively competent at php and html, but it's the Curl and the JQuery side of things which is new to me, and would appreciate any thoughts, links, or code suggestions. I've attempted reading the API - but even that seems sparse - and several links are broken to their own tutorials, so that's put me off a bit for now.

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  • replacing variables in output in php

    - by Thorpe Obazee
    Right now I have this code. <?php error_reporting(E_ALL); require_once('content_config.php'); function callback($buffer) { // replace all the apples with oranges foreach ($config as $key => $value) { $buffer = str_replace($key, $value, $buffer); } return $buffer; } ob_start("callback"); ?> some content <?php ob_end_flush(); ?> in the content_config.php file: $config['SiteName'] = 'MySiteName'; $config['SiteAuthor'] = 'thatGuy'; What I want to do is that I want to replace the placeholders that with the key of the config array with its value. Right now, it doesn't work :(

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