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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 & 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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  • Launch a new page in Zend/php on button click

    - by BlueMonster
    PHP & Zend Noob here I've downloaded the skeleton project from here: https://github.com/zendframework/ZendSkeletonApplication Say i want to open a new page that simply displays "hello world" text if click on the "ZF2 Development Portal" button(bottom left green button) on the page that launches --- how do i do this? See image: I've read through this tutorial, but i'm not sure how the model, view, or controller are actually launched? See tutorial: http://blog.wilgucki.pl/2012/07/tworzenie-modulw-w-zend-framework-2.html From looking at the code, i know that i will have to change this line of code: <div class="span4"> <h2><?php echo $this->translate('Follow Development') ?></h2> <p><?php echo sprintf($this->translate('Zend Framework 2 is under active development. If you are interested in following the development of ZF2, there is a special ZF2 portal on the official Zend Framework website which provides links to the ZF2 %swiki%s, %sdev blog%s, %sissue tracker%s, and much more. This is a great resource for staying up to date with the latest developments!'), '<a href="http://framework.zend.com/wiki/display/ZFDEV2/Home">', '</a>', '<a href="http://framework.zend.com/zf2/blog">', '</a>', '<a href="http://framework.zend.com/issues/browse/ZF2">', '</a>') ?></p> <p><a class="btn btn-success" href="http://framework.zend.com/zf2" target="_blank"><?php echo $this->translate('ZF2 Development Portal') ?> &raquo;</a></p> </div> More specifically this line: <p><a class="btn btn-success" href="http://framework.zend.com/zf2" target="_blank"><?php echo $this->translate('ZF2 Development Portal') ?> &raquo;</a></p> but i'm really confused as to what i'm supposed to change it to in order to launch a new page. Any ideas? Thanks in advance!

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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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  • 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 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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  • 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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  • Providing updating suggestions list with javascript, php and ajax

    - by user1104854
    I'm trying to modify this example on making a live updating list to integrate it with my API. So, instead of using GET on the page with the form, I'd like to send it to that page via a function call. So, here's my form // message.php //function to display the hint sent from gethint.php function message_hint($hint){ echo $hint; } //displays the form for sending messages function send_message_form($to_user,$title,$message){ include 'gethint.php'; ?> <table> <form name = "send_message" method="post"> <td>Send A Message</td> <tr><td>To:</td><td><input type = "text" size="50" name="to_user" id = "to_user" value ="<? echo $to_user; ?>" onkeyup="showHint(this.value)"></td></tr> <tr><td>Title:</td><td><input type = "text" size="50" name="message_title"></td></tr> <tr><td>Message:</td><td><textarea rows="4" cols="50" name="message_details"></textarea></td></tr> <tr><td><input type="submit" name="submit_message"></td></tr> </table> </form> <? } Here's the head of message.php <head> <script> function showHint(str){ var to_user = document.getElementById("to_user").value //to_user is the id of the textbox if (str.length==0){ to_user.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){ alert(to_user) //properly displays the name via alert box to_user.innerHTML=xmlhttp.responseText; } } xmlhttp.open("GET","gethint.php?q="+to_user,true); xmlhttp.send(); } </script> </head> The page gethint.php is exactly the same, aside from this at the bottom. //echo $response //this was the original output $message = new messages; $message->message_hint($response);

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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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  • 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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  • 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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  • 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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  • 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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  • PHP OAuth Twitter

    - by Sandhurst
    I have created a twitter app which I am using to post tweets. The problem that I am not able to resolve is everytime I have to allow access to my application. so lets say I need to tweet three messages, so all the three times I have to allow access to my app. I just need that once user has allowed access to my app, next time he should only be asked to allow acces is that when he/she relogins. Here's my code that I am using Share content on twitter"; include 'lib/EpiCurl.php'; include 'lib/EpiOAuth.php'; include 'lib/EpiTwitter.php'; include 'lib/secret.php'; $twitterObj = new EpiTwitter($consumer_key, $consumer_secret); $oauth_token = $_GET['oauth_token']; if($oauth_token == '') { $url = $twitterObj-getAuthorizationUrl(); echo ""; echo "Sign In with Twitter"; echo ""; } else { $twitterObj-setToken($_GET['oauth_token']); $token = $twitterObj-getAccessToken(); $twitterObj-setToken($token-oauth_token, $token-oauth_token_secret); $_SESSION['ot'] = $token-oauth_token; $_SESSION['ots'] = $token-oauth_token_secret; $twitterInfo= $twitterObj-get_accountVerify_credentials(); $twitterInfo-response; $username = $twitterInfo-screen_name; $profilepic = $twitterInfo-profile_image_url; include 'update.php'; } if(isset($_POST['submit'])) { $msg = $_REQUEST['tweet']; $twitterObj-setToken($_SESSION['ot'], $_SESSION['ots']); $update_status = $twitterObj-post_statusesUpdate(array('status' = $msg)); $temp = $update_status-response; header("Location: MessageStatus.html"); exit(); } ?

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  • As3 & PHP URLEncoding problem!

    - by Jk_
    Hi everyone, I'm stuck with a stupid problem of encoding. My problem is that all my accentuated characters are displayed as weird iso characters. Example : é is displayed %E9 I send a string to my php file : XMLLoader.load(new URLRequest(online+"/query.php?Query=" + q)); XMLLoader.addEventListener(Event.COMPLETE,XMLLoaded); When I trace q, I get : "INSERT INTO hello_world (message) values('éàaà');" The GOOD query My php file look like this : <?php include("conection.php");//Conectiong to database $Q = $_GET['Query']; $query = $Q; $resultID = mysql_query($query) or die("Could not execute or probably SQL statement malformed (error): ". mysql_error()); $xml_output = "<?xml version=\"1.0\"?>\n"; // XML header $xml_output .= "<answers>\n"; $xml_output .= "<lastID id=".'"'.mysql_insert_id().'"'." />\n"; $xml_output .= "<query string=".'"'.$query.'"'." />\n"; $xml_output .= "</answers>"; echo $xml_output;//Output the XML ?> When I get back my XML into flash the $query looks like this : "INSERT INTO hello_world (message) values('%E9%E0a%E0');" And these values are then displayed into my DB which is annoying. Any help would be appreciated! Cheers. Jk_

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  • Php and python regexp difference?

    - by Ajel
    I need to parse a string 'Open URN: 100000 LA: ' and get 100000 from it. on python regexp (?<=Open URN: )[0-9]+(?= LA:) works fine but in php it gives following error: preg_match(): Unknown modifier '[' I need it working php, so please help me to solve this problem and tell about difference in python and php regexps.

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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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  • 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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  • 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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  • 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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