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  • Running code/script as a result of a form submission in ASP.NET

    - by firmbeliever
    An outside vendor did some html work for us, and I'm filling in the actual functionality. I have an issue that I need help with. He created a simple html page that is opened as a modal pop-up. It contains a form with a few input fields and a submit button. On submitting, an email should be sent using info from the input fields. I turned his simple html page into a simple aspx page, added runat=server to the form, and added the c# code inside script tags to create and send the email. It technically works but has a big issue. After the information is submitted and the email is sent, the page (which is supposed to just be a modal pop-up type thing) gets reloaded, but it is now no longer a pop-up. It's reloaded as a standalone page. So I'm trying to find out if there is a way to get the form to just execute those few lines of c# code on submission without reloading the form. I'm somewhat aware of cgi scripts, but from what I've read, that can be buggy with IIS and all. Plus I'd like to think I could get these few lines of code to run without creating a separate executable. Any help is greatly appreciated.

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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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  • 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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  • Capture form fields and repopulate the form with them

    - by Joel Cunningham
    I am currently testing a large web form and would like to be able to easily populate the form with several different lots of test data without having to type them each time. Is there a generic way to capture form inputs on a web page and have them repopulated on a different page load? I thought a tool like greasemonkey might be able to do something like this.

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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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  • 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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  • 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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  • URL routing in an MVC framework - PHP

    - by Walderman
    I'm developing an MVC framework in PHP from scratch; mostly for the learning experience but this could easily end up in a live project. I went through this tutorial as a base and I've expanded from there. Requests are made like this: examplesite.com/controller/action/param1/param2/ and so on... And this is my .htaccess file: RewriteEngine on RewriteCond %{REQUEST_FILENAME} !-f RewriteCond %{REQUEST_FILENAME} !-d RewriteRule ^(.*)$ index.php?rt=$1 [L,QSA] So all requests go to index.php and they are routed to the correct controller and action from there. If no controller or action is given, then the default 'index' is assumed for both. I have an index controller with an index action, which is supposed to be the home page of my site. I can access it by going to examplesite.com (since the index part is assumed). It has some images, a link to a stylesheet, and some scripts. They are linked with paths relative to index.php. I thought this would be fine since all request go to index.php and all content is simply included in this page using php. This works if I go to examplesite.com. I will see all of the images and styles, and scripts will run. However, if I go to examplesite.com/index, I am routed to the correct part of the site, but all of the links don't work. Does the browser think I am in a different folder? I would like to be able to use relative paths for all of the content in my site, because otherwise I need to use absolute paths everywhere to make sure things will show up. Is this possible?

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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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  • trouble connecting to MySql DB (PHP)

    - by user332817
    Hi I have the following PHP code to connect to my db. <?php ob_start(); $host="localhost"; // Host name $username="root"; // Mysql username $password=""; // Mysql password $db_name="test"; // Database name $tbl_name="members"; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); ?> however I get the following error: Warning: mysql_connect() [function.mysql-connect]: [2002] A connection attempt failed because the connected party did not (trying to connect via tcp://localhost:3306) in C:\Program Files (x86)\EasyPHP-5.3.2i\www\checklogin.php on line 11 Warning: mysql_connect() [function.mysql-connect]: A connection attempt failed because the connected party did not properly respond after a period of time, or established connection failed because connected host has failed to respond. in C:\Program Files (x86)\EasyPHP-5.3.2i\www\checklogin.php on line 11 Fatal error: Maximum execution time of 30 seconds exceeded in C:\Program Files (x86)\EasyPHP-5.3.2i\www\checklogin.php on line 11 I am able to add a db/tables via phpmyadmin but I cant connect using php. here is a screenshot of my phpmyadmin page: http://img294.imageshack.us/img294/1589/sqls.jpg any help would be appreciated, thanks in advance.

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  • Flex + PHP + ValueObjects

    - by Tempname
    I have a php/flex value object that I am using to transmit data to/from in my application. Everything works great php-flex, but I am having an issue with flex-php. In my MergeTemplateService.php service I have the following code. This is the method that flex hits directly: function updateTemplate($valueObject){ $object = DAOFactory::getMergeTemplateDAO()->update($valueObject); return $object; } I am passing a value object that from flex looks like this: (com.rottmanj.vo::MergeTemplateVO)#0 communityID = 0 creationDate = (null) enterpriseID = 0 lastModifyDate = (null) templateID = 2 templateName = "My New Test Template" userID = 0 The issue I am having is that my updateTemplate method sees the value object as an array and not an object. In my amfphp globals.php I have set my voPath as: $voPath = "services/class/dto/"; Any help with this is greatly appreciated Here are my two value objects: AS3 VO: package com.rottmanj.vo { [RemoteClass(alias="MergeTemplate")] public class MergeTemplateVO { public var templateID:int; public var templateName:String; public var communityID:int; public var enterpriseID:int; public var userID:int; public var creationDate:String; public var lastModifyDate:String public function MergeTemplateVO(data:Object = null):void { if(data != null) { templateID = data.templateID; templateName = data.templateName; communityID = data.communityID; enterpriseID = data.enterpriseID; userID = data.userID; creationDate = data.creationDate; lastModifyDate = data.lastModifyDate; } } } } PHPVO: <?php class MergeTemplate{ var $templateID; var $templateName; var $communityID; var $enterpriseID; var $userID; var $creationDate; var $lastModifyDate; var $_explictType = 'MergeTemplate'; } ?>

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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 active page code - I can't figure out parse error

    - by dmschenk
    I'm trying to build an active page menu with PHP and MySQL and am having a difficult time fixing the error. In the while statement I have an if statement that is giving me fits. Basically I think I'm saying that "thispage" is equal to the "title" based on pageID and as the menu is looped through if "thispage" is equal to "title" then echo id="active". Thanks <?php mysql_select_db($database_db_connection, $db_connection); $query_rsDaTa = "SELECT * FROM pages WHERE pagesID = 4"; $rsDaTa = mysql_query($query_rsDaTa, $db_connection) or die(mysql_error()); $row_rsDaTa = mysql_fetch_assoc($rsDaTa); $totalRows_rsDaTa = mysql_num_rows($rsDaTa); $query_rsMenu = "SELECT * FROM menu WHERE online = 1 ORDER BY menuPos ASC"; $rsMenu = mysql_query($query_rsMenu, $db_connection) or die(mysql_error()); $thisPage = ($row_rsDaTa['title']); ?> <link href="../css/MainStyle.css" rel="stylesheet" type="text/css" /> <h2><?php echo $thisPage; ?></h2> <div id="footcontainer"> <ul id="footlist"> <?php while($row_rsMenu = mysql_fetch_assoc($rsMenu)) { echo (" <li" . <?php if ($thisPage==$row_rsDaTa['title']) echo id="active"; ?> . "<a href=\"../" . $row_rsMenu['menuURL'] . "\">" . $row_rsMenu['menuName'] . "</a></li>\n"); } echo "</ul>\n"; ?> </div> <?php mysql_free_result($rsMenu); mysql_free_result($rsDaTa); ?>

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  • Overriding form submit based on counting elements with jquery.each

    - by MrGrigg
    I am probably going about this all wrong, but here's what I'm trying to do: I have a form that has approximately 50 select boxes that are generated dynamically based on some database info. I do not have control over the IDs of the text boxes, but I can add a class to each of them. Before the form is submitted, the user needs to select at least one item from the select box, but no more than four. I'm a little bit sleepy, and I'm unfamiliar with jQuery overall, but I'm trying to override $("form").submit, and here's what I'm doing. Any advice or suggestions are greatly appreciated. $("form").submit(function() { $('.sportsCoachedValidation').each(function() { if ($('.sportsCoachedValidation :selected').text() != 'N/A') { sportsSelected++ } }); if (sportsSelected >= 1 && sportsSelected <= 4) { return true; } else if (sportsSelected > 4) { alert('You can only coach up to four sports.'); sportsSelected = 0; return false; } else { alert('Please select at least one coached sport.'); sportsSelected = 0; return false; } });

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  • jQuery Validate on form submitted by JavaScript

    - by Daniel
    My form is submitted by a link using JavaScript, but I am also trying to validate the from justing jQuery validate. The validation doesn't work when submitted by the link, but it does if I change the link to a submit button. What am I doing wrong? My form: <form id="findmatch" method="post" action="search"> <div> <label class="formlabel">Match Type <input type="text" name="matchtype" id="matchtype" class="forminput" /> </label> <label class="formlabel">Location (postcode) <input type="text" name="location" id="location" class="forminput" /> </label> <label class="formlabel">Radius (miles) <input type="text" name="Radius" id="Radius" class="forminput" /> </label> <label class="formlabel">Keywords <input type="text" onblur="javascript:usePointFromPostcode(document.getElementById('location').value, showCompleteLatLng)" onchange="javascript:usePointFromPostcode(document.getElementById('location').value, showCompleteLatLng)" name="keywords" id="keywords" class="forminput" /> </label> <input id="lat" class="hidden" name="lat" type="text" value="" /> <input id="lon" class="hidden" name="lon" type="text" value="" /> <a href="javascript:document.getElementById('findmatch').submit();" onmouseover="javascript:usePointFromPostcode(document.getElementById('location').value, showCompleteLatLng)" class="submit">Search</a> </div> </form> And my jQuery is <script type="text/javascript"> $(document).ready(function () { $("#findmatch").validate({ rules: { location: "required", Radius: { required: true, digits: true }, keywords: "required" }, messages: { location: "Please enter your postcode", Radius: { required: "Please enter a radius", digits: "Please only enter numbers" }, keywords: "Please enter the keywords you wish to search for" } }); }); </script>

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  • PHP While loop seperating unique categories from multiple 'Joined' tables

    - by Hob
    I'm pretty new to Joins so hope this all makes sense. I'm joining 4 tables and want to create a while loop that spits out results nested under different categories. My Tables categories id | category_name pages id | page_name | category *page_content* id | page_id | image_id images id | thumb_path My current SQL join <?php $all_photos = mysql_query(" SELECT * FROM categories JOIN pages ON pages.category = categories.id JOIN image_pages ON image_pages.page_id = pages.id JOIN images ON images.id = image_pages.image_id ");?> The result I want from a while loop I would like to get something like this.... Category 1 page 1 Image 1, image 2, image 3 page 2 Image 2, image 4 Category 2 page 3 image 1 page 4 image 1, image 2, image 3 I hope that makes sense. Each image can fall under multiple pages and each page can fall under multiple categories. at the moment I have 2 solutions, one which lists each category several times according to the the amount of pages inside them: eg. category 1, page 1, image 1 - category 1, page 1, image 2 etc One that uses a while loop inside another while loop inside another while loop, resulting in 3 sql queries. <?php while($all_page = mysql_fetch_array($all_pages)) { ?> <p><?=$all_page['page_name']?></p> <?php $all_images = mysql_query("SELECT * FROM images JOIN image_pages ON image_pages.page_id = " . $all_page['id'] . " AND image_pages.image_id = images.id"); ?> <div class="admin-images-block clearfix"> <?php while($all_image = mysql_fetch_array($all_images)) { ?> <img src="<?=$all_image['thumb_path']?>" alt="<?=$all_image['title']?>"/> <?php } ?> </div> <?php } } ?

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  • Extracting DCT coefficients from encoded images and video

    - by misha
    Is there a way to easily extract the DCT coefficients (and quantization parameters) from encoded images and video? Any decoder software must be using them to decode block-DCT encoded images and video. So I'm pretty sure the decoder knows what they are. Is there a way to expose them to whomever is using the decoder? I'm implementing some video quality assessment algorithms that work directly in the DCT domain. Currently, the majority of my code uses OpenCV, so it would be great if anyone knows of a solution using that framework. I don't mind using other libraries (perhaps libjpeg, but that seems to be for still images only), but my primary concern is to do as little format-specific work as possible (I don't want to reinvent the wheel and write my own decoders). I want to be able to open any video/image (H.264, MPEG, JPEG, etc) that OpenCV can open, and if it's block DCT-encoded, to get the DCT coefficients. In the worst case, I know that I can write up my own block DCT code, run the decompressed frames/images through it and then I'd be back in the DCT domain. That's hardly an elegant solution, and I hope I can do better. Presently, I use the fairly common OpenCV boilerplate to open images: IplImage *image = cvLoadImage(filename); // Run quality assessment metric The code I'm using for video is equally trivial: CvCapture *capture = cvCaptureFromAVI(filename); while (cvGrabFrame(capture)) { IplImage *frame = cvRetrieveFrame(capture); // Run quality assessment metric on frame } cvReleaseCapture(&capture); In both cases, I get a 3-channel IplImage in BGR format. Is there any way I can get the DCT coefficients as well?

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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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  • Pass a variable from the source file to an included file in PHP

    - by Alpha1
    For my website I want to store the general format of the site in a single PHP file in a single location, and each of the different pages content in the local location of the page. I then want to pass the title and content address to the included file via a variable. However I can't get the included format file to read the variables storing the title and content data. AKA, the called file for the individual page would be: <?php $title = 'Some Title'; $source_file = 'content.php'; readfile('http:...../format.php'); ?> The format file would be: <html> ... <title> <?php echo $title; ?> </title> ... <?php include($source_file); ?> ... I recall reading somewhere I need to include something to get the variables at the start of the format file, however I can't remember what it is or find where I found that information.

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  • Select multi-select form from array

    - by Budove
    Here's the issue. I have a database column called pymnt_meth_pref, which contains a comma separated string of payment methods chosen from a multiselect form. <td>Payment Methods Used:<br /> <input type="checkbox" name="pyment_meth_pref[]" value="Cash">I can pay with cash.<br /> <input type="checkbox" name="pyment_meth_pref[]" value="Check">I can pay by check.<br /> <input type="checkbox" name="pyment_meth_pref[]" value="Credit Card">I can pay by credit card.<br /> <input type="checkbox" name="pyment_meth_pref[]" value="Paypal">I can pay with Paypal.<br /> </td> This array is posted to a variable and turned into a comma separated string if (isset($_POST['pyment_meth_pref'])) $pymntmethpref = implode(", ", $_POST['pyment_meth_pref']); if (isset($_POST['pyment_meth_acc'])) $pymntmethacc = implode(", ", $_POST['pyment_meth_acc']); This is then inserted into the database as a comma separated string. What I would like to do, is take this string and apply the values to the original form when the user goes back to the form as 'pre-selected' checkboxes, indicated that the user has already selected those values previously, and keeping those values in the database if they choose to edit any other information in the form. I'm assuming this would need to be done with javascript but if there is a way to do it with PHP I'd rather do it that way.

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  • PHP Form multiple buttons

    - by Ken
    I have a form with 2 buttons, that depending on which is selected will either be deleted or edited from the database. Those are each individual pages using SQL statements (questionedit and questiondelete). However, when i press a button, nothing happens...Any Ideas Here is my javascript <script type="text/javascript"> function SelectedButton(button) { if(button == 'edit') { document.testedit_questionform.action ="testedit_questionedit.php"; } else if(button == 'delete') { document.testedit_questionform.action ="testedit_questiondelete.php"; } document.forms[].testedit_questionform.submit(); } </script> Here is my form (being echoed from a loop) <form name=\"testedit_questionform\" action=\"SelectedButton\" method=\"POST\"> <span class=\"grid_11 prefix_1\" id=\"\" > Question:<input type=\"text\" name=\"QuestionText\" style=\"width:588px; margin-left:10px;\" value=\"$row[0]\"/> <input type=\"button\" value=\"Edit\" name=\"Operation\"onclick=\"submitForm(\'edit\')\" /> <input type=\"button\" value=\"Delete\" name=\"Operation\"onclick=\"submitForm(\'delete\')\" /> <input type=\"hidden\" name=\"QId\" value=\"$row[3]\" /><br />"); </form>

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  • validate form view and gridview at the same form

    - by Saeed
    i have formview when with tow required validator and gridview with reqired validator when i click insert on form view it fires the validation on the gridview i want when i click inserton form view just validate the tow validators on the form and doesnt fire the validator on gridview

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  • Include php code within echo from a random text

    - by lisa
    I want to display a php code at random and so for I have <?php // load the file that contain thecode $adfile = "code.txt"; $ads = array(); // one line per code $fh = fopen($adfile, "r"); while(!feof($fh)) { $line = fgets($fh, 10240); $line = trim($line); if($line != "") { $ads[] = $line; } } // randomly pick an code $num = count($ads); $idx = rand(0, $num-1); echo $ads[$idx]; ?> The code.txt has lines like <?php print insert_proplayer( array( "width" => "600", "height" => "400" ), "http://www.youtube.com/watch?v=xnPCpCVepCg"); ?> Proplayer is a wordpress plugin that displays a video. The codes in code.txt work well, but not when I use the pick line from code.txt. Instead of the full php line I get: "width" => "600", "height" => "400" ), "http://www.youtube.com/watch?v=xnPCpCVepCg"); ?> How can I make the echo show the php code, rather than a txt version of the php code?

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  • How to avoid notice in php when one of the conditions is not true

    - by user225269
    I've notice that when one of the two conditions in a php if statement is not true. You get an undefined index notice for the statement that is not true. And the result in my case is a distorted web page. For example, this code: <?php session_start(); if (!isset($_SESSION['loginAdmin']) && ($_SESSION['loginAdmin'] != '')) { header ("Location: loginam.php"); } else { include('head2.php'); } if (!isset($_SESSION['login']) && ($_SESSION['login'] != '')) { header ("Location: login.php"); } else { include('head3.php'); } ?> If one of the if statements is not true. The one that is not true will give you a notice that it is undefined. In my case it says that the session 'login' is not defined. If session 'LoginAdmin' is used. What can you recommend that I would do in order to avoid these undefined index notice.

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