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  • asp.net membership db. Adding branches to a user

    - by diver-d
    Hi everyone, I am playing around with the membership database and have a question. Lets say I have a company called company ABC. There are 10 branches that belong to comapany ABC. IS there anyway to create a relationship within the database using company ABC as the parent and having the 10 branches as childs? I hope that makes sense :)

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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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  • 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: 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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  • asp.net dropdown iniside datagrid

    - by harold-sota
    I'm inserting a dropdwon list in datagrid on row editing. When i run the project the datasource is not rekognized. The asp.net part is there: <asp:TemplateField HeaderText="Lookup 1"> <EditItemTemplate> <asp:DropDownList ID="Loocup1DropDownList" Width="100%" runat="server" DataSource ="<%GetValueForDropDownCombinationContent()%>" DataValueField="LOOKUP_ID" DataTextField="lookup_name" > </asp:DropDownList> </EditItemTemplate> <ItemTemplate> <asp:Label ID="LOOKUP1_NAME" runat="server" Text='<%# Bind("LOOKUP1_NAME") %>'></asp:Label> </ItemTemplate> This is the vb.net function: Protected Function GetValueForDropDownCombinationContent() As DataSet Dim dsProductLookups As New DataSet dsProductLookups = DocumentManager.Data.DataRepository.Provider.ExecuteDataSet("sp_GetCombinationsLookups", productCombo.SelectedValue) Return dsProductLookups End Function any ideas???

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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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  • 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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  • Progressively stream the output of an ASP.NET page - or render a page outside of an HTTP request

    - by Evgeny
    I have an ASP.NET 2.0 page with many repeating blocks, including a third-party server-side control (so it's not just plain HTML). Each is quite expensive to generate, in terms of both CPU and RAM. I'm currently using a standard Repeater control for this. There are two problems with this simple approach: The entire page must be rendered before any of it is returned to the client, so the user must wait a long time before they see any data. (I write progress messages using Response.Write, so there is feedback, but no actual results.) The ASP.NET worker process must hold everything in memory at the same time. There is no inherent needs for this: once one block is processed it won't be changed, so it could be returned to the client and the memory could be freed. I would like to somehow return these blocks to the client one at a time, as each is generated. I'm thinking of extracting the stuff inside the Repeater into a separate page and getting it repeatedly using AJAX, but there are some complications involved in that and I wonder if there is some simper approach. Ideally I'd like to keep it as one page (from the client's point of view), but return it incrementally. Another way would be to do something similar, but on the server: still create a separate page, but have the server access it and then Response.Write() the HTML it gets to the response stream for the real client request. Is there a way to avoid an HTTP request here, though? Is there some ASP.NET method that would render a UserControl or a Page outside of an HTTP request and simply return the HTML to me as a string? I'm open to other ideas on how to do this as well.

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  • Control button Post Back when click it in ASP.NET with VB.NET

    - by Lefteris Gkinis
    In my Default.aspx web page i use the following code in order to save some Attributes to an xml file <asp:Button ID="Button1" Text="Save Attributes" CssClass="Button01" runat="server" OnClientClick="SaveAttributesButton_Click(); return, false;" > </asp:Button> when i Click on the button that starts to make postback all the code right from the page load I have already use the HTML Button but this need a java script which is run when the page load And sub without my command Please is there anybody which can assist me on this issue?

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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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  • 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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  • 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 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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  • .NET make a copy of an embedded file resource to the local drive

    - by Matt H.
    Hi, i'm new to the realm to working with Files in .NET I'm creating a WPF application in VB.NET with the 3.5 Framework. (If you provide an example in C#, that's perfectly fine.) In my project I have a Template for an MS Access database. My desired behavior is that when the users clicks File--New, they can create a new copy of this template, give it a filename, and save it to their local directory. The database already has the tables and some starting data needed to interface with my application (a user-friendly data editor) I'm thinking the approach is to include this "template.accdb" file as a resource in the project, and write it to a file somehow at runtime? Any guidance will be very, very appreciated. Thanks!

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  • How to adjust asp.net URL routing based on domain/host?

    - by DrewF
    What's the best way to adjust the path destination for a routing table created in the global.asax Application_Start event based on the domain/sub domain/host? The following worked in IIS6, but with IIS7 the request object is decoupled from the Application_Start event and therefore does not work anymore: Dim strHost As String = Context.Request.Url.Host Dim strDir As String = "" If strHost.Contains("domain1.com") Then strDir = "area1/" Else strDir = "area2/" End If routes.MapPageRoute("Search", "Search", "~/" & strDir & "search.aspx")

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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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  • How to perform an event on Textbox focus in ASP MVC?

    - by NewbieProgrammer
    I have three textboxes A, B and C in create user view. When user enters some text in textbox A and B, and then when he enters textbox C, I want to display the text of textbox A + text of textbox B in C with "-" in between. For example, He enters "ABC" in textBox A and then he enters "123" in textBox B. Now upon entering textBox C (focus event), I want to display "ABC - 123 - " in textBox C. "-" are added through code. How do I do that in MVC ?

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