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  • submittingthe form even if the form is in invalid state

    - by Abu Hamzah
    i am using asp.net web forms and i am using bassistance.de jquery validation.. i have bunch of fields in the form and its validatating how it suppose to but its still executing my code behind code, why is it doing that? here is my code: form: <script type="text/javascript"> $(document).ready(function() { $("#aspnetForm").validate({ rules: { <%=txtVisitName.UniqueID %>: { maxlength:1, //minlength: 12, required: true } ................. }, messages: { <%=txtVisitName.UniqueID %>: { required: "Enter visit name", minlength: jQuery.format("Enter at least {0} characters.") } ............ }, success: function(label) { label.html("&nbsp;").addClass("checked"); } }); $("#aspnetForm").validate(); }); </script> <div > <h1> Page</h1> <asp:Label runat="server" ID='Label1'>Visit Name:</asp:Label> <asp:TextBox ID="txtVisitName" runat='server'></asp:TextBox> ............ ............... <button id="btnSubmit" name="btnSubmit" type="submit"> Submit</button> </div> and i have a external .js file referencing to my web form page. $(document).ready(function() { $("#btnSubmit").click(function() { SavePage(); }); }); i have a WebMethod .cs and i have bookmark and it does hitting that line of code even thoug my page is in invalid state. how would i fix this? thanks.

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  • Unzip .zip file uploaded from Android via PHP to IIS 7

    - by HaOx
    I'm sending compressed file in .zip extension from Android via PHP to IIS server. Almost is working everything, but I cannot achieve unzip files with php. I've this code: <?php $target_path1 = "C:/Windows/Temp/"; $target_path1 = $target_path1 . basename( $_FILES['uploaded_file']['name']); /* I'm making the folder */ $directorio = substr($target_path1, 0, 32); if (!is_dir($directorio)) { mkdir($directorio); } /* I declare a path */ $barra = "/"; $target_path1 = $directorio . $barra . basename( $_FILES['uploaded_file']['name']); $target_path2 = $directorio . $barra; if(move_uploaded_file($_FILES['uploaded_file']['tmp_name'], $target_path1)) { /* Here I want to unzip the uploaded file */ } else{ echo "There was an error uploading the file, please try again!"; echo "filename: " . basename( $_FILES['uploaded_file']['name']); echo "target_path: " .$target_path1; } ?> So, how can I unzip uploaded file? I tried so many methods but no one worked. I'd be appreciated if someone could tell what I have to do to unzip the uploaded file. And if I have to configure some parameters in php.ini or IIS server. Thanks in advance.

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  • PHP static objects giving a fatal error

    - by Webbo
    I have the following PHP code; <?php component_customer_init(); component_customer_go(); function component_customer_init() { $customer = Customer::getInstance(); $customer->set(1); } function component_customer_go() { $customer = Customer::getInstance(); $customer->get(); } class Customer { public $id; static $class = false; static function getInstance() { if(self::$class == false) { self::$class = new Customer; } else { return self::$class; } } public function set($id) { $this->id = $id; } public function get() { print $this->id; } } ?> I get the following error; Fatal error: Call to a member function set() on a non-object in ....../classes/customer.php on line 9 Can anyone tell me why I get this error? I know this code might look strange, but it's based on a component system that I'm writing for a CMS. The aim is to be able to replace HTML tags in the template e.g.; <!-- component:customer-login --> with; <?php component_customer_login(); ?> I also need to call pre-render methods of the "Customer" class to validate forms before output is made etc. If anyone can think of a better way, please let me know but in the first instance, I'd like to know why I get the "Fatal error" mentioned above. Cheers

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  • larger file upload problem with php

    - by chris
    I need to upload a csv file to a server. works fine for smaller files but when the file is 3-6 meg its not working. $allowedExtensions = array("csv"); foreach ($_FILES as $file) { if ($file['tmp_name'] > '') { if (!in_array(end(explode(".", strtolower($file['name']))), $allowedExtensions)) { die($file['name'].' is an invalid file type!<br/>'. '<a href="javascript:history.go(-1);">'. '&lt;&lt Go Back</a>'); } if (move_uploaded_file($file['tmp_name'], $uploadfile)) { echo "File is valid, and was successfully uploaded.\n"; } else { echo "Possible file upload attack!\n"; } echo "File has been uploaded"; } //upload form <form name="upload" enctype="multipart/form-data" action="<? echo $_SERVER['php_self'];?>?action=upload_process" method="POST"> <!-- MAX_FILE_SIZE must precede the file input field --> <input type="hidden" name="MAX_FILE_SIZE" value="31457280" /> <!-- Name of input element determines name in $_FILES array --> Send this file: <input name="userfile" type="file" /> <input type="submit" value="Send File" /> </form> I have also added this to htaccess php_value upload_max_filesize 20M php_value post_max_size 20M php_value max_execution_time 200 php_value max_input_time 200 Where am i going wrong?

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  • How To Call Javascript In Ajax Response? IE: Close a form div upon success...

    - by B.Gordon
    I have a form that when you submit it, it sends the data for validation to another php script via ajax. Validation errors are echo'd back in a div in my form. A success message also is returned if validation passes. The problem is that the form is still displayed after submit and successful validation. I want to hid the div after success. So, I wrote this simple CSS method which works fine when called from the page the form is displayed on. The problem is that I cannot seem to call the hide script via returned code. I can return html like echo "<p>Thanks, your form passed validation and is being sent</p>"; So I assumed I could simply echo another line after that echo "window.onload=displayDiv()"; inside script tags (which I cannot get to display here)... and that it would hide the form div. It does not work. I am assuming that the problem is that the javascript is being returned incorrectly and not being interpreted by the browser... How can I invoke my 'hide' script on the page via returned data from my validation script? I can echo back text but the script call is ineffective. Thanks! This is the script on the page with the form... I can call it to show/hide with something like onclick="displayDiv()" while on the form but I don't want the user to invoke this... it has be called as the result of a successful validation when I write the results back to the div... function displayDiv() { var divstyle = new String(); divstyle = document.getElementById("myForm").style.display; if(divstyle.toLowerCase()=="block" || divstyle == "") { document.getElementById("myForm").style.display = "none"; } else { document.getElementById("myForm").style.display = "block"; } } PS: I am using the mootools.js library for the form validation if this matters for the syntax..

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  • I'm trying to make a php Curl call to formstack api, but I get nothing

    - by Oskar Calvo
    This is one of my first curls code, so it can have mistakes I'm trying to be able to make calls to form/:id/submissions https://www.formstack.com/developers/api/resources/submission#form/:id/submission_GET If I load: https://www.formstack.com/api/v2/form/1311091/submission.json?oauth_token=44eedc50d015b95164897f5e408670f0&min_time=2012-09-01%2000:01:01&max_time=2012-10-27%2000:01:01 If works great. If I try this code: <?php $host = 'https://www.formstack.com/api/v2/'; // TODO this should manage dinamics values or build an action in every method. $action = 'form/1311091/submission.json'; $url = $host . $action; // TODO this values will arrive like an array with values $postData['oauth_token']= '44eedc50d015b95164897f5e408670f0'; $postData['min_time'] ='2012-09-01 00:01:01'; $postData['max_time'] ='2012-10-27 00:01:01'; // TODO make a method with this action function getElements($postData) { $elements = array(); foreach ($postData as $name=>$value) { $elements[] = "{$name}=".urlencode($value); } } $curl = curl_init(); curl_setopt($curl, CURLOPT_URL, $url); curl_setopt($curl, CURLOPT_HTTPGET, true); curl_setopt($curl, CURLOPT_POSTFIELDS, $elements); $result = curl_exec($curl) ; curl_close($curl); var_dump($result); ?>

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  • php, image dosn't show

    - by lolalola
    Hi, whats wrong with my code. image dosn't show in test2.php file File: test2.php: <img src = "test.php" /> File: test.php session_start(); $md5_hash = md5(rand(0,999)); $security_code = substr($md5_hash, 15, 5); $_SESSION["security_code"] = $security_code; $width = 100; $height = 20; header("Content-type: image/png"); $image = ImageCreate($width, $height); $white = ImageColorAllocate($image, 255, 255, 255); $black = ImageColorAllocate($image, 0, 0, 0); $grey = ImageColorAllocate($image, 204, 204, 204); ImageFill($image, 0, 0, $black); //Add randomly generated string in white to the image ImageString($image, 3, 30, 3, $security_code, $white); //Throw in some lines to make it a little bit harder for any bots to break imageRectangle($image,0,0,$width-1,$height-1,$grey); imageline($image, 0, $height/2, $width, $height/2, $grey); imageline($image, $width/2, 0, $width/2, $height, $grey); imagepng($image); imagedestroy($image);

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  • PHP IIS problems downloading file says it is corrupt

    - by Matt
    Hi, I am running PHP on IIS 6 with mssql. I have uploaded a file to my webserver through a php script. Upon checking the file on the server the file is ok and not corrupt. However, when i then have a link on my website to try and download the file, it says the file is corrupt. I know the file isnt corrupt as i can view it perfectly if i look at the file on the server. Is seems like this is a common problem as a similar problem was posted here: http://www.bigresource.com/Tracker/Track-php-1pAakBhT/ Any help would be much appreciated. Thanks, M My download code is as follows: $filesize = $rows->filesize; $filepath = $rows->filepath; header("Content-Disposition: attachment; filename=$filename"); header("Content-length: $filesize"); header("Content-type: application/pdf"); header("Cache-control: must-revalidate"); header("Content-Description: PHP Generated Data"); readfile($filepath);

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  • Image generated with PHP dosn't show

    - by lolalola
    Hi, whats wrong with my code. image dosn't show in test2.php file File: test2.php: <img src = "test.php" /> File: test.php session_start(); $md5_hash = md5(rand(0,999)); $security_code = substr($md5_hash, 15, 5); $_SESSION["security_code"] = $security_code; $width = 100; $height = 20; header("Content-type: image/png"); $image = ImageCreate($width, $height); $white = ImageColorAllocate($image, 255, 255, 255); $black = ImageColorAllocate($image, 0, 0, 0); $grey = ImageColorAllocate($image, 204, 204, 204); ImageFill($image, 0, 0, $black); //Add randomly generated string in white to the image ImageString($image, 3, 30, 3, $security_code, $white); //Throw in some lines to make it a little bit harder for any bots to break imageRectangle($image,0,0,$width-1,$height-1,$grey); imageline($image, 0, $height/2, $width, $height/2, $grey); imageline($image, $width/2, 0, $width/2, $height, $grey); imagepng($image); imagedestroy($image);

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  • php script randomly hangs up

    - by sergdev
    I install php 5 (more precisely 5.3.1) as apache module. After this one of my application becomes randomly hang up on mysql_connect - sometimes works, sometimes no, sometimes reload of page helps. How can this be fixed? I use Windows Vista, Apache/2.2.14 (Win32) PHP/5.3.1 with php module, MySql 5.0.67-community-nt. After a minute I obtain the error message: Fatal error: Maximum execution time of 60 seconds exceeded in path\to\mysqlcon.php on line 9 I run MySql locally and heavy load could not be a reason: SHOW PROCESSLIST shows about 3 process SHOW VARIABLES LIKE 'MAX_CONNECTIONS' is 100. UPDATE: At first I thought that this is connected with mysql_connect. But now I can't say for certain. More difficult thing is when I insert the line to debug: $fh = fopen("E://_debugLog", 'a'); fwrite($fh, __FILE__ . " : " . __LINE__ . "\n"); fclose($fh); script start working near that location as a rule.

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  • Adding with PHP to a MySQL database

    - by shinjuo
    I am pretty new to PHP and I am trying to make an inventory database. I have been trying to make it so that a user can enter a card ID and then amount the want to add to the inventory and have it update the inventory. For example someone could type in test and 2342 and it would update test. Here is what I have been trying with no success: add.html <body> <form action="add.php" method="post"> Card ID: <input type="text" name="CardID" /> Amount to Add: <input type="text" name="Add" /> <input type="submit" /> </form> </body> </html> add.php <?php $link = mysql_connect('tulsadir.ipowermysql.com', 'cbouwkamp', '!starman1'); if (!$link){ die('Could not connect: ' . mysql_error()); } mysql_select_db("tdm_inventory", $link); $add = $_POST[Add] mysql_query("UPDATE cardLists SET AmountLeft = '$add' WHERE cardID = 'Test'"); echo "test successful"; mysql_close($link); ?>

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  • Any PHP MVC framework planning to use 5.3 features?

    - by alexandrul
    I would like to get started with PHP, and 5.3 release seems to bring many nice features (namespaces, lambda functions, and many others). I have found some MVC frameworks, and some of them support only PHP 5: PHP Frameworks PHP MVC Frameworks Model–view–controller on Wikipedia but can anyone recommend one of those MVC frameworks that plans to actively use PHP 5.3 features, not just being compatible with PHP 5.3? Update Results so far: Zend Framework 2.0 (in development) Lithium (in development, based on CakePHP) Symfony (in development) FLOW3 (in development, alpha)

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  • how to close a windows form in c# when that form transfer control to an other form?

    - by moon
    i have encountered a problem in my application i have two windows forms one that is loaded when my application is started and ask for a password and other is shown when click log in with correct password , problem is how can i close the first form that is log in form when user proceed to next from that is actual application, currently i hide the first form but the requirement is to close that form to prevent extra processing when i close that log in form my application is closed and when i close the actual application form that log in form remain open. my application is not fully closed, log in form is running in background how can in fix this....

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  • Post valuse and upload Image to php server in android

    - by lawat
    I am trying to upload image from android phone to php server with additional values,the method is post my php file look like this if($_POST['val1']){ if($_POST['val2']){ if($_FILE['image']){ ...... } } }else{ echo "Value not found"; } I am doing is URL url=new URL("http://www/......../myfile.php"); HttpURLConnection con=(HttpURLConnection) url.openConnection(); con.setDoInput(true); con.setDoOutput(true); con.setUseCaches(false); con.setRequestMethod("POST");//Enable http POST con.setRequestProperty("Connection", "Keep-Alive"); con.setRequestProperty("Content-Type", "multipart/form-data;boundary="+"****"); connection.setRequestProperty("uploaded_file", imagefilePath); DataOutputStream ostream = new DataOutputStream( con.getOutputStream()); String res=("Content-Disposition: form-data; name=\"val1\""+val1+"****"+ "Content-Disposition: form-data; name=\"val2\""+val2+"****" "Content-Disposition: form-data; name=\"image\";filename=\"" + imagefilePath +"\""+"****"); outputStream.writeBytes(res); my actual problem is values are not posting so first if condition get false and else section is executed that is it give value not found please help me

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  • PHP - How can I check if return() was called from an include()'d file?

    - by John Himmelman
    How can I tell if return() was called from within the included file. The problem is that include() returns 'int 1', even if return() wasn't called. Here is an example... included_file_1.php <?php return 1; included_file_2.php <?php echo 'no return here, meep'; main.php <?php $ret = include('included_file_1.php'); // This file DID return a value, int 1, but include() returns this value even if return() wasn't called in the included file. if ($ret === 1) { echo 'file did not return anything'; } var_dump($ret); $ret = include('included_file_2.php'); // The included file DID NOT return a value, but include() returns 'int 1' if ($ret === 1) { echo 'file did not return anything'; } var_dump($ret);

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  • PHP class to C# class?

    - by LordSauron
    I work for a company that makes application's in C#. recently we got a customer asking us to look in to rebuilding an application written in PHP. This application receives GPS data from car mounted boxes and processes that into workable information. The manufacturer for the GPS device has a PHP class that parses the received information and extracts coordinates. We were looking in to rewriting the PHP class to a C# class so we can use it and adapt it. And here it comes, on the manufacturers website there is a singel line of text that got my skin krawling: "The encoding format and contents of the transmitted data are subject to constant changes. This is caused by implementations of additional features by new module firmware versions which makes it virtually impossible to document it and for you to properly decode it yourself." So i am now looking for a option to use the "constantly changing" PHP class and access it in C#. Some thing link a shell only exposing some function's i need. Except i have no idea how i can do this. Can any one help me find a solution for this.

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  • PHP Foreach statement issue. Multiple rows are returned

    - by Daniel Patilea
    I'm a PHP beginner and lately i've been having a problem with my source code. Here it is: <html> <head> <title> Bot </title> <link type="text/css" rel="stylesheet" href="main.css" /> </head> <body> <form action="bot.php "method="post"> <lable>You:<input type="text" name="intrebare"></lable> <input type="submit" name="introdu" value="Send"> </form> </body> </html> <?php //error_reporting(E_ALL & ~E_NOTICE); mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db("robo") or die(mysql_error()); $intrebare=$_POST['intrebare']; $query = "SELECT * FROM dialog where intrebare like '%$intrebare%'"; $result = mysql_query($query) or die(mysql_error()); $row = mysql_fetch_array($result) or die(mysql_error()); ?> <div id="history"> <?php foreach($row as $rows){ echo "<b>The robot says: </b><br />"; echo $row['raspuns']; } ?> </div> It returns me the result x6 times. This problem appeared when I've made that foreach because I wanted the results to stuck on the page one by one after every sql querry. Can you please tell me what seems to be the problem? Thanks!

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  • JQuery-AJAX: No further request after timeout and delay in form post

    - by Nogga
    I got a form containing multiple checkboxes. This form shall be sent to the server to receive appropriate results from a server side script. This is already working. What I would achieve now: 1) Implementing a timeout: This is already working, but as soon as a timeout occurs, a new request is not working anymore. 2) Implementing a delay in requesting results: A delay shall be implemented so that not every checkbox is resulting in a POST request. This is what I have right now: function update_listing() { // remove postings from table $('.tbl tbody').children('tr').remove(); // get the results through AJAX var request = $.ajax({ type: "POST", url: "http://localhost/hr/index.php/listing/ajax_csv", data: $("#listing_form").serialize(), timeout: 5000, success: function(data) { $(".tbl tbody").append(data); }, error: function(objAJAXRequest, strError) { $(".tbl tbody").append("<tr><td>failed " + strError + "</td></tr>"); } }); return true; } Results are for now passed as HTML table rows - I will transform them to CSV/JSON in the next step. Thanks so much for your advice.

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  • Write to PDF using PHP from android device

    - by Brent Mitchell
    I am trying to write to a pdf file on php server. I have sent variables to the server, create the pdf document, then have my phone download the document to view on device. The variables seem not to write on the php file. I have my code below public void postData() { try { Calculate calc = new Calculate(); HttpClient mClient = new DefaultHttpClient(); StringBuilder sb=new StringBuilder("myurl.com/pdf.php"); HttpPost mpost = new HttpPost(sb.toString()); String value = "1234"; List nameValuepairs = new ArrayList(1); nameValuepairs.add(new BasicNameValuePair("id",value)); mpost.setEntity(new UrlEncodedFormEntity(nameValuepairs)); } catch (UnsupportedEncodingException e) { Log.w(" error ", e.toString()); } catch (Exception e) { Log.w(" error ", e.toString()); } } And my php code to write the variable "value" onto the pdf document: //code to reverse the string if($_POST[] != null) { $reversed = strrev($_POST["value"]); $this->SetFont('Arial','u',50); $this->Text(52,68,$reversed); } I am just trying to write the variable in a random spot, but the variable the if statement is always null and I do not know why. Thanks. Sorry if it is a little sloppy.

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  • how to pass session_id() throught out the php pages?

    - by Piyush
    when user clicks on login button(index.php) I am calling chechlogin.php where I am checking loginId an password as- if($count==1) { // Register $myusername, $mypassword and redirect to file "login_success.php" session_register("myusername"); session_register("mypassword"); $_SESSION['UserId'] = $myusername; $_session['SessionId'] = session_id(); header("location:LoggedUser.php"); } in LiggedUser.php <?php session_start(); //starting session if (!isset($_SESSION['SessionId']) || $_SESSION['SessionId'] == '') { header("location:index.php"); } ? Problem: It is always going back to index.php page although I am entering right userid and password.I think session_id() is not working properly or ??

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  • Collect all fields in form to an array

    - by Industrial
    Hi everyone, Each div with the class "row" is added upon request from the user, to be able to add multiple items at once. So now is the question how I'll collect all the forms in to an array that PHP can read (like JSON for instance). I'll guess that there's already some easy and effective way of doing this? <div class="container"> <div class="row"> <input type="text" name="value1" id="textfield" /> <input type="text" name="value2" id="textfield" /> <input type="text" name="value3" id="textfield" /> </div> </div> Here's what I would like to achieve out of the shown example: array( array ('value1' => '', 'value2' => '', 'value3' => '') ); Thanks! Update: The form will be handled with PHP and it would be super to be able to do something like a foreach loop on the specific container-div content.

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  • Upload to a PHP Server, using Ajax ( XMLHttp POST)

    - by Krishnanunni
    Right now i'm using the below method to Upload a file to PHP <form enctype="multipart/form-data" action="http://sserver.com/fileupload.php" method="POST"> <input type="hidden" name="MAX_FILE_SIZE" value="30000000" /> <input type="hidden" name="filename" value="file_uploaded.gif" /> <input type="hidden" name="username" value="foobar"/> Please choose a file: <input name="uploaded" type="file" /><br /> <input type="submit" value="Upload" /> </form> I read the $_POST and $_FILE in php to complete upload like this. $target = $_SERVER['DOCUMENT_ROOT']."/test/upload/"; $target = $target . basename( $_FILES['uploaded']['name']) ; echo $target; $ok=1; if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target)) { echo "The file ". basename( $_FILES['uploaded']['name']). " has been uploaded"; } else { echo "Sorry, there was a problem uploading your file."; } My questions is , can i change the above said code (HTML) to an Ajax XMLHttpRequest without changes in PHP.

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  • PHP classes, parse syntax errors when using 'var' to declare variables

    - by jon
    I am a C# guy trying to translate some of my OOP understanding over to php. I'm trying to make my first class object, and are hitting a few hitches. Here is the beginning of the class: <?php require("Database/UserDB.php"); class User { private var $uid; private var $username; private var $password; private var $realname; private var $email; private var $address; private var $phone; private var $projectArray; public function _construct($username) { $userArray = UserDB::GetUserArray($username); $uid = $userArray['uid']; $username = $userArray['username']; $realname = $userArray['realname']; $email = $userArray['email']; $phone = $userArray['phone']; $i = 1; $projectArray = UserDB::GetUserProjects($this->GetID()); while($projectArray[$i] != null) { $projectArray[$i] = new Project($projectArray[$i]); } UserDB.php is where I have all my static functions interacting with the Database for this User Class. I am getting errors using when I use var, and I'm getting confused. I know I don't HAVE to use var, or declare the variables at all, but I feel it is a better practice to do so. the error is "unexpected T_VAR, expecting T_VARIABLE" When I simply remove var from the declarations it works. Why is this?

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  • PHP & form validation

    - by user887515
    I have a form that I'm submitting via ajax, and I want to return a message of the list of fields that were empty. I've got that all done and dusted, but it just seems really long winded on the PHP side of things. How can I do the below in a less convoluted way? <?php if(empty($_POST["emailaddress"])){ $error = 'true'; $validation_msg = 'Country missing.'; if(empty($error_msg)){ $error_msg .= $validation_msg; } else{ $error_msg .= '\n' . $validation_msg; } } if(empty($_POST["password"])){ $error = 'true'; $validation_msg = 'Country missing.'; if(empty($error_msg)){ $error_msg .= $validation_msg; } else{ $error_msg .= '\n' . $validation_msg; } } if(empty($_POST["firstname"])){ $error = 'true'; $validation_msg = 'First name missing.'; if(empty($error_msg)){ $error_msg .= $validation_msg; } else{ $error_msg .= '\n' . $validation_msg; } } if(empty($_POST["lastname"])){ $error = 'true'; $validation_msg = 'Last name missing.'; if(empty($error_msg)){ $error_msg .= $validation_msg; } else{ $error_msg .= '\n' . $validation_msg; } } if($error){ header('HTTP/1.1 500 Internal Server Error'); header('Content-Type: application/json'); die($error_msg); } ?>

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  • Increasing understanding of validating a string with PHP string functions

    - by user1554264
    I've just started attempts to validate data in PHP and I'm trying to understand this concept better. I was expecting the string passed as an argument to the $data parameter for the test_input() function to be formatted by the following PHP functions. trim() to remove white space from the end of the string stripslashes() to return a string with backslashes stripped off htmlspecialchars() to convert special characters to HTML entities The issue is that the string that I am echoing at the end of the function is not being formatted in the way I desire at all. In fact it looks exactly the same when I run this code on my server - no white space removed, the backslash is not stripped and no special characters converted to HTML entities. My question is have I gone about this in the wrong approach? Should I be creating the variable called $santised_input on 3 separate lines with each of the functions trim(), stripslashes() and htmlspecialchars()? By my understanding surely I am overwriting the value of the $santised_input variable each time I recreate it on a new line of code. Therefore the trim() and stripslashes() string functions will never be executed. What I am trying to achieve is using the "$santised_input" variable to run all of these PHP string functions when the $data argument is passed to my test_input() function. In other words can these string functions be chained together so that I only need to create $santised_input once? <?php function test_input($data) { $santised_input = trim($data); $santised_input = stripslashes($data); $santised_input = htmlspecialchars($data); echo $santised_input; } test_input("%22%3E%3Cscript%3Ealert('hacked')%3C/script%3E\ "); //Does not output desired result "&quot;&gt;&lt;script&gt;alert('hacked')&lt;/script&gt;" ?>

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