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POhow to get http response code
 

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    <p>I have a form which sends the values to the variables $_POST['person_id'] , $_POST['accident_loc'], $_POST['message'] and $_POST['name']. </p> <p>The php file inserts the values as a record in the database, i dont have any problem with that. But i want to return an error when the form doesn't send any values to the variables. </p> <p>I mean when i submit a form without any values. so i made an if statement <code>if(isset($_POST['person_id']) &amp;&amp; isset($_POST['accident_loc']) &amp;&amp; isset($_POST['message']) &amp;&amp; isset($_POST['name'])){}</code> if this statement fails call the function "sendResponse()". </p> <p>now i want this file to return the status code 400 to my javascript file so that i can write some error conditions. But when i tried to display the value of the status code <code>request.status</code> it gave me 0 as its values </p> <p>Below is my php file.. can you tell me how to send the status code ??</p> <pre><code> &lt;?php //allowing access to the server which contains the following host-origin if($_SERVER[HOST_ORIGIN]=="http:\\localhost") header('Access-Control-Allow-Origin:http:\\http:localhost'); function sendResponse(){ header('HTTP/1.1 400 invalid request'); header('Content-type: text/html'); echo "invalid request"; } if(isset($_POST['person_id']) &amp;&amp; isset($_POST['accident_loc']) &amp;&amp; isset($_POST['message']) &amp;&amp; isset($_POST['name'])) { //php mysql database info require ("phpMysql_dbInfo.php"); //getting the arguments from the url $person_id=$_POST['person_id']; $accident_loc=$_POST['accident_loc']; $message=$_POST['message']; $name=$_POST['name']; //connecting to the database $connection= mysql_connect($local_host,$username,$password); if(!$connection) die(mysql_error()."&lt;/br&gt;"); //selecting the db $select_db= mysql_select_db($database_name); if(!$select_db) die(mysql_error()."&lt;/br&gt;"); //inserting the values into the database $query= sprintf("insert into messages_to_people(name,accident_location,message,person_id) values('%s','%s','%s','%s')", mysql_real_escape_string($name), mysql_real_escape_string($accident_loc), mysql_real_escape_string($message), mysql_real_escape_string($person_id)); $result=mysql_query($query); if(!$result) echo mysql_error().'&lt;/br&gt;'; } else sendResponse(); ?&gt; </code></pre> <p><strong>UPDATE:</strong> here is my javascript </p> <pre><code>function downloadXml (url,params,callback) { var request= window.XMLHttpRequest ? new XMLHttpRequest(): new ActiveXObject('Microsoft.XMLHTTP'); request.onreadystatechange= function(){ if(request.readyState==4){ //gives me a zero when i try to display it through alert alert(request.status); callback(request, request.status); } }; request.open('POST',url,true); request.setRequestHeader("Content-type", "application/x-www-form-urlencoded"); request.send(params); } </code></pre> <p>if you find any problem in my javascript pls correct me</p>
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    1. COLook into how to enable error / warning reporting in PHP, then inspect the PHP's response with Fiddler or Firebug. I stongly suspect it will give you a warning that headers can't be altered after output has been written - you are echoing your MySQL error (which in itself might be a bad idea) before setting the response header.
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    2. COWhy do you always execute `sendResponse()`? Don't you want an `else` there, so that it's only called when the `if` condition fails?
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    3. COya you are right i am sorry while pasting it i did a mistake will correct it now
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