PHP POST variables are empty

Question

My problem is that I am trying to push one parameter via POST/GET, this parameter is passed to js function in coode bellow, all this files are in same directory.

Thy for help, answers. Best regards.

<div id="sidebar"> <?php include('showContent.js'); ?>
 <ul>
   <li>
   <h2>TITLE</h2>
      <ul>
    <li><a onclick="showContent('1');">Link1</a></li>
    <li><a onclick="showContent('2');">Link2</a></li>
    <li><a onclick="showContent('3');">Link3</a></li>
      </ul>
    </li>
  </ul>
</div>

showContent.js

<script>function showContent(cId)
 {
 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)
     {
     document.getElementById("contentArea").innerHTML=xmlhttp.responseText;
     }
   }
 xmlhttp.open("POST","contact/sendContent.php?cId="+cId,true);
 xmlhttp.send();
 }

sendContent.php

<?php
    $cId=$_POST["cId"];
    $tmp='error.php';

    switch ($cId) {
        case 1:{
            $tmp='contact.php';
            break;
        }
        case 2:{
            $tmp='idCard.php';
            break;
        }
        case 3:{
            $tmp='location.php';
            break;
        }

    }

    ob_start();
    include($tmp);
    echo ob_get_clean();

?>

P.S.: You should add hint on text editor's buttons how to use them, I spent a lot of time to figure out how to use this sucking editor for code formating.

Hint: select code pres this button!

Very easy when you know, very annoying if something do not behave like it should!

How are you expecting a $_POST to contain anything if you do not send any POSTed data with your request? — CodeZombie
– CodeZombie, Commented Feb 17, 2013 at 21:41
You are sending request to contact/sendContent.php?q="+contentId whereas you are accessing CId on other page? It should be $_POST['q'] — Ghazanfar Mir
– Ghazanfar Mir, Commented Feb 17, 2013 at 21:41
plus the previous comment, you are missing the id='contentArea' somewhere else — Mariano Montañez Ureta
– Mariano Montañez Ureta, Commented Feb 17, 2013 at 21:46

scartag · Accepted Answer · 2013-02-17 21:41:44Z

1

Although you are using a POST request, you aren't really posting anything.

Since you have a query string appended you could access it.

$cId=$_REQUEST["q"];

Or

$cId=$_GET["q"];

answered Feb 17, 2013 at 21:41

scartag

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1 Comment

ojc Over a year ago

When I pasted code in this forum I was rechecking several different thins (this is for incompatible* names) sorry for bad eng.

Kevin Garman · Accepted Answer · 2013-02-17 22:06:25Z

You've got several problems:
-as mentioned you need $_GET or $_REQUEST
-close your script tags in the JS file
-I'm not familiar with ob_start() but echo works great
-if the files are all in the same folder, then your AJAX request path is wrong

Here are working files...

<html>
<body>
<div id="sidebar"> <?php include('showContent.js'); ?>
 <ul>
   <li>
   <h2>TITLE</h2>
      <ul>
    <li><a onclick="showContent('1');">Link1</a></li>
    <li><a onclick="showContent('2');">Link2</a></li>
    <li><a onclick="showContent('3');">Link3</a></li>
      </ul>
    </li>
  </ul>
</div>
<div id="contentArea">content area</div>
</body>
</html>



<script type="text/javascript">function showContent(cId)
 {
 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)
     {
     document.getElementById("contentArea").innerHTML=xmlhttp.responseText;
     }
   }
 xmlhttp.open("POST","sendContent.php?cId="+cId,true);
 xmlhttp.send();
 }
</script>


<?php
    $cId=$_REQUEST["cId"];
    $tmp='error.php';

    switch ($cId) {
        case 1:
            $tmp='contact.php';
            break;

        case 2:
            $tmp='idCard.php';
            break;

        case 3:
            $tmp='location.php';
            break;
    }

    echo $tmp;
?>

script is closed (bad copy paste). it works now :) Thx anyway.

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