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Question

How to receive XMLHttpRequest with PHP?

Answer 1

+1 A:

fopen with input:// wrapper or $HTTP_RAW_POST_DATA

Col. Shrapnel 2010-05-24 05:35:36

What the Col is referring to is here: http://us2.php.net/manual/en/ini.core.php#ini.always-populate-raw-post-data

Joseph 2010-05-24 19:26:16

Answer 2

+3 A:

You will read it exactly the sme way you read normal request vars.

$_GET['varname'] and $_POST['varname']

chris12892 2010-05-24 05:40:09

Unless I'm doing something wrong, I get nothing but an empty array when I print_r $_POST. In my ajax call, my parameter should only be the xml string in the format '<xml>...</xml>, right? Thank you for your input.

Adrian 2010-05-24 12:05:31

You might try to `print_r($_GET)`, just to see if somehow it's getting send via GET.

chris12892 2010-05-24 16:44:07

Answer 3

A:

When you use "method: post" and you want to send a post body, you need the parameter postBody. So something like this might work:

var myAjax = new Ajax.Request('http://localhost:8080/post2.php',{
    contentType: 'text/xml',
    postBody: xml,
    method: 'post',
    onSuccess: function(transport){
        alert(transport.status); alert(transport.responseText);
    }
});

But why did you build a XML doc around your content? You can simply sent the message "heya there" with the postBody without the XML arround.

Edit: Here you'll find all the Ajax.Request options: http://www.prototypejs.org/api/ajax/options

Kau-Boy 2010-05-30 21:01:41

Answer 4

A:

php://input allows you to read raw POST data like

Welcome <?php print(file_get_contents('php://input')); ?>!<br />

Note: php://input is not available with enctype="multipart/form-data".

Ramesh Tabarna 2010-05-31 11:09:05

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How to receive XMLHttpRequest with PHP?

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