PHP script not receiving posted JSON

Question

I'm having trouble with the following JavaScript, where data is a JavaScript object:

var ajax_send = function(data) {
        var httpRequest;

        makeRequest('/prototype/test.php', data);

        function makeRequest(url, data) {
            if (window.XMLHttpRequest) {
                httpRequest = new XMLHttpRequest();
            } else if (window.ActiveXObject) {
                try {
                    httpRequest = new ActiveXObject("Msxml2.XMLHTTP");
                } 
                catch (e) {
                    try {
                        httpRequest = new ActiveXObject("Microsoft.XMLHTTP");
                    } 
                    catch (e) {}
                }
            }

            if (!httpRequest) {
                alert('Giving up :( Cannot create an XMLHTTP instance');
                return false;
            }

            httpRequest.onreadystatechange = alertContents;
            httpRequest.open('POST', url);
            httpRequest.setRequestHeader('Content-Type', 'application/json');
            httpRequest.send(JSON.stringify(data));
        }

        function alertContents() {
            if (httpRequest.readyState === 4) {
                if (httpRequest.status === 200) {
                    alert(httpRequest.responseText);
                } else {
                    alert('There was a problem with the request.');
                }
            }
        }
    };

The PHP is:

$data = $_POST['data'];
$obj = json_decode($data);
echo $obj;

In dev tools the request payload looks ok but it doesn't seem to be what the PHP is looking for. Nothing gets through to the PHP script and the response is empty.

What am I doing wrong?

It is impossible to tell what the PHP is looking for since you didn't show us the PHP code. — Quentin
– Quentin, Commented Nov 11, 2014 at 14:01
Not directly related, but using jquery to do ajax calls is quite a lot nicer imo. — DanielST
– DanielST, Commented Nov 11, 2014 at 14:24
Damn, I forgot the "no use jQuery replies"! You're right of course, it is much easier and nicer. This is a learning exercise for me as much as anything else and I like to understand the vanilla. — Derek Johnson
– Derek Johnson, Commented Nov 11, 2014 at 14:27

casraf · Accepted Answer · 2014-11-11 14:14:50Z

3

When you send POST requests with Content-Type: application/json, things work a bit differently with PHP.

You are going to need to access it like so:

$postData = json_decode(file_get_contents('php://input'));

Instead of the usual $_POST.

If you want to send it as a regular form that will be populated to $_POST, you need to set your header like so:

httpRequest.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');

And populate your fields like so:

key=value&key2=value2&key3=value3

etc

edited Nov 11, 2014 at 14:14

answered Nov 11, 2014 at 14:08

casraf

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8 Comments

casraf Over a year ago

I updated the answer, try setting the request header differently. You'll see how the data looks then. Also, what does your error say exactly?

Derek Johnson Over a year ago

Yep, that works, thank you very much :) I would still be interested in knowing how to post JSON without converting to form values though.

casraf Over a year ago

If you could tell me more in detail about your error, I can help you figure out why it's not working. Did you try look at the contents of file_get_contents('php://input') before trying to parse it?

Derek Johnson Over a year ago

Hmm. Not much details on the error, just 500 internal server error. The contents of file_get_content('php://input') is valid JSON though.

Derek Johnson Over a year ago

Ok, things go wrong when I use the json_decode() function. Is this likely to be a localhost setup problem?

|

fabpico · Accepted Answer · 2022-05-13 21:55:23Z

0

On Symfony you can use $request->getContent(); which internally makes file_get_contents('php://input')). Then decode the result.

$requestData = $request->getContent(); // this makes file_get_contents('php://input')
$data = json_decode($requestData, true);

answered May 13, 2022 at 21:55

fabpico

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