Empty Response from a webhook using php - php

Question

When a new record is created or updated, the Webhook is triggered and it posts three keys in the post body. One of the keys named payload contains encoded JSON with attributes of the record.
This is an example of the POST body message:See the Payload here
I want when the webhook fires it creates a file and posts the date to the file but I'm geting empty payload.
Here is the code.
<?php
/**
* Created by PhpStorm.
* User: Lee N
* Date: 16/07/2018
* Time: 14:46
*/
$data = file_get_contents('php://input');
//decode JSON data to PHP array
$content = json_decode($data, true);
if($content ==""){
$data = "Payload fired but no datkhkjhkjhjka";
}else{
$data = $content;
}
$pagename = 'from_vend';
$newFileName = $pagename.".txt";
if (file_put_contents($newFileName, $data) !== false) {
echo "File created (" . basename($newFileName) . ")";
} else {
echo "Cannot create file (" . basename($newFileName) . ")";
}

Answer 1

Your POST request is x-www-form-urlencoded. Quoting Wikipedia (emphasis mine):
When a web browser sends a POST request from a web form element, the default Internet media type is "application/x-www-form-urlencoded". This is a format for encoding key-value pairs with possibly duplicate keys. Each key-value pair is separated by an '&' character, and each key is separated from its value by an '=' character. Keys and values are both escaped by replacing spaces with the '+' character and then using URL encoding on all other non-alphanumeric[9] characters.
When you send a POST request like this to PHP, it will automatically populate $_POST with the key/value pairs. So in order to access the JSON payload, use
$decodedJson = json_decode($POST['payload']);
Whereas php://input is (quoting the PHP Manual):
a read-only stream that allows you to read raw data from the request body.
Since this is raw data (as in full post body), you would need to run it through parse_str first in order to decode the JSON payload (which is what PHP does automatically for you in $_POST).
If you want to write the full post body, you can just remove the json_decode. Because the full post body is not JSON and hence, will not decode as JSON.

Answer 2

Try using
parse_str($str, $output);
instead of
json_decode($data, true);
the post request is x-www-form-urlencoded. So json_decode will not work here as it is not a json string. If we check the payload we can see that it is a query string and parse_str will parse it into variables.
so the code will be
<?php
/**
* Created by PhpStorm.
* User: Lee N
* Date: 16/07/2018
* Time: 14:46
*/
$data = file_get_contents('php://input');
//decode JSON data to PHP array
//$content = json_decode($data, true);
parse_str($data, $content );
if($content ==""){
$data = "Payload fired but no datkhkjhkjhjka";
}else{
$data = $content;
}
$pagename = 'from_vend';
$newFileName = $pagename.".txt";
if (file_put_contents($newFileName, $data) !== false) {
echo "File created (" . basename($newFileName) . ")";
} else {
echo "Cannot create file (" . basename($newFileName) . ")";
}