How do I use received JSON with PHP [closed] - php

Question

Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
Questions asking for code must demonstrate a minimal understanding of the problem being solved. Include attempted solutions, why they didn't work, and the expected results. See also: Stack Overflow question checklist
Closed 9 years ago.
Improve this question
I'm reading an integration API and it says it will send requests in this format:
content-type: application/json
accept: application/json
x-gsc-api-key: lRDs8NfsVTmHD8SC1234
{
"name" : "testuser",
"password" : "098f6bcd4621d373cade4e832627b4f6",
"reference" : "b86de61-af20-4c47-af9a-6f2edeebc4fe"
}
I want to server the requests with PHP. I know some other posts also ask something along the same lines, but they were sending the request via AJAX method. I don't know which method the request is going to be sent.
Can I just reference the variables $_POST['name'] and $_POST['password'] to get the values? Or do I need to call json_decode($_POST)?
Is this information enough the write a PHP script that server the request?
Any help would be appreciated.

Answer 1

No, you cannot use $_POST directly. You will have to manually extract the response body, json_decode it and work on the results. But it's easy to do:
$data = json_decode(file_get_contents('php://input'));
echo $data->name;

Answer 2

$json = file_get_contents('php://stdin');

Answer 3

You have to use the json_decode() function for this (http://php.net/manual/en/function.json-decode.php)

Answer 4

No - you can't - however you can convert the JSON to $_POST like so:
<?php
$_POST = array_merge($_POST, json_decode(file_get_contents("php://stdin"), true));
?>

Answer 5

Not sure if i read the question correctly but >.>
$json = '{"name":"testuser","password":"098f6bcd4621d373cade4e832627b4f6","reference":"b86de61-af20-4c47-af9a-6f2edeebc4fe"}';
$data = #json_decode($json);
$name = $data->name;
$password = $data->password;
$ref = $data->reference;
echo "${name} ${password} ${ref}";
or use the true like below. . .
$json = '{"name":"testuser","password":"098f6bcd4621d373cade4e832627b4f6","reference":"b86de61-af20-4c47-af9a-6f2edeebc4fe"}';
$data = #json_decode($json, true);
$name = $data['name'];
$password = $data['password'];
$ref = $data['reference'];
echo "${name} ${password} ${ref}";