PHP使用（$ _GET ['url']）然后显示json数据？

<?php

function get_stuff()
{
    ($_GET['http://****RemovedForSecurityPurposes****.repairshopr.com/api/v1/tickets?api_key=****RemovedForSecurityPurposes****']);
    echo $_GET;
}

get_stuff();

?> /* Yes, Pointless closing tag, I'm cool like that. */

For some reason when I run this code the output I'm getting is "Array" and I can't figure out why? I know it is an array that I'm getting from the URL but I thought it would just print in whatever format its in? Am I missing something?

Thanks In advance!

The Message of OP when I run this code the output I'm getting is "Array".

You need to use file_get_contents to read a file from a remote server. If your file response an array then you have to use print_r or var_export or var_dump. Or if your file response is a string(json) then you need to store it in a variable and apply a decode method.

function get_stuff(){
    $response = file_get_contents('http://****RemovedForSecurityPurposes****.repairshopr.com/api/v1/tickets?api_key=****RemovedForSecurityPurposes****');
    print_r($response); // if array
    $arr = json_decode($response); // decode of json
    print_r($arr);
}

get_stuff();

I think you will understand what i mean. Let me know if you are useful or need some help.

You cannot pass website URL to $_GET.

You should use file_get_contents()

$content = file_get_contents('http://YOUR_URL');

If you have valid json then you can convert it to an array.

$data = json_decode($content, true);

then print it,

echo '<pre>'; print_r($data);