php如何响应ajax请求

var container = document.getElementsByClassName('container')[0];
container.onclick = function()
{
    var xhr = new XMLHttpRequest();
    xhr.open('GET','/learn/php_ajax/responseExample.php',true);
    xhr.send();
 
    xhr.onload = function(ev)
    {
        var data = JSON.parse(ev.currentTarget.responseText);
        console.log(data);
    };
}

$array = array("a" => "视频", "b"=>"教2程", "c" => "六");
exit(json_encode($array));

{
    a: "视频",
    b: "教2程",
    c: "六"
}