将php obj返回到javascript回调

Question

How can I get the print_r() output in the done() so I can check on data.email and data.url

jQuery

    $.post("php/checkinput.php", {
        email: contactEmail.val(),
        url: contactWebsite.val()
    }).done(function(data){
        console.log(data)
        doCheck(data)
    });

PHP

$email = $_POST["email"];
$url = $_POST["url"];
$obj = new stdClass();
$arr = array();
if (!filter_var($email, FILTER_VALIDATE_EMAIL)) {
    $obj->email = "invalid";
} else {
    $obj->email = "valid";
}
//array_push($obj, $arr)
if (!filter_var($url, FILTER_VALIDATE_URL)) {
    $obj->url = "invalid";
} else {
    $obj->url = "valid";
}
print_r($obj);
//Object
//(
    //[email] => invalid
    //[url] => invalid
//)

Answer 1

you have to return a valid javascript object, so, you should use a JSON String...

echo json_encode($obj);
exit;

and, in your javascript:

$.post().then(function(data) { console.log(data); });

I think that data is already decoded, but, if it isn't, you can do:

data = JSON.parse(data);
console.log(data)