JQUERY AJAX 호출에서 PHP의 JSON 응답을 올바르게 처리하는 방법은 무엇입니까?

<code class="php"><?php

header('Content-Type: application/json');
touch('phpTouch.txt');
// process email
$email = 1;
if ($email) {
    $value = array('return' => 1, 'msg1' => 'Message sent OK, we will be in touch ASAP');
} else {
    $value = array('return' => 0, 'msg1' => 'Message Failed, please try later');
}
$output = json_encode($value);
echo $output;

?></code>

<code class="javascript">success: function (data) {
    alert("SUCCESS:");
    for (var key in data) {
        $('#msgid').append(key);
        $('#msgid').append('=' + data[key] + '<br />');
    }
},
error: function (data) {
    alert("ERROR: ");
    for (var key in data) {
        $('#msgid').append(key);
        $('#msgid').append('=' + data[key] + '<br />');
    }
}</code>

<code class="javascript">success: function (data) {
    alert("SUCCESS:");
    $('#msgid').append('Return: ' + data.return + '<br />');
    $('#msgid').append('Message: ' + data.msg1 + '<br />');
}</code>