parseJSON抛出意外的令牌错误

Question

我的代码适用于除一个以外的所有文件。它会在

处引发unexpected token错误

data = $.parseJSON(msg);

我真的不知道是什么导致了这个问题......

使用Javascript：

function submitform() {
    var pass = document.getElementById("password").value;
    var nologinsession = document.getElementById("Username");

    if (nologinsession == null) {
        var userid = "<?php echo $_SESSION['SESS_MEMBER_ID']; ?>";
        var username = "none";
    } else {
        var username = document.getElementById("Username").value;
        var userid = "none";
    }                       
    $.ajax({
        type: "POST",
        url: "php/call.php",
        data: { pass: pass, username: username, userid: userid }
    }).done(function( msg ) {
        data = $.parseJSON(msg);
        var permission = data['permission'];
        var reason = data['reason'];
        if (permission == 'okay') {
            location.reload;
        } else if (permission == 'nope') {
            alert(reason);
        } else {
            alert('There appears to be something wrong. Beam me up, Scotty.');
        }
    });
};

call.php ：

<?php

//Start session
session_start();

//Include database connection details
require_once('Connection.php');

$password = $_POST['pass'];
$username = $_POST['username'];
$userid = $_POST['userid'];
$requester = $_POST['requester'];

$permission;
$reason;

if($userid == 'none') {

    $query = "SELECT something FROM table WHERE email='$username'";

    // the result of the query
    $query = mysql_query("$query") or die("Invalid query: " . mysql_error());
    $row = mysql_fetch_assoc($query);

    $result = $row["something"];

    if($result == $password) {
        $permission = 'okay';
        $reason = 'empty';
    } else {
        $permission = 'nope';
        $reason = 'Password is incorrect.';
    }
} else {
    $query = "SELECT something FROM table WHERE ID='$userid'";

    // the result of the query
    $query = mysql_query("$query") or die("Invalid query: " . mysql_error());
    $row = mysql_fetch_assoc($query);
    $result = $row["something"];

    if($result == $password) {
        $permission = 'okay';
        $reason = 'empty';
    } else {
        $permission = 'nope';
        $reason = 'Password is incorrect.';
    }
};

$json['permission'] = $permission;
$json['reason'] = $reason;
print(json_encode($json));
?>

修正了一些评论后让我走上正轨：

原来这是建立数据库连接的问题。将新IP地址放在白名单上，问题就是修复。谢谢你们！