Javascript 未捕获的语法错误：意外标记<；

我明白了

编辑这里是登录php页面，如果您需要javascript，请告诉我。您是否认为这比设置标准表单和页面更有效：

var json = jQuery.parseJSON(data);

---

从jQuery 1.9开始，将抛出JavaScript错误，而不是返回
null
。因此，您的json无法显示该错误

从jQuery文档：

在jQuery 1.9之前，$.parseJSON返回null，而不是在传递空字符串、null或未定义时抛出错误，即使这些字符串不是有效的JSON

要防止JS抛出错误，请执行以下操作：

<?php

require("./../config.php");

if (!isset($_POST['req']))
{
    die("No request provided.");
}
else 
{ 
    if ($_POST['req'] == "login")
    {
        $json = array("success"=>false, "msg"=>"", "req"=>"login");

        if (!isset($_POST['user']) || empty($_POST['user']))
        {
            $json['success'] = false;
            $json['msg'] = "No user provided";
        }
        else
        { 
            if (!isset($_POST['pass']) || empty($_POST['pass']))
            {
                $json['success'] = false;
                $json['msg'] = "No password provided";
            }
            else
            { 
                $user = mysql_real_escape_string($_POST['user']);
                $password = mysql_real_escape_string($_POST['pass']);

                $password = hash("sha512", $password);

                $query = mysql_query("SELECT `id` FROM `users` WHERE `username` = '" . $user . "' AND `password` = '" . mysql_real_escape_string($password) . "' LIMIT 1");

                if (mysql_num_rows($query)) 
                {
                    $sessID = mysql_real_escape_string(session_id());
                    $hash = mysql_real_escape_string(hash("sha512", $sessID.$_SERVER['HTTP_USER_AGENT']));

                    $userData = mysql_fetch_assoc($query);
                    $expires = time() + (60 * 15);

                    mysql_query("INSERT INTO `active_users` (`user`, `session_id`, `hash`, `expires`) VALUES (" . (int) $userData['id'] . ", '" . $sessID . "', '" . $hash . "', " . $expires . ")");
                    $json['success'] = true;
                    $json['msg'] = "Logged in";
                }
                else
                {
                    $json['success'] = false;
                    $json['msg'] = "Username or password are incorrect.";
                }
            }
        }

        print(json_encode($json));
    }
}
?>


附言：发布你的JSON，让我们知道问题出在哪里。您可以通过添加
console.log（数据）之前尝试
（如果您使用我的代码）。
正在解析的数据看起来像…？将您的代码与JSON数据一起发布到此处。显然
数据
不是有效的JSON。什么是
console.log（数据）显示？听起来好像
数据
可能是从web服务器发送的基于HTML的错误页面，而不是您真正想要的json资源。正如其他人所说，将您要返回的数据发布。我看到过这种类型的错误，因为服务器返回404或500错误页面，ajax调用试图解析返回的HTML，而它希望返回JSON。
<?php

require("./../config.php");

if (!isset($_POST['req']))
{
    die("No request provided.");
}
else 
{ 
    if ($_POST['req'] == "login")
    {
        $json = array("success"=>false, "msg"=>"", "req"=>"login");

        if (!isset($_POST['user']) || empty($_POST['user']))
        {
            $json['success'] = false;
            $json['msg'] = "No user provided";
        }
        else
        { 
            if (!isset($_POST['pass']) || empty($_POST['pass']))
            {
                $json['success'] = false;
                $json['msg'] = "No password provided";
            }
            else
            { 
                $user = mysql_real_escape_string($_POST['user']);
                $password = mysql_real_escape_string($_POST['pass']);

                $password = hash("sha512", $password);

                $query = mysql_query("SELECT `id` FROM `users` WHERE `username` = '" . $user . "' AND `password` = '" . mysql_real_escape_string($password) . "' LIMIT 1");

                if (mysql_num_rows($query)) 
                {
                    $sessID = mysql_real_escape_string(session_id());
                    $hash = mysql_real_escape_string(hash("sha512", $sessID.$_SERVER['HTTP_USER_AGENT']));

                    $userData = mysql_fetch_assoc($query);
                    $expires = time() + (60 * 15);

                    mysql_query("INSERT INTO `active_users` (`user`, `session_id`, `hash`, `expires`) VALUES (" . (int) $userData['id'] . ", '" . $sessID . "', '" . $hash . "', " . $expires . ")");
                    $json['success'] = true;
                    $json['msg'] = "Logged in";
                }
                else
                {
                    $json['success'] = false;
                    $json['msg'] = "Username or password are incorrect.";
                }
            }
        }

        print(json_encode($json));
    }
}
?>

try {
    var json = $.parseJSON(data);
    //if code below here runs, means json is valid


} catch(e){
    //if code below here runs, means json is invalid

}