JSON输入的意外结尾-登录系统出现语法错误

Question

我正在尝试创建一个登录系统。但是，我收到JSON输入的意外结束，并且不确定为什么会因为它在我正在使用的另一个页面上起作用。就是说JS第12行的JSON输入意外结束。

我正在使用的JS是：

$('#formUserLogin').submit(function(event){ // If data is submitted from the form called formUserLogin
    formData = $('#formUserLogin').serialize(); // Serialize data

    event.preventDefault();

    $.ajax({
        type: "POST", // Defines type for ajax function
        url: "policeLoginDAO.php", // URL to send data to
        data: formData+"&phpFunction=login", // Data to send + PHP function to use it with
        datatype: 'json', // Datatype to send
        success: function(msg){
            dataJson = JSON.parse(msg);
            if(dataJson['result']=='false') { // If result is false do this:
                $("divMessage").html("Wrong username or password"); // Display message in divMessage div in html page
            } else { // Else if it's correct, do this:
                window.location="../policeHome.php"; // Open page, policeHome.php and set the value of badgeno to the stored badgeno in session storage
            }
        }
    });
});

我正在使用的PHP是：

<?php
if(isset($_POST['phpFunction'])) {
    if($_POST['phpFunction'] == 'login') {
        login();
    }
}

function login() {

    session_start(); // Starts the session for the user
    include "../../config.php"; // Includes the config file to connect the the database
    $uName = strip_tags($_POST['userName']); // Strips the tags for the username
    $pWord = strip_tags($_POST['password']); // Strips the tags for the password
    $userID = $_SESSION['uName']; // Sets the username to the session
    $uName1 = strip_tags(mysqli_real_escape_string($connection, $uName)); // Strips the tags and escapes the string for the username
    $pWord1 = strip_tags(mysqli_real_escape_string($connection, $pWord)); // Strips the tags and escapes the string for the password


    $sql = "SELECT * FROM `tbl_police` WHERE Username='".$uName1."' "; // Selects all data from the tbl_user where email is equal to the user input
    $result = $connection->query($sql); // Result is equal to the queries outcome
    $pass = $row['Password']; // Gets the password
    if ($result->num_rows === 1) { // If the result is identical to 1
        $row = mysqli_fetch_array($result); // Fetch result
        if (password_verify($pWord1, $row['Password'])) { // Verify the password against the users input by using the password_verify function. If it's correct... Do this:

            echo json_encode($row);
            $_SESSION['uName'] = $uName;

        }else{ // If the password is not correct... Do this:
            echo("Hello");

        }
        mysqli_close($connection); // Close the connection to the database
    }
}
?>

任何帮助将不胜感激，谢谢

Answer 1

伙计，我认为您需要稍微修改代码。您需要在成功函数中指定目标元素的属性，即jQuery函数调用的元素是类还是ID。