Question

我正在使用AJAX调用函数提交如下的登录表单，但是它被定向到的页面仅在发现用户不存在（正常工作）时响应，但是当登录信息被认为是正确的时，警报或其他方面没有任何回报，我错过了什么？

 <script>
        $(document).ready(function() {
            $("#loginbutton").click(function() {
                var user = $("#user").val();
                var pass = $("#pass").val();
                var dataString = "user="+user+"&pass="+pass;
                if (user == "" || pass == "") {
                    alert("Please fill in all fields");
                }
                else {
                    $.ajax({
                        type: "POST",
                        url: "login.php",
                        data: dataString,
                        cache: false,
                        success: function(result) {
                            alert(result)
                        },
                    });
                }
                return false;
            });
        });
        </script>

这是PHP：

<?php

include_once($_SERVER['DOCUMENT_ROOT'].'/wp-config.php');
include_once($_SERVER['DOCUMENT_ROOT'].'/wp-load.php');
include_once($_SERVER['DOCUMENT_ROOT'].'/wp-includes/wp-db.php');

global $userdata;
global $wpdb; 

//get the posted values

$posted_username = 'matt-inovate';
$posted_password = 'Qazplm10';

$user_name = htmlspecialchars($posted_username,ENT_QUOTES);

$pass_word = wp_hash_password($posted_password);

$pass_md5 = md5($posted_password);

$pass = $pass_word;

$userinfo = get_userdatabylogin($user_name);

$passcheck = $userinfo->user_pass;

$conn = mysql_connect("127.0.0.1", "fw1359063030", "75ozg3cMxM47kRlJEPdLl2iOPb3f4R") or mysql_error();
    $conn = mysql_select_db("db8760018806", $conn);

    $sql = "SELECT * FROM wp_users WHERE user_login = '$posted_username'";
    $result = mysql_query($sql) or mysql_error();
    $rows = mysql_num_rows($result);

    if ($rows == 1) {
        $row = mysql_fetch_array($result);
        echo $passcheck."<br>";
        echo $row[2]."<br>";
        if ($passcheck == $row[2]) { echo "Login Success."; } else { echo "Login Failed."; }
    }
    else { echo "Could not find user."; }
?>

jQuery / AJAX表单提交和结果没有返回

0 个答案: