Question

我现在一直在开发一个网站，我已经坚持了很长一段时间。看，我的索引有一个普通的PHP页面，在该页面上有一个注册用户登录的按钮。我有一个JavaScript函数，通过AJAX处理请求，然后将其发送到PHP页面， login.php中。我发现该函数有效，它将我登录，分配我的会话变量，如后面的PHP页面所示，但不显示我已设置的通知。我还要提一下xmlhttp.responseText的选项是注册期间唯一可用的选项。管理员当然除外，它将在数据库本身中被更改。最后一件事，如果我输入了错误的详细信息，xmlhttp.responseText = 'Wrong'然后按预期显示通知。

JavaScript函数代码：

function login(name, pass) {
var xmlhttp;
var check = 0;
if(name == "") {
    alert("Please enter your username");
    check = 1;
}
if(pass == "") {
    alert("Please enter your password");
    check = 1;
}
if(check != 0) {
    return;
}
if(window.XMLHttpRequest) { // Code for IE7+, Firefox, Chrome, Opera, Safari
    xmlhttp = new XMLHttpRequest();
} else { // Code for IE5, IE6
    xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
    if(xmlhttp.readyState == 4 && xmlhttp.status == 200) {
        var resp = xmlhttp.responseText;
        if(resp == "Wrong") {
            alert("The details you have entered are incorrect");
            return;
        }
        if(resp == "user") {
            alert("Welcome back!");
            return;
        }
        if(resp == "developer") {
            alert("Welcome back, developer!");
            return;
        }
        if(resp == "admin") {
            alert("Welcome back, Admin!");
            return
        }
    }
    if(xmlhttp.readyState == 4 && xmlhttp.status == 404) {
        alert("The login page was not found, please try again later");
    }
}
xmlhttp.open("GET", "login.php?name="+name+"&pass="+pass, true);
xmlhttp.send();
}

这是PHP页面，login.php：

<?php
include "base.php";
$username=mysql_real_escape_string($_GET['name']);
$password=md5(mysql_real_escape_string($_GET['pass']));
$checkLogin=mysql_query("SELECT * FROM Users WHERE Username = '".$username."' AND Password = '".$password."'");
if(mysql_num_rows($checkLogin) == 1) {
    $login=mysql_fetch_array($checkLogin);
    $_SESSION['ID'] = $login['ID'];
    $_SESSION['Username'] = $username;
    $_SESSION['Email'] = $login['Email'];
    $_SESSION['First'] = $login['Fist'];
    $_SESSION['Last'] = $login['Last'];
    $_SESSION['Type'] = $login['Type'];
    echo $_SESSION['Type'];
} else {
    echo "Wrong";
}
?>

通知在AJAX之后没有运行

0 个答案: