abstract：<!DOCTYPE html><html><head> <meta charset="UTF-8"> <title>Ajax实战:表单验证</title></head><body><h3>用户登录</h3>&
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<title>Ajax实战:表单验证</title>
</head>
<body>
<h3>用户登录</h3>
<form>
<p>邮箱: <input type="email" name="email"></p>
<p>密码: <input type="password" name="password"></p>
<p><button type="button">提交</button></p>
</form>
<script>
let btn = document.getElementsByTagName('button')[0];
btn.onclick = function () {
//1.创建xhr对象
let xhr = new XMLHttpRequest();
//2.监听响应状态
xhr.onreadystatechange = function(){
if (xhr.readyState === 4) { // 准备就绪
// 判断响应结果:
if (xhr.status === 200) {
// 响应成功,通过xhr对象的responseText属性可以获取响应的文本,此时是html文档内容
let p = document.createElement('p'); //创建新元素放返回的内容
p.style.color = 'red';
let json = JSON.parse(xhr.responseText);
if (json.status === 1) {
p.innerHTML = json.msg;
} else if (json.status == 0) {
p.innerHTML = json.msg;
}
// 将响应文本添加到新元素上
document.forms[0].appendChild(p); // 将新元素插入到当前页面中
btn.disabled = true;
setTimeout(function(){
document.forms[0].removeChild(p);
btn.disabled = false;
if (json.status == 1) {
location.href = 'admin.php';
}
},2000);
} else {
// 响应失败,并根据响应码判断失败原因
alert('响应失败'+xhr.status);
}
} else {
// http请求仍在继续,这里可以显示一个一直转来转去的图片
}
}
//3.设置请求参数
xhr.open('post','inc/check.php',true);
//4. 设置头信息,将内容类型设置为表单提交方式
xhr.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
//4.发送请求
let data = {
email: document.getElementsByName('email')[0].value,
password: document.getElementsByName('password')[0].value
};
// data = 'email='+data.email+'&password='+data.password;
let data_json=JSON.stringify(data);
xhr.send('data='+data_json);
}
</script>
</body>
</html>
<?php
//print_r($_POST['data']);
//echo $data['email'];
$user = json_decode($_POST['data']);
//echo $user->email;
$email = $user->email;
$password = sha1($user->password);
$pdo = new PDO('mysql:host=localhost;dbname=php','root','root');
$sql = "SELECT COUNT(*) FROM `user` WHERE `email`='{$email}' AND `password`='{$password}' ";
$stmt = $pdo->prepare($sql);
$stmt->execute();
if ($stmt->fetchColumn(0) == 1) {
echo json_encode(['status'=>1,'msg'=>'登录成功,正在跳转...']) ;
exit;
} else {
echo json_encode(['status'=>0,'msg'=>'邮箱或密码错误,登录失败!']) ;
exit;
}
Correcting teacher：韦小宝Correction time：2019-02-22 10:44:32
Teacher's summary：使用jQuery来结合ajax的确是很方便啊 这在以后的实际开发中会去使用的 但是基础的js结合ajax也要理解啊