表单登录验证

原创2019-03-17 12:21:19134
摘要:index.html<!DOCTYPE html><html lang="en"><head><meta charset="UTF-8"><title>Ajax实战:表单验证</title></head><body><h3>用户登录</h3>

index.html

<!DOCTYPE html>

<html lang="en">

<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>

check.php

<?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;

}


批改老师:查无此人批改时间:2019-03-18 09:22:41
老师总结:完成的不错。ajax请求可以做到不刷新页面,更换数据。所以现在很常用。继续加油

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