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Ajax的工作原理分析
HTTP请求对应一个页面时，ajax代 理当前页面的请求，使得当前页面不跳转，等js收到服务器端响应的数据后,再由js来刷新页面更新数据
<!doctype html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta name="viewport"
content="width=device-width, user-scalable=no, initial-scale=1.0, maximum-scale=1.0, minimum-scale=1.0">
<meta http-equiv="X-UA-Compatible" content="ie=edge">
<title>Ajax实战:表单验证</title>
</head>
<body>
<h3>用户登录</h3>
<form action="">
<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.getElementsByName('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.readyStatue===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的文档内容
let p =document.createElement('p');//创建新元素放返回的内容
p.style.color='red';
let JS json =ON.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.form[0].appendChild(p);//将新元素插入到当前页面中
btn.disabled=true;
setTimeout(function () {
document.form[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
xhr.setRequestHeader('Content-type','application/x-www-form-urlencoded')
//5.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>点击 "运行实例" 按钮查看在线实例
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'=>'登陆成功,正在跳转...']);
}else{
echo json_encode(['status=>0','msg'=>'邮箱或密码错误,登陆失败!']);
exit;
}点击 "运行实例" 按钮查看在线实例
