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不会写作文,不废话,不矫情,上代码:
HTML部分:
<div class="login">
<p class="log-err"> </p>
<input type="text" name="user_name" placeholder="用户名" />
<input type="password" name="user_pass" placeholder="密码" />
<button class="submit">登陆</button>
</div>
JS部分,本文使用JQ:
var _input = $(".login-main").find("input");
//获取输入内容
var username = _input.eq(0).val();
var userpass = _input.eq(1).val();
//执行ajax
$.ajax({
type: "post",
url: "login.php",
data: "au_name=" + username + "&" + "au_pass=" + userpass,
//请求执行前的操作
beforeSend: function(){
$(".submit").text("正在登陆...");
},
//执行成功回调函数,接收后台返回值
success: function(data){
if(data){
//返回用户昵称,生成cookie并跳转
$.cookie('im_au', data);
window.location.href = "../";
}else{
//返回false,提示登陆失败
$(".log-err").text("*用户名或密码错误!");
_input.eq(i).val("");
}
vcode();
$(".submit").text("登陆");
}
}
});
PHP部分:
<?php
//获取post数据
$au_name = $_POST['au_name'];
$au_pass = $_POST['au_pass'];
//执行sql查询
$sql = "SELECT * FROM im_au where au_name = '$au_name' and au_pass = '$au_pass'";
$res = $mysqli -> query($sql);
//查询到记录返回用户昵称,否则返回false
if($res -> fetch_array()){
echo $res['au_nick'];
}else{
echo false;
}
mysqli_close($mysqli);
?>
新手学习笔记,不足支出欢迎补充,欢迎大神指点!
原文地址:http://blog.xuxiangbo.com/im-23.html