php ajax user login code_PHP tutorial

An entry-level ajax user login code. Beginners in need can refer to it,

login.html

代码如下	复制代码
用户名 密码

js code

_chk.php

The code is as follows

代码如下

复制代码

function chk_login(form){     var username=form.username;     var password=form.password;     if(username.value == ""){         alert("用户名不能为空");         username.focus();         return false;     }     if(password.value == ""){         alert("密码不能为空");         password.focus();         return false;     }     var url = 'login_chk.php?username='+username.value+'&password='+password.value;     xmlhttp.open('get',url,true); //为什么下面加了下面注释掉的代码后，会没反应了，没加就有反应，我的php文件路径都是正确的，也有引入xmlhttprequest.js xmlhttp.onreadystatechange = function(){         if(xmlhttp.readyState == 4){             if(xmlhttp.status == 200){                 var msg = xmlhttp.responseText;                 if(msg == "1"){                     alert("登陆成功");                     window.location='index.php?name='+username.value;                 }else if((msg == "2"){                     alert("登陆用户名或密码错误");                 }else{                     alert(msg);                 }             }         }     }     xmlhttp.send(null);*/ } ogin

Copy code

代码如下	复制代码
echo 1 ?>

function chk_login(form){
var username=form.username;
var password=form.password;
If(username.value == ""){
alert("Username cannot be empty");
         username.focus();
         return false;
}
If(password.value == ""){
alert("Password cannot be empty");
           password.focus();
         return false;
}

var url = 'login_chk.php?username='+username.value+'&password='+password.value;
xmlhttp.open('get',url,true);
//Why after adding the commented out code below, there will be no response, but there will be response without adding it. My php file paths are all correct, and xmlhttprequest.js has also been introduced
xmlhttp.onreadystatechange = function(){
If(xmlhttp.readyState == 4){
If(xmlhttp.status == 200){
            var msg = xmlhttp.responseText;
                    if(msg == "1"){
alert("Login successful");
                        window.location='index.php?name='+username.value;
                      }else if((msg == "2"){
alert("Login user name or password is wrong");
                     }else{
alert(msg);
                   }
             }
         }
}
xmlhttp.send(null);*/
}

ogin

Because it is a test file, the database has not been read. I will not write it here. Friends who need it can add the user name record for reading the database by themselves. If it exists, just return 1. Please refer to the tutorial http://www.bKjia.c0m/phper/php-cy/34654.htm http://www.bKjia.c0m/phper/21/46169c59dabdf0501de3d9ac9653e096.htm

http://www.bkjia.com/PHPjc/631323.htmlwww.bkjia.comtruehttp: //www.bkjia.com/PHPjc/631323.htmlTechArticleAn entry-level ajax user login code. Beginners in need can refer to it. The login.html code is as follows Copy the code input name=username type=text / username input name=passw...