form表单1的ajax验证

form表单的ajax验证1:

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<!DOCTYPE html>
<html>
    <head>
        <meta charset="UTF-8">
        <title></title>
    </head>
    <body>
        <div id="form">
            账号:<input type="text" /><br />
            密码:<input type="text" /><br />
            <button id="btn">登录</button>
        </div>
        <script src="ajax.js" type="text/javascript" charset="utf-8"></script>
        <script type="text/javascript">
            var btn = document.getElementById('btn');
            var user = document.getElementById('form').getElementsByTagName('input')[0];
            var pass = document.getElementById('form').getElementsByTagName('input')[1];
            btn.onclick = function() {
                ajax('form.php?user='+user.value+'&pass='+pass.value,function(str) {
                    alert(str);
                });
            }
        </script>
    </body>
</html>

 PHP代码:

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<?php
/*
****************************************************
 
    请求方式: get
 
    url:    form.php?user=用户名&pass=密码
         
    return: '登陆成功'
            '账号密码不能为空'
            '账号错误'
            '密码错误'
 
****************************************************
*/
$user=$_GET["user"];
$pass=$_GET["pass"];
 
// $user=$_POST["user"];
// $pass=$_POST["pass"];
 
if ($user=="laowang"&&$pass=="12345"){
    echo '登陆成功';
}else{
    if ($user==""||$pass=="") {
        echo '账号密码不能为空';
    } else if ($user!="laowang"){
        echo '账号错误';
    }else if ($pass!="12345"){
        echo '密码错误';
    };
};
 
?>

  

posted @   HandsomeHan  阅读(229)  评论(0编辑  收藏  举报
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