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  • PHP使用AJAX返回登录成功信息完整参考代码

    以下是完整参考代码,index.php为登录页面,ajax.php为处理ajax无刷新请求页面。

    index.php

    <!DOCTYPE html>
    <html>
        <head>
            <meta charset="utf-8" />
            <title>登录</title>
            <script type="text/javascript" src="http://cdn.bootcss.com/jquery/2.1.4/jquery.min.js"></script>
        </head>
        <body>
                            帐号:<input type="text" id="account" />
            <br><br>
                            密码:<input type="password" id="password" />
            <br />
            <input type="button" value="登录" id="btnlogin" />
            <script type="text/javascript">
            $(function(){
                $("#btnlogin").click(function(){
                    $.ajax({
                        type:"post",
                        url:"ajax.php",
                        data:{account:$("#account").val(),password:$("#password").val()},
                        dataType:"json",
                        success:function(data){
                            if(data.type==1){
                                alert("登录成功");
                            }else{
                                alert("登录失败");
                            }
                        },
                        error:function(){
                            alert("请求异常");
                        }
                    });
                });
            });
            </script>
        </body>
    </html>
    

     

    ajax.php

    <?php
    header("Content-Type:text/html; charset=utf-8");
    $account = $_POST['account'];
    $password = $_POST['password'];
    $result = array();
    if ($account != '' && $password != '') {
        //$row = $db->query("SELECT * FROM account where user = '".$account."' and password = '".$password."'");
        $row = true;//这里去查数据库,假设这里返回true
        if($row){
            $result['type'] = 1;
            $result['msg'] = '登录成功';
        }else{
            $result['type'] = 0;
            $result['msg'] = '用户名或密码不正确';
        }
    } else {
        $result['type'] = 0;
        $result['msg'] = '参数传输不正确';
    }
    echo json_encode($result);
    ?>
    
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  • 原文地址:https://www.cnblogs.com/woniu666/p/9929404.html
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