Question

我尝试搜索，但没有任何答案像我想要的那样..好吧，我的问题。

我有一个名为index.php的登录页面，而index.php里面有div用于登录信息。

的index.php

<html>

<head>
<meta http-equiv="Content-Language" content="en-us">
<meta http-equiv="Content-Type" content="text/html; charset=windows-1252">

<script type="text/javascript" src="js/login.js"></script>

<title></title>
</head>

<body>

<form method="POST">
    <p>Username <input type="text" name="cUsername" size="20"></p>
    <p>Password <input type="text" name="cPassword" size="20"></p>
    <p><input type="submit" value="Login" name="B1"></p>
</form>

<div id="msg"></div>
</body>

</html>

这里我的Jquery没有刷新页面登录。

login.js

$(document).ready(function() {
    $('#Loading4').hide();    
});

function ajax-login(){

    var cUusername = $("#cUsername").val();
    var password = $("#password").val();

    if(cUsername.length > 2){
        $('#Loading4').show();
        $.post("login.php", {
            cUsername: $('#cUsername').val(),
            password: $('#password').val(),
        }, function(response){
            $('#Info4').fadeOut();
             $('#Loading4').hide();
            setTimeout("finishAjax4('Info4', '"+escape(response)+"')", 450);
        });
        return false;
    }
}

function finishAjax4(id, response){

  $('#'+id).html(unescape(response));
  $('#'+id).fadeIn(1000);
}

这里我的登录页面叫

的login.php

<?php

if($_REQUEST)
{
    $username   = $_REQUEST['username'];
    $query = "select * from tbl_user where username = '".strtolower($username)."'";
    $results = mysql_query( $query) or die('Error to connect to the database');

    if(mysql_num_rows(@$results) > 0) // not available
    {
        echo '<div id="msg">Login Successful !</div>';
    }
    else
    {
        echo '<div id="msg">Not Register Yet !</div>';
    }

}?>

我的意思是，我希望显示消息成功或失败来自 “login.php” 使用ID msg调用DIV，其中DIV位于“index.php”文件中。在index.php中将显示何时使用ajax登录成功或失败时显示。但是我想从login.php页面调用这个DIV。

我可以这样做吗？谢谢

Answer 1

检查一下。

$.post('index.php', $('form').serialize(), function(data){
    $("#msg").html(data)
});

了解更多details

Answer 2

您可以将您的login.php编辑为响应为JSON

<?php

if($_REQUEST)
{
    $username   = $_REQUEST['username'];
    $query = "select * from tbl_user where username = '".strtolower($username)."'";
    $results = mysql_query( $query) or die('Error to connect to the database');

    if(mysql_num_rows(@$results) > 0) // not available
    {
        $res = array('id'=>'msg', 'data'=>'Login Successful !');//here you can pass the id you need to change it and the new html data 
    }
    else
    {
        $res = array('id'=>'msg', 'data'=>'Not Register Yet !');//here also you can pass the id you need to change it and the new html data 
    }

}
header('Content-type: application/json');// to response as JSON code in the header
echo json_encode($res);// to encode the result string as JSON code
?>

login.js就像这样：

$(document).ready(function() {
    $('#Loading4').hide();    
});

function ajax-login(){
    var cUusername = $("#cUsername").val();
    var password = $("#password").val();
    if(cUsername.length > 2){
        $('#Loading4').show();
        $.post("login.php", {
            cUsername: $('#cUsername').val(),
            password: $('#password').val(),
        }, function(response){
            $('#Info4').fadeOut();
             $('#Loading4').hide();
             setTimeout(function(){
                   finishAjax4(response['id'], response['data'])
             },450);
        });
        return false;
    }
}

function finishAjax4(id, response){

  $('#'+id).html(unescape(response));
  $('#'+id).fadeIn(1000);
}

从另一个PHP页面调用DIV

2 个答案: