Jquery 从其他PHP页面调用DIV
我试着搜索,但没有找到我想要的答案。。好的,这是我的问题 我有一个名为index.php的登录页面,index.php内部有用于信息登录的div index.phpJquery 从其他PHP页面调用DIV,jquery,ajax,html,login,Jquery,Ajax,Html,Login,我试着搜索,但没有找到我想要的答案。。好的,这是我的问题 我有一个名为index.php的登录页面,index.php内部有用于信息登录的div index.php <html> <head> <meta http-equiv="Content-Language" content="en-us"> <meta http-equiv="Content-Type" content="text/html; charset=windows-1252">
<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>
这里我的登录页面叫做
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>';
}
}?>
检查一下这个
$.post('index.php', $('form').serialize(), function(data){
$("#msg").html(data)
});
有关更多信息,您可以编辑login.php以响应JSON
<?php
if($_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
?>
旁注:停止使用不推荐的mysql.*
函数。请改用MySQLi或PDO。正如@Shivan Raptor所说,“停止使用mysql”,它已经被弃用,并且充满了威胁。。。。改用dbo或mysqli。
<?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
?>
$(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);
}