我有一个登录框......
当用户开始输入时..我想检查输入的LOGIN NAME是否存在于数据库中......
如果登录名存在,我将设置登录按钮激活...如果它不存在我将设置登录按钮deactive ...
offcourse我需要 AJAX 来通过 PHP 来执行我的 mySQL 我很难知道它将如何完成...
让我们说这是我的查询
<?php
$result = mysql_query("SELECT * FROM accounts WHERE name='mytextboxvalue'");
?>
怎么做
答案 0 :(得分:1)
您可以使用JSON-RPC,这里是implementation in php 在JQuery中,您可以使用此代码。
var id = 1;
function check_login(){
var request = JSON.stringify({'jsonrpc': '2.0',
'method': 'login_check',
'params': [$('#login_box').val()],
'id': id++});
$.ajax({url: "json_rpc.php",
data: request,
success: function(data) {
if (data) {
$('#login_button').removeAttr('disabled');
} else {
$('#login_button').attr('disabled', true);
}
},
contentType: 'application/json',
dataType: 'json',
type:"POST"});
}
并在php中
<?php
include 'jsonRPCServer.php';
//mysql_connect
//mysql_select_db
class Service {
public function login_check($login) {
$login = mysql_real_escape_string($login);
$id = mysql_query("SELECT * FROM accounts WHERE name='$login'");
return mysql_num_rows($id) != 0;
}
}
$service = new Service();
jsonRPCServer::handle($service);
?>
答案 1 :(得分:1)
保持简单:
$(document).ready(function(){
var Form = $('#myForm');
var Input = $('input.username',Form)
Input.change(function(event){
Value = Input.val();
if(Value.length > 5)
{
$.getJSON('/path/to/username_check.php',{username:Value},function(response){
if(response.valid == true)
{
Form.find('input[type*=submit]').attr('disabled','false');
}else
{
Form.find('input[type*=submit]').attr('disabled','true');
}
});
}
});
});
然后是PHP方..
<?php
//Load DB Connections etc.
if(!empty($_REQUEST['username']))
{
$username = mysql_real_escape_string($_REQUEST['username']);
if(isset($_SESSION['username_tmp'][$username]))
{
echo json_encode(array('valid' => (bool)$_SESSION['username_tmp'][$username]));
die();
}
//Check the database here... $num_rows being a validation var from mysql_result
$_SESSION['username_tmp'][$username] = ($num_rows == 0) ? true : false;
echo json_encode(array('valid' => (bool)$_SESSION['username_tmp'][$username]));
die();
}
?>
答案 2 :(得分:0)