我有簡單的Ajax代碼:得到阿賈克斯數據和發送到mysql
function showMe(data) {
$("body").append();
if(data.success == true) {
$("body").append("<img src="+data.data.link+" height=180 /><br /><a href="+data.data.link+">"+data.data.link+"</a>");
$.ajax ({
type: "POST",
url: "sql.php",
data: "y=+data.data.link+",
});
我需要得到「+ data.data.link +」值,發送給MySQL數據庫,但它發出data.data。鏈接而不是真正的鏈接。 如何獲得真正的價值併發送給數據庫? 這裏是sql.php:
<?php
define('IN_PHPBB', true);
$phpbb_root_path = (defined('PHPBB_ROOT_PATH')) ? PHPBB_ROOT_PATH : 'forum/';
$phpEx = substr(strrchr(__FILE__, '.'), 1);
include($phpbb_root_path . 'common.' . $phpEx);
include($phpbb_root_path . 'includes/functions_display.' . $phpEx);
include("$phpbb_root_path/includes/functions_user.php");
$user->session_begin();
$auth->acl($user->data);
$user->setup('viewtopic');
include "forum/config.php";
$link = mysql_connect("$dbhost", "$dbuser", "$dbpasswd");
$db_selected = mysql_select_db("$dbname", $link);
$y = @$_POST['y'];
$date = date('d.m.y');
$name = $user->data['username'];
mysql_query("INSERT INTO `gallery` (name, createdate, piclink) VALUES('$name', '$date', '".$y."')");
unlink("gallery/$imagename");
?>
感謝所有幫助:)
您需要顯示sql.php的響應 –