我不確定當我使用AJAX向PHP頁面發送一些信息時,如果我已經完成了正確的工作,有些代碼連接到數據庫並存儲一段文本。 AJAX調用的PHP頁面必須使PHP頁面與普通的PHP頁面相比有所不同?這不起作用,我收到了404 Not Found消息?AJAX調用PHP頁面,需要幫助
這裏是一個PHP頁面:
<?php
session_start();
$mysqli = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_DATABASE);
$replyArticleId = isset($_POST['replyArticleId']) ? $_POST['replyArticleId'] : '';
$replyText = isset($_POST['replyText']) ? $_POST['replyText'] : '';
$replySign = $_SESSION['accountId'];
date_default_timezone_set("Europe/Stockholm");
$date = new DateTime();
$replyDate = $date->format('Y-m-d H:i:s');
$tableUser = DB_PREFIX . WS_DB_USER;
$tablePost = DB_PREFIX . WS_DB_POST;
$tableComment = DB_PREFIX . WS_DB_COMMENT;
$tableArticle = DB_PREFIX . WS_DB_ARTICLE;
$tableReply = DB_PREFIX . WS_DB_REPLY;
// Add new comment
$query1 = "INSERT INTO {$tableReply} (replyArticleId, replyText,replyUserId,
replyDate) VALUES ('{$replyArticleId}','{$replyText}','{$replySign}', '{$replyDate}');";
$query2 = "UPDATE {$tableArticle} SET articleDateUpdated = NOW() WHERE articleId = {$replyArticleId};";
$res = $mysqli->query($query1) or die($mysqli->error);
$res = $mysqli->query($query2) or die($mysqli->error);
$mysqli->close();
這是我用來打電話和發一些測試內容的AJAX代碼:
$.ajax({
url: "PAddReplyProcessAJAX.php?replyArticleId=1",
type: "POST",
dataType: "text",
data: "replyText=" + "test"
});
?>
檢查您使用的網址..嘗試訪問瀏覽器中的相同網址。 – KillerFish 2011-06-16 19:38:28