所以我試圖插入一行到我的數據庫。我打電話給一個類似Ajax的函數來在我的表中插入一個新行。但它不插入一行。Ajax函數沒有運行
function showResult(first, last)
{
var First = first;
var Last = last;
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.open("POST","http://www.website.ca/portal/MyChapter2/cgi-bin/DetermineUser.php?FirstName="+First+"&LastName="+Last,true);
xmlhttp.send();
}
這裏是它的文件,以便將行插入表中。
<?php
require_once (dirname(__FILE__) . '/../../include/Initialization.php');
require_once (PORTAL_PATH . '/include/FormLibrary.php');
require_once (PORTAL_PATH . '/include/SingleRowQuery.php');
require_once (PORTAL_PATH . '/include/Functions.php');
require_once (PORTAL_PATH . '/include/VolunteerInterests.php');
require_once (PORTAL_PATH . '/TaskManager/cgi-bin/AutoTaskFunctions.php');
$FirstName = $_POST['FirstName'];
$LastName = $_POST['LastName'];
$sql="INSERT INTO `Track_Notification`(`Track_ID`, `Track_UserID`) VALUES ('$FirstName','$LastName')";
echo ("success");
?>
它實際上做了什麼? – Archer
爲什麼使用jQuery進行標記? –
**'Ajax Function Not Running' ** ...它在網絡瀏覽器中失敗了嗎?由於語法錯誤,它沒有運行嗎?你的PHP文件是否精美和華麗?請按照以下步驟正確調試AJAX問題,然後您可以正確地指責哪些代碼被破壞:http://stackoverflow.com/a/21617685/2191572 – MonkeyZeus