所以我试图插入一行到我的数据库。我打电话给一个类似Ajax的函数来在我的表中插入一个新行。但它不插入一行。Ajax函数没有运行
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();
}
这里是它的文件,以便将行插入表中。
<?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");
?>
它实际上做了什么? – Archer
为什么使用jQuery进行标记? –
**'Ajax Function Not Running' ** ...它在网络浏览器中失败了吗?由于语法错误,它没有运行吗?你的PHP文件是否精美和华丽?请按照以下步骤正确调试AJAX问题,然后您可以正确地指责哪些代码被破坏:http://stackoverflow.com/a/21617685/2191572 – MonkeyZeus