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這是我第一次使用jQuery,我不認爲它似乎工作。這裏是一些代碼...JQuery位置正確嗎?
<head>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.5/jquery.min.js"></script>
<script type = "text/javascript">
function insertComment(comment, assignment, pid){
alert(comment);
alert(assignment);
alert(pid);
var url = "addComment.php";
$.post(url, {comment: comment, assignment: assignment, pid: pid});
}
</script>
繼承人addComment.php
<?php
include ("viewComments.php");
//Connection string to get to database
$host = 'host';
$user = 'user';
$password = 'pw';
$dbh = new mysqli($localhost, $user, $password, "kao17_CS242Portfolio");
//Prpared statement for user inpur, prevent sql-injection attacks
$stmt = $dbh->prepare("INSERT INTO tbl_Comment (comment, project_title, parent_comment_id) VALUES (? , ? , ?)");
$stmt->bind_param('ssi', $prevComment, $prevAssignment, $parentID);
$prevAssignment = $_POST['assignment'];
$parentID = $_POST['pid'];
$prevComment = profanityCheck($_POST['comment']);
$stmt->execute();
?>
它沒有做任何事情。是否有提醒,如果它實際上去addComments?或者我只是使用後錯誤(SQL應該是正確的)。我知道我得到的參數信息正確(警報)任何幫助將是偉大的,謝謝!
檢查螢火蟲/控制檯,看到任何錯誤..任何404? –
檢查您的瀏覽器控制檯。你應該在那裏看到一個請求。在大多數瀏覽器中,它是'F12'鍵。 – PeeHaa