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这是我第一次使用jQuery,我不认为它似乎工作。这里是一些代码...JQuery位置正确吗?
<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>
继承人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();
?>
它没有做任何事情。是否有提醒,如果它实际上去addComments?或者我只是使用后错误(SQL应该是正确的)。我知道我得到的参数信息正确(警报)任何帮助将是伟大的,谢谢!
检查萤火虫/控制台,看到任何错误..任何404? –
检查您的浏览器控制台。你应该在那里看到一个请求。在大多数浏览器中,它是'F12'键。 – PeeHaa