Question

这是我第一次使用jquery，我认为它似乎没有用。这是一些代码...

<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。如果真的去了addComments，它是否会提醒？或者我只是使用错误的帖子（SQL应该是正确的）。我知道我正在获取参数的信息（警报）任何帮助都会很棒，谢谢！

Answer 1

您可以在调用$.post：

时添加回调功能

$.post(url, {comment: comment, assignment: assignment, pid: pid}, function(data, textStatus, jqXHR) {
  alert(data);
});

然后在PHP中执行SQL语句后添加die('OK')。如果在运行OK时在警告框中显示insertComment，那么您知道调用已成功，并且您的PHP存在任何问题。

JQuery位置正确吗？

1 个答案: