我有一個(gè)喜歡按鈕，它允許用戶喜歡我網(wǎng)站上的帖子。如果用戶喜歡他們之前不喜歡的帖子，它將 +1，如果他們?cè)俅伟聪孪嗤南矚g按鈕，它將 -1。這在我筆記本電腦上的虛擬服務(wù)器上工作。但是，相同的代碼在我的實(shí)時(shí)站點(diǎn)上不起作用。在我的實(shí)時(shí)網(wǎng)站上，用戶可以多次喜歡同一個(gè)帖子，這不是我想要的。我正在使用 JQuery Ajax 調(diào)用來(lái)觸發(fā)一些 MySQL 代碼的 PHP 文件。任何人都可以看到下面的 PHP 有什么明顯的錯(cuò)誤嗎？include ("../con/config.php");$postid = $_POST['postid'];$userid = $_POST['userid'];$query = $con->prepare("SELECT COUNT(*) AS CntPost FROM Likes WHERE UserID = ? AND PostID = ?");$query->bind_param('ss',$userid,$postid);$query->execute();$result = $query->get_result();$fetchdata = $result->fetch_assoc();$count = $fetchdata['CntPost'];if($count == 0){     $stmt = $con->prepare("INSERT INTO Likes(UserID,PostID) VALUES(?,?)");     $stmt->bind_param("ss", $userid, $postid);     $stmt->execute();} else {     $stmt = $con->prepare("DELETE FROM Likes WHERE UserID = ? AND PostID = ?");     $stmt->bind_param("ss", $userid, $postid);     $stmt->execute();}// count numbers of likes in post$query = $con->prepare("SELECT COUNT(*) AS CntLike FROM Likes WHERE PostID = ?");$query->bind_param('s', $postid);$query->execute();$result = $query->get_result();$fetchlikes = $result->fetch_assoc();$totalLikes = $fetchlikes['CntLike'];$return_arr = array("likes"=>$totalLikes,"type"=>$count);echo json_encode($return_arr);