尝试添加一个“;像“;函数,没有显示错误,但无法工作


Trying to add a "like" function, no errors displayed, but wont work

我正试图在我的社交网络中添加一个"赞"函数,但我的代码不起作用,也不会打印任何错误
函数使用$_GET来确定是否应该"点赞"评论或帖子,并检查用户是否已经点赞了该帖子或评论。如果是这样,函数将"不同"。

有人能帮我找出我做错了什么吗?

代码:

<?php
ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);
session_start();
include "./db-connect.php";
$memberID= $_SESSION['memberID'];
if(isset($_GET['post_id'])){
$postID=$_GET['post_id'];
$sqlCheck="SELECT * from cs_likes WHERE post_id = $postID AND member_id = $memberID";
$sqlInsert="INSERT INTO cs_likes (post_id, member_id) VALUES ('$postID','$memberID')";
$sqlDelete="DELETE FROM cs_likes WHERE post_id= $postID AND member_id = $memberID";
}
elseif(isset($_GET['comment_id'])){
$commentID=$_GET['comment_id'];
$sqlCheck="SELECT * from cs_likes WHERE comment_id = $commentID AND member_id = $memberID";
$sqlInsert="INSERT INTO cs_likes (comment_id, member_id) VALUES ('$postID','$memberID')";
$sqlDelete="DELETE FROM cs_likes WHERE comment_id= $commentID AND member_id = $memberID";
}
$checkResult=mysqli_query($link, $sqlCheck);
if(mysqli_num_rows($checkResult)=0)
$result=mysqli_query($link,$sqlInsert);
else
$result=mysqli_query($link,$sqlDelete);

?>

您的代码。。。

if(mysqli_num_rows($checkResult)=0)

应该是二重相等。。。

if(mysqli_num_rows($checkResult)==0)