dongying6659 2016-04-30 19:24
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试图添加“喜欢”功能,没有显示错误,但不会工作

I am trying to add a "like" function to my social network, however my code does not work and wont print any errors.
The functions uses $_GET to find out if a comment or a post should be "liked" and procedes to check if the user has already liked the post or comment. If so, the function will "unlike".

Can anyone help me figure out what I've done wrong?

Code:

<?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);



?>
  • 写回答

1条回答 默认 最新

  • duanfen9090 2016-04-30 19:30
    关注

    Your code...

    if(mysqli_num_rows($checkResult)=0)
    

    Should be a double equal...

    if(mysqli_num_rows($checkResult)==0)
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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