计算jQuery upvote插件中的总票数

Question

我真的很困惑解决这个问题。目前我的项目中有一个jQuery upvote插件。因此，每次我点击upvote它都会将db中的值保存为true，每次投票都将它保存在数据库中，并将值保存为false，并将另一列中的值保存为0.我所要做的就是计算总票数就像堆栈溢出一样。即使是现在，我对如何解决这个问题感到困惑。我最终在php中制作了这段代码，这也使我的整个程序变得缓慢：

$sample1 = $this->db->prepare("SELECT * from Ratings WHERE TopicID = :current");
$sample1->bindParam(':current', $id);
$sample1->execute();
$RES1 = $sample1->fetchAll(PDO::FETCH_ASSOC); 
$upVote = 0;

foreach ($RES1 as $mk){
     if(($mk['Upvote'] === 'true') && ($mk['Downvote'] ==='false')){
             $upVote++;
                                    }
     else if(( $mk['Upvote'] ==='false') && ($mk['Downvote'] === 'true')){
             $upVote--;
                                }
     else if(($mk['Upvote'] === 'false') && ($mk['Downvote'] === 'false')){
            $upVote--;
                                }
     else if(($mk['Downvote'] === 'false')){
             $upVote++;
                                }
     else if(($mk['Downvote'] === 'true') && ($mk['Upvote'] ==='0') || ($mk['Upvote'] === 'false')){
             $upVote--;
     }

 }

我的表结构：

Answer 1

我认为您应该在数据库端执行此操作并重写您的查询：

SELECT ((SELECT COUNT(*) FROM Ratings WHERE TopicID = :current AND Upvote = true) - (SELECT COUNT(*) FROM Ratings WHERE TopicID = :current AND Downvote = true)) AS total_votes

并且在PHP端只需获取total_votes：

$sample1 = $this->db->prepare("SELECT ((SELECT COUNT(*) FROM Ratings WHERE TopicID = :current AND Upvote = true) - (SELECT COUNT(*) FROM Ratings WHERE TopicID = :current AND Downvote = true)) AS total_votes");
$sample1->bindParam(':current', $id);
$sample1->execute();
$result = $sample1->fetch(); 
$total_votes = $result->total_votes;