按投票排序评论?

时间:2010-11-15 02:58:37

标签: php sql

我正在创建一个类似于stackoverflow的应用程序,因为它有问题和答案的评级,我也有标签显示最早,最新和投票的评论。我无法按投票排序。

这是我的功能:

 /**
     *
     * @param int $threadid
     * @param string $tab
     * @param object $voting referencing Voting.class.php (may not be needed)
     * @return database query/array 
     */
    public function getComments($threadid, $tab = 'oldest', $voting = null) {
        if ($tab == 'oldest') {
            $sql = "SELECT * FROM comments WHERE threadid = :threadid ORDER BY date ASC";
        } else if ($tab == 'newest') {
            $sql = "SELECT * FROM comments WHERE threadid = :threadid ORDER BY date DESC";
        } else if ($tab == 'votes') {
            //i dont know what to do here? read below for more explanation
        } else {
            $sql = "SELECT * FROM comments WHERE threadid = :threadid ORDER BY date ASC";
        }

        $stmt = $this->db->prepare($sql);
        $stmt->bindParam(':threadid', $threadid);
        $stmt->execute();
        $row = $stmt->fetchAll(PDO::FETCH_ASSOC);
        return $row;
    }

这是我的数据库设计:

**comments:** | id | userid | threadid | message | date |

**commentsrating:** | userid | commentid | voteup | votedown |

如果评级与评论分开,则可以对符合其余代码的$tab == 'votes'进行查询吗?

最后是HTML:

<?php //Get comments
if (isset($_GET['tab'])) {
    $getComments = $thread->getComments($threadid, $_GET['tab'], $voting);
} else {
    $getComments = $thread->getComments($threadid);
} ?>

    <?php for ($i = 0; $i < count($getComments); $i++) { ?>
               <p>
    <?php echo $getComments[$i]['message']; ?>
               </p>
               <p>  
                   <span class="bid_votes_count" id="bid_votes_count<?php echo $getComments[$i]['id'] ?>">
    <?php echo $voting->getEffectiveCommentVotes($getComments[$i]['id']) . " votes"; ?>
                   </span>

                   <span class="bid_vote_buttons" id="bid_vote_buttons<?php echo $getComments[$i]['id'] ?>">
                      <a href="javascript:;" class="bid_vote_up" id="<?php echo $getComments[$i]['id'] ?>"></a>
                      <a href="javascript:;" class="bid_vote_down" id="<?php echo $getComments[$i]['id'] ?>"></a>
                  </span>
              </p>
    <?php } ?>

Thanks in advance!

4 个答案:

答案 0 :(得分:3)

rating字段添加到comments表格并手动维护或使用commentsrating表格上的触发器进行维护。

显然,将预先计算的评级值放在那里,现在你可以有一个简单而可怕的快速问题来执行。

答案 1 :(得分:1)

试一试:

$sql = "SELECT * FROM comments c JOIN (SELECT commentid, ". 
       "(SUM(voteup) - SUM(votedown)) votes, " . 
       "FROM commentsrating GROUP BY commentid) i ON " .
       "c.id = i.commentid WHERE c.threadid = :threadid " . 
       "ORDER BY i.votes DESC";

显然不是疯狂JOIN的最佳表现,但应该让你“有效”,之后你可以担心性能。

答案 2 :(得分:1)

最快的方法是将投票的运行记录存储在评论表中作为字段(如另一个答案中所述)。但是,您还应该考虑修改评论表以使其更有意义,并帮助解决统计问题。

评论: | userid | commentid |投票|

您应该只有投票的原因是评论评分只能进行一次投票。它不能同时进行投票和投票。

投票应为int(1)+/-。例如。可以是1或-1(理论上为0)。这样您就可以执行以下SQL:

SELECT c.id, c.message, c.date, SUM(cr.vote) AS 'votes'
FROM `comments` c
JOIN `commentsrating` cr ON c.id=cr.commentid
WHERE c.threadid=':threadid'
GROUP BY c.id
ORDER BY `votes`

查询将花费比order by total_votes字段更长的时间,但无论如何都应该这样做。

答案 3 :(得分:0)

SELECT * FROM `comments` WHERE `threadid`=:threadid ORDER BY (SELECT `voteup`-`votedown` FROM `commentsrating` WHERE `commentid`=`comments`.`id`) DESC

试试吗?

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