Question

我需要一个SQL查询，它提供一个表的内容和第二个表的计数。我试过了，不能。

我有一个＆＃34;评论＆＃34;表。在另一个名为＆＃34; likings＆＃34;的表中，有许多评论喜欢。

在我的算法中，在＆＃34; likings＆＃34;表格中有一个名为＆＃34; likeType＆＃34;的列，如果它是＆＃34; 1＆＃34;这意味着喜欢，否则（如果它是＆＃34; 0＆＃34;）意味着不喜欢。我需要发表评论来自＆＃34; comments＆＃34;表，来自＆＃34; likeType=1＆＃34;的likeType=0和likings的计数一个查询中的表。

这是我最好的尝试，但没有成功：

PHP方面：

   $getFirst8Comments = "SELECT
                            episodecomments.cmtID,
                            episodecomments.cmtConEpisode,
                            episodecomments.cmtOwner,
                            episodecomments.cmtDate,
                            episodecomments.cmtContent,
                            episodecomments.cmtSpoiler,
                            SUM(IF(episodecommentsliking.liType='1', 1, 0)) AS likes,
                            SUM(IF(episodecommentsliking.liType='0', 1, 0)) AS dislikes
                        FROM episodecomments
                        LEFT JOIN episodecommentsliking
                        ON episodecomments.cmtID = episodecommentsliking.liCmtID
                        GROUP BY
                            episodecomments.cmtID,
                            episodecomments.cmtConEpisode,
                            episodecomments.cmtOwner,
                            episodecomments.cmtDate,
                            episodecomments.cmtContent,
                            episodecomments.cmtSpoiler
                        WHERE episodecomments.cmtConEpisode='$epID'
                        ORDER BY episodecomments.cmtID
                        DESC LIMIT 8";

   while ($getF8C = mysqli_fetch_array($getFirst8Comments))
   {
      echo "Something coming through-<br>";
   }

   $getFirst8Comments = "SELECT
                            episodecomments.cmtID,
                            episodecomments.cmtConEpisode,
                            episodecomments.cmtOwner,
                            episodecomments.cmtDate,
                            episodecomments.cmtContent,
                            episodecomments.cmtSpoiler,
                            SUM(IF(episodecommentsliking.liType='1', 1, 0)) AS likes,
                            SUM(IF(episodecommentsliking.liType='0', 1, 0)) AS dislikes
                        FROM episodecomments
                        LEFT JOIN episodecommentsliking
                        ON episodecomments.cmtID = episodecommentsliking.liCmtID
                        GROUP BY
                            episodecomments.cmtID,
                            episodecomments.cmtConEpisode,
                            episodecomments.cmtOwner,
                            episodecomments.cmtDate,
                            episodecomments.cmtContent,
                            episodecomments.cmtSpoiler
                        WHERE episodecomments.cmtConEpisode='$epID'
                        ORDER BY episodecomments.cmtID
                        DESC LIMIT 8";

   while ($getF8C = mysqli_fetch_array($getFirst8Comments))
   {
      echo "Something coming through-<br>";
   }

弹出错误：

提前致谢。

Answer 1

MySQL不支持Window。您还需要告知它COUNT对这些聚合（和）函数的内容。

请尝试使用GROUP BY：{SUM如果您计算的是COUNT与非NULL值，则可以正常工作 - 请参阅MySQL documentation）< / p>

NULL

另外 - 如果可能存在没有相似数据的评论记录，请使用“SELECT comments.cmtID, comments.cmtConEpisode, comments.cmtOwner, comments.cmtDate, comments.cmtContent, comments.cmtSpoiler, SUM(IF(commentsliking.liType='1', 1, 0)) AS likes, SUM(IF(commentsliking.liType='0', 1, 0)) AS dislikes FROM comments INNER JOIN commentsliking ON comments.cmtID = commentsliking.liCmtID GROUP BY comments.cmtID, comments.cmtConEpisode, comments.cmtOwner, comments.cmtDate, comments.cmtContent, comments.cmtSpoiler ORDER BY comments.cmtID DESC LIMIT 8”。

我已经进一步更新了我的答案，以解决自LEFT JOIN命令引发的错误。

检查PHP documentation以获取有关如何正确使用mysqli_*函数的完整详细信息 - 特别是关于它需要mysqli_fetch_array()类型作为参数，而不是{{1} [SQL]）

在一个SQL查询中从另一个表中获取一个正常表和计数（JOIN）

1 个答案: