Question

SELECT  a.gid, a.qid,
        SUM(a.result = 1) first_attempt_correct,
        SUM(a.result = 0) first_attempt_incorrect
        FROM maf_game_stats a
                LEFT JOIN
                (
                    SELECT USERID, gid, QID, MIN(ACCESSTIME) min_date
                    FROM maf_game_stats
                    GROUP BY USERID, gid, qid
                )  b ON a.USERID = b.USERID AND
                    a.gid = b.gid AND
                    a.qid = b.qid AND
                    a.ACCESSTIME = b.min_date
        WHERE   a.gid ='2'
        AND a.ACCESSTIME >= '2014-09-01' AND a.ACCESSTIME < '2014-09-10' + INTERVAL 1 DAY 
        GROUP BY a.gid, a.qid

以上给定的查询执行时间40到50秒我怎样才能打破这些查询

Answer 1

添加

在哪里gid =＆＃39; 2＆＃39;

到内部左连接查询。没有它，你正在查询所有＆＃34; gid＆＃34;值。所以，因为你的外部查询只关心＆＃34; gid = 2＆＃34;，谁在乎你是否有＆＃34; gid = 1,3,2903，甚至389433＆＃34;在内部结果集中。有相同的，它应该是瞬间的。

此外，您可能需要将日期条件应用于内部查询，但不确定您要从中获取的确切内容。

Answer 2

我建议使用not exists写一下，然后添加一些索引。查询是：

SELECT  gs.gid, gs.qid,
        SUM(gs.result = 1) first_attempt_correct,
        SUM(gs.result = 0) first_attempt_incorrect
FROM maf_game_stats gs
WHERE gs.gid ='2' AND
      gs.ACCESSTIME >= '2014-09-01' AND gs.ACCESSTIME < '2014-09-10' + INTERVAL 1 DAY AND
      NOT EXISTS (SELECT 1
                  FROM maf_game_stats gs2
                  WHERE gs2.gid = gs.gid and gs2.qid = gs.qid and gs2.AccessTime > gs2.AccessTime
                 )
GROUP BY gs.gid, gs.qid;

索引为：maf_game_starts(gid, accesstime, qid, result)和maf_game_stats(gid, qid, AccessTime)。

Answer 3

最终解决方案如下：

SELECT  a.gid, a.qid,
        SUM(a.result = 1) first_attempt_correct,
        SUM(a.result = 0) first_attempt_incorrect
        FROM maf_game_stats a
                INNER JOIN
                (
                    SELECT USERID, gid, QID, MIN(ACCESSTIME) min_date
                    FROM maf_game_stats
                    WHERE   gid ='2' 
                    AND ACCESSTIME >= '2014-09-01' AND ACCESSTIME < '2014-09-12' + INTERVAL 1 DAY
                    GROUP BY USERID, gid, qid
                )  b ON a.USERID = b.USERID AND
                    a.gid = b.gid AND
                    a.qid = b.qid AND
                    a.ACCESSTIME = b.min_date
        WHERE   a.gid ='2'
        AND a.ACCESSTIME >= '2014-09-01' AND a.ACCESSTIME < '2014-09-12' + INTERVAL 1 DAY 
        GROUP BY a.gid, a.qid

mysql查询内连接需要40到45秒

3 个答案: