这是整个查询...
SELECT s.*, (SELECT url FROM show_medias WHERE show_id = s.id AND is_primary = 1) AS media_url
FROM (shows As s)
WHERE `s`.`id` IN (
SELECT DISTINCT st.show_id
FROM show_time_schedules AS sts
LEFT JOIN show_times AS st ON st.id = sts.show_time_id
WHERE sts.schedule_date BETWEEN CAST('2012-01-10' AS date) AND CAST('2012-01-14' AS date)
)
AND `s`.`is_active` = 1
ORDER BY s.name asc
如果...
SELECT url FROM show_medias WHERE show_id = s.id AND is_primary = 1
(0.0004 sec)
和...
SELECT DISTINCT st.show_id
FROM show_time_schedules AS sts
LEFT JOIN show_times AS st ON st.id = sts.show_time_id
WHERE sts.schedule_date BETWEEN CAST('2012-01-10' AS date) AND CAST('2012-01-14' AS date)
(0.0061 sec)
有明显的原因......
SELECT s.*, (inner query 1) AS media_url
FROM (shows As s)
WHERE `s`.`id` IN ( inner query 2 )
AND `s`.`is_active` = 1
ORDER BY s.name asc
正在使用5.7245 sec？
EXPLAIN EXTENDED
id select_type table type possible_keys key key_len ref rows filtered Extra
1 PRIMARY s ALL NULL NULL NULL NULL 151 100.00 Using where; Using filesort
3 DEPENDENT SUBQUERY sts ALL NULL NULL NULL NULL 26290 100.00 Using where; Using temporary
3 DEPENDENT SUBQUERY st eq_ref PRIMARY PRIMARY 4 bvcdb.sts.show_time_id 1 100.00 Using where
2 DEPENDENT SUBQUERY show_medias ALL NULL NULL NULL NULL 159 100.00 Using where
答案 0 :(得分：3)
您始终可以使用EXPLAIN or EXPLAIN EXTENDED查看MySql使用查询执行的操作
你也可以用稍微不同的方式编写查询，试试以下内容吗？
SELECT s.*,
sm.url AS media_url
FROM shows AS s
INNER JOIN show_medias AS sm ON s.id = SM.show_id
WHERE `s`.`id` IN (
SELECT DISTINCT st.show_id
FROM show_time_schedules AS sts
LEFT JOIN show_times AS st ON st.id = sts.show_time_id
WHERE sts.schedule_date BETWEEN CAST('2012-01-10' AS date) AND CAST('2012-01-14' AS date)
)
AND `s`.`is_active` = 1
AND sm.is_primary = 1
ORDER BY s.name asc
看看它的效果会很有趣。我希望它会更快，因为目前我认为MySql将为你拥有的每个节目运行内部查询1（这样一个查询将运行多次。连接应该更有效。）
如果您想要所有在show_medias中没有行的节目，请使用LEFT JOIN替换INNER JOIN。
编辑：
我很快就会看看您的EXPLAIN EXTENDED，我也想知道您是否想要尝试以下内容;它删除所有子查询：
SELECT DISTINCT s.*,
sm.url AS media_url
FROM shows AS s
INNER JOIN show_medias AS sm ON s.id = SM.show_id
INNER JOIN show_times AS st ON (s.id = st.show_id)
RIGHT JOIN show_time_schedules AS sts ON (st.id = sts.show_time_id)
WHERE `s`.`is_active` = 1
AND sm.is_primary = 1
AND sts.schedule_date BETWEEN CAST('2012-01-10' AS date) AND CAST('2012-01-14' AS date)
ORDER BY s.name asc
（也可以看看EXPLAIN EXTENDED对你来说很好 - 你可以将它添加到这个评论中。）
进一步编辑：
在您的EXPLAIN EXTENDED（a good start on how to read these is here）
上USING FILESORT和USING TEMPORARY都是关键指标。希望我推荐的第二个查询应该删除任何TEMPORARY表（在子查询中）。然后尝试关闭ORDER BY以查看是否会产生影响（我们可以将其添加到目前为止的结果中： - ）
我还可以看到查询可能在许多索引查找中丢失;所有id列都是索引匹配的主要候选者（通常为index caveats）。我还尝试添加这些索引，然后再次运行EXPLAIN EXTENDED以查看现在的差异（正如我们已经从上面的评论中了解到的那样编辑！）
答案 1 :(得分：2)
这是CTE解决方案:(我的不好，mysql没有CTE，但问题太通用了）
WITH RECURSIVE tree AS (
SELECT t0.id
, t0.study_start_time
, t0.study_end_time
FROM tab t0
WHERE NOT EXISTS (SELECT * FROM tab nx
WHERE nx.id=t0.id
AND nx.study_end_time = t0.study_start_time
)
UNION
SELECT tt.id
,tt.study_start_time
,t1.study_end_time
FROM tab t1
JOIN tree tt ON t1.id=tt.id
AND t1.study_start_time = tt.study_end_time
)
SELECT * FROM tree
WHERE NOT EXISTS (SELECT * FROM tab nx
WHERE nx.id=tree.id
AND tree.study_end_time = nx.study_start_time
)
ORDER BY id
;
结果：
CREATE TABLE
INSERT 0 15
id | study_start_time | study_end_time
------+------------------+----------------
1234 | 168 | 480
2345 | 175 | 233
2345 | 400 | 425
4567 | 200 | 225
4567 | 250 | 289
4567 | 300 | 310
4567 | 320 | 340
4567 | 360 | 390
(8 rows)
QUERY计划（在添加明显的PK和索引之后）：
DROP TABLE
NOTICE: CREATE TABLE / PRIMARY KEY will create implicit index "tab_pkey" for table "tab"
CREATE TABLE
CREATE INDEX
INSERT 0 15
QUERY PLAN
-------------------------------------------------------------------------------------------------------------------------------------------
Merge Anti Join (cost=16209.59..16292.13 rows=6386 width=12) (actual time=0.189..0.193 rows=8 loops=1)
Merge Cond: ((tree.id = nx.id) AND (tree.study_end_time = nx.study_start_time))
CTE tree
-> Recursive Union (cost=0.00..15348.09 rows=8515 width=12) (actual time=0.022..0.136 rows=15 loops=1)
-> Merge Anti Join (cost=0.00..175.04 rows=1455 width=12) (actual time=0.019..0.041 rows=8 loops=1)
Merge Cond: ((t0.id = nx.id) AND (t0.study_start_time = nx.study_end_time))
-> Index Scan using tab_pkey on tab t0 (cost=0.00..77.35 rows=1940 width=12) (actual time=0.010..0.018 rows=15 loops=1)
-> Index Scan using sssss on tab nx (cost=0.00..77.35 rows=1940 width=8) (actual time=0.003..0.008 rows=14 loops=1)
-> Merge Join (cost=1297.04..1500.28 rows=706 width=12) (actual time=0.010..0.012 rows=1 loops=6)
Merge Cond: ((t1.id = tt.id) AND (t1.study_start_time = tt.study_end_time))
-> Index Scan using tab_pkey on tab t1 (cost=0.00..77.35 rows=1940 width=12) (actual time=0.001..0.004 rows=9 loops=6)
-> Sort (cost=1297.04..1333.42 rows=14550 width=12) (actual time=0.006..0.006 rows=2 loops=6)
Sort Key: tt.id, tt.study_end_time
Sort Method: quicksort Memory: 25kB
-> WorkTable Scan on tree tt (cost=0.00..291.00 rows=14550 width=12) (actual time=0.000..0.001 rows=2 loops=6)
-> Sort (cost=726.15..747.44 rows=8515 width=12) (actual time=0.166..0.169 rows=15 loops=1)
Sort Key: tree.id, tree.study_end_time
Sort Method: quicksort Memory: 25kB
-> CTE Scan on tree (cost=0.00..170.30 rows=8515 width=12) (actual time=0.025..0.149 rows=15 loops=1)
-> Sort (cost=135.34..140.19 rows=1940 width=8) (actual time=0.018..0.018 rows=15 loops=1)
Sort Key: nx.id, nx.study_start_time
Sort Method: quicksort Memory: 25kB
-> Seq Scan on tab nx (cost=0.00..29.40 rows=1940 width=8) (actual time=0.003..0.004 rows=15 loops=1)
Total runtime: 0.454 ms
(24 rows)