怎么做?
这个问题的前标题是“在带有子查询的复杂查询中使用排名 (@Rank := @Rank + 1) - 会起作用吗?”因为我一直在寻找使用排名的解决方案,但是现在我看到 Bill 发布的解决方案要好得多。
原问题:
我正在尝试编写一个查询,该查询将在给定特定顺序的情况下从每个组中获取最后一条记录:
SET @Rank=0;
select s.*
from (select GroupId, max(Rank) AS MaxRank
from (select GroupId, @Rank := @Rank + 1 AS Rank
from Table
order by OrderField
) as t
group by GroupId) as t
join (
select *, @Rank := @Rank + 1 AS Rank
from Table
order by OrderField
) as s
on t.GroupId = s.GroupId and t.MaxRank = s.Rank
order by OrderField
表达式 @Rank := @Rank + 1
通常用于排名,但对我来说,在 2 个子查询中使用它看起来很可疑,但只初始化了一次。会这样吗?
其次,它是否适用于一个被多次评估的子查询?就像 where (or having) 子句中的子查询(上面的另一种写法):
SET @Rank=0;
select Table.*, @Rank := @Rank + 1 AS Rank
from Table
having Rank = (select max(Rank) AS MaxRank
from (select GroupId, @Rank := @Rank + 1 AS Rank
from Table as t0
order by OrderField
) as t
where t.GroupId = table.GroupId
)
order by OrderField
提前致谢!
最佳答案
所以您想获得每组中 OrderField
最高的行吗?我会这样做:
SELECT t1.*
FROM `Table` AS t1
LEFT OUTER JOIN `Table` AS t2
ON t1.GroupId = t2.GroupId AND t1.OrderField < t2.OrderField
WHERE t2.GroupId IS NULL
ORDER BY t1.OrderField; // not needed! (note by Tomas)
(Tomas 编辑:如果同一组中有更多具有相同 OrderField 的记录,而您恰好需要其中之一,则可能需要扩展条件:
SELECT t1.*
FROM `Table` AS t1
LEFT OUTER JOIN `Table` AS t2
ON t1.GroupId = t2.GroupId
AND (t1.OrderField < t2.OrderField
OR (t1.OrderField = t2.OrderField AND t1.Id < t2.Id))
WHERE t2.GroupId IS NULL
编辑结束。)
换句话说,返回没有其他行 t2
的行 t1
具有相同的 GroupId
和更大的 OrderField
。当 t2.*
为 NULL 时,表示左外连接没有找到这样的匹配,因此 t1
具有最大的 OrderField
值团体。
没有等级,没有子查询。如果您在 (GroupId, OrderField)
上有一个复合索引,这应该运行得很快并使用“使用索引”优化对 t2 的访问。
关于性能,请参阅我对 Retrieving the last record in each group 的回答.我尝试了使用 Stack Overflow 数据转储的子查询方法和连接方法。差异是显着的:在我的测试中,join 方法的运行速度提高了 278 倍。
拥有正确的索引以获得最佳结果很重要!
关于使用@Rank 变量的方法,它不会像您编写的那样工作,因为在查询处理完第一个表后@Rank 的值不会重置为零。我给你举个例子。
我插入了一些虚拟数据,除了我们知道每组最大的行之外,还有一个为空的额外字段:
select * from `Table`;
+---------+------------+------+
| GroupId | OrderField | foo |
+---------+------------+------+
| 10 | 10 | NULL |
| 10 | 20 | NULL |
| 10 | 30 | foo |
| 20 | 40 | NULL |
| 20 | 50 | NULL |
| 20 | 60 | foo |
+---------+------------+------+
我们可以证明第一组的排名增加到三,第二组的排名增加到六,并且内部查询正确返回了这些:
select GroupId, max(Rank) AS MaxRank
from (
select GroupId, @Rank := @Rank + 1 AS Rank
from `Table`
order by OrderField) as t
group by GroupId
+---------+---------+
| GroupId | MaxRank |
+---------+---------+
| 10 | 3 |
| 20 | 6 |
+---------+---------+
现在运行没有连接条件的查询,强制所有行的笛卡尔积,我们还获取所有列:
select s.*, t.*
from (select GroupId, max(Rank) AS MaxRank
from (select GroupId, @Rank := @Rank + 1 AS Rank
from `Table`
order by OrderField
) as t
group by GroupId) as t
join (
select *, @Rank := @Rank + 1 AS Rank
from `Table`
order by OrderField
) as s
-- on t.GroupId = s.GroupId and t.MaxRank = s.Rank
order by OrderField;
+---------+---------+---------+------------+------+------+
| GroupId | MaxRank | GroupId | OrderField | foo | Rank |
+---------+---------+---------+------------+------+------+
| 10 | 3 | 10 | 10 | NULL | 7 |
| 20 | 6 | 10 | 10 | NULL | 7 |
| 10 | 3 | 10 | 20 | NULL | 8 |
| 20 | 6 | 10 | 20 | NULL | 8 |
| 20 | 6 | 10 | 30 | foo | 9 |
| 10 | 3 | 10 | 30 | foo | 9 |
| 10 | 3 | 20 | 40 | NULL | 10 |
| 20 | 6 | 20 | 40 | NULL | 10 |
| 10 | 3 | 20 | 50 | NULL | 11 |
| 20 | 6 | 20 | 50 | NULL | 11 |
| 20 | 6 | 20 | 60 | foo | 12 |
| 10 | 3 | 20 | 60 | foo | 12 |
+---------+---------+---------+------------+------+------+
从上面我们可以看出,每个组的最大排名是正确的,但是随着它处理第二个派生表,@Rank 会继续增加,达到 7 或更高。所以第二个派生表的排名永远不会与第一个派生表的排名重叠。
您必须添加另一个派生表以强制 @Rank 在处理两个表之间重置为零(并希望优化器不会更改它评估表的顺序,或者使用 STRAIGHT_JOIN 来防止这种情况发生):
select s.*
from (select GroupId, max(Rank) AS MaxRank
from (select GroupId, @Rank := @Rank + 1 AS Rank
from `Table`
order by OrderField
) as t
group by GroupId) as t
join (select @Rank := 0) r -- RESET @Rank TO ZERO HERE
join (
select *, @Rank := @Rank + 1 AS Rank
from `Table`
order by OrderField
) as s
on t.GroupId = s.GroupId and t.MaxRank = s.Rank
order by OrderField;
+---------+------------+------+------+
| GroupId | OrderField | foo | Rank |
+---------+------------+------+------+
| 10 | 30 | foo | 3 |
| 20 | 60 | foo | 6 |
+---------+------------+------+------+
但是这个查询的优化很糟糕。它不能使用任何索引,它创建两个临时表,对它们进行硬排序,甚至使用连接缓冲区,因为它在连接临时表时也不能使用索引。这是 EXPLAIN
的示例输出:
+----+-------------+------------+--------+---------------+------+---------+------+------+---------------------------------+
| id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra |
+----+-------------+------------+--------+---------------+------+---------+------+------+---------------------------------+
| 1 | PRIMARY | <derived4> | system | NULL | NULL | NULL | NULL | 1 | Using temporary; Using filesort |
| 1 | PRIMARY | <derived2> | ALL | NULL | NULL | NULL | NULL | 2 | |
| 1 | PRIMARY | <derived5> | ALL | NULL | NULL | NULL | NULL | 6 | Using where; Using join buffer |
| 5 | DERIVED | Table | ALL | NULL | NULL | NULL | NULL | 6 | Using filesort |
| 4 | DERIVED | NULL | NULL | NULL | NULL | NULL | NULL | NULL | No tables used |
| 2 | DERIVED | <derived3> | ALL | NULL | NULL | NULL | NULL | 6 | Using temporary; Using filesort |
| 3 | DERIVED | Table | ALL | NULL | NULL | NULL | NULL | 6 | Using filesort |
+----+-------------+------------+--------+---------------+------+---------+------+------+---------------------------------+
而我使用左外连接的解决方案优化得更好。它不使用临时表,甚至报告 "Using index"
这意味着它可以仅使用索引来解析连接,而无需接触数据。
+----+-------------+-------+------+---------------+---------+---------+-----------------+------+--------------------------+
| id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra |
+----+-------------+-------+------+---------------+---------+---------+-----------------+------+--------------------------+
| 1 | SIMPLE | t1 | ALL | NULL | NULL | NULL | NULL | 6 | Using filesort |
| 1 | SIMPLE | t2 | ref | GroupId | GroupId | 5 | test.t1.GroupId | 1 | Using where; Using index |
+----+-------------+-------+------+---------------+---------+---------+-----------------+------+--------------------------+
您可能会读到有人在他们的博客上声称“联接使 SQL 变慢”,但这是无稽之谈。优化不佳导致 SQL 变慢。
关于mysql - 获取每组最高/最小 <whatever> 的记录,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/8748986/