因此,您想获得OrderField
每组最高的行吗?我会这样:
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 EDIT: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,以相同或更大的值返回t1
没有其他行t2
存在的行。当为NULL时,表示左外部联接未找到此类匹配项,因此在组中具有最大值。GroupId
OrderField
t2.*
t1
OrderField
没有等级,没有子查询。如果您在上拥有复合索引,这应该可以快速运行并使用“使用索引”优化对t2的访问(GroupId, OrderField)
。
关于性能,请参阅我对检索每个组中的最后一个记录的回答。我尝试了使用堆栈溢出数据转储的子查询方法和联接方法。区别非常明显:在我的测试中,join方法的运行速度快了278倍。
重要的是您必须具有正确的索引以获得最佳结果!
关于使用@Rank变量的方法,它不能像您编写的那样起作用,因为在查询处理完第一个表之后,@ Rank的值不会重置为零。我给你看一个例子。
我插入了一些虚拟数据,其中一个额外的字段为null,除了在我们知道每组最大的行上:
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,并且内部查询正确地返回了这些:
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,在没有连接条件的情况下运行查询,以强制所有行的笛卡尔积,并且我们还获取所有列:
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 |
+---------+---------+---------+------------+------+------+
从上面我们可以看到每组的最大排名是正确的,但是@Rank在处理第二个派生表时继续增加,直到7或更高。因此,第二个派生表中的等级根本不会与第一个派生表中的等级完全重叠。
您必须添加另一个派生表,以在处理两个表之间强制@Rank重置为零(并希望优化程序不要更改其对表的求值顺序,否则希望使用STRAIGHT_JOIN来防止这种情况):
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,此查询的优化很糟糕。它不能使用任何索引,它会创建两个临时表,对它们进行艰难的排序,甚至使用连接缓冲区,因为它在连接临时表时也无法使用索引。这是来自EXPLAIN
以下示例的输出:
+----+-------------+------------+--------+---------------+------+---------+------+------+---------------------------------+
| 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 |
+----+-------------+------------+--------+---------------+------+---------+------+------+---------------------------------+
而我使用左外部联接的解决方案优化得更好。它不使用临时表,甚至不使用报告"Using index"
,这意味着它可以仅使用索引来解决联接,而无需处理数据。
+----+-------------+-------+------+---------------+---------+---------+-----------------+------+--------------------------+
| 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 |
+----+-------------+-------+------+---------------+---------+---------+-----------------+------+--------------------------+
您可能会读到人们在他们的博客上声称“加入会使SQL变慢”的说法,但这是无稽之谈。最差的优化会使SQL变慢。