我需要根据2个字段计算行数进行分组。
id group_id strain_id death_date death_cause status
-----------------------------------------------------------------------
1 512 164 2015-12-01 Culled P
2 512 164 2015-12-02 Culled A
3 512 164 2015-12-02 Surplus B
4 512 230 2015-12-06 Culled A
5 512 164 2015-12-28 Culled A
6 512 230 2016-01-20 Culled B
7 512 230 2016-01-20 Surplus P
8 512 164 NULL NULL P
9 512 230 NULL NULL B
10 512 230 NULL NULL A
11 512 164 2016-01-25 Culled B
12 512 164 2016-02-29 Culled A
13 512 230 2016-02-03 Surplus P
14 512 230 2016-02-03 Culled A
id group_name
--------------
512 Mice
id strain_name
----------------
164 Strain 1
230 Strain 2
id total_animals alive_count dead_count
----------------------------------------------------------------------
512 14 3 11
id animal_id history_type history_date
--------------------------------------------------------
1001 2 MA 2015-11-20
1002 2 MR 2015-12-01
1003 3 MA 2015-12-01
1004 6 FA 2015-12-21
1005 9 FA 2016-02-07
1006 10 MA 2016-01-27
1007 11 FA 2015-12-12
因此,当我按照strain_id
和death_cause
对它们进行分组时,这就是它们看起来应该是直观的:
1 512 164 2015-12-01 Culled P
2 512 164 2015-12-02 Culled A
5 512 164 2015-12-28 Culled A
11 512 164 2016-01-25 Culled B
12 512 164 2016-02-29 Culled A
3 512 164 2015-12-02 Surplus B
4 512 230 2015-12-06 Culled A
6 512 230 2016-01-20 Culled B
14 512 230 2016-02-03 Culled A
7 512 230 2016-01-20 Surplus P
13 512 230 2016-02-03 Surplus P
我想从SQL查询中得到以下结果:
g_name s_name d_cause a_total c_alive c_dead c_pup c_breeder c_total
------------------------------------------------------------------------------
Mice Strain 1 Culled 12 3 9 1 2 5
Mice Strain 1 Surplus 12 3 9 0 1 1
Mice Strain 2 Culled 12 3 9 0 1 3
Mice Strain 2 Surplus 12 3 9 2 0 2
基本上我想计算使用2个类别的动物数量,在这种情况下是strain_name
和death_cause
请注意,对于要被视为繁殖者(c_breeder
)的动物,我必须查看“交配历史记录”表并检查animal_id
是否曾有过任何这些代码{{1 }或MA
。
我在FA
,INNER JOIN
和groups
上使用group_animal_count
。我对strains
使用LEFT JOIN
,因为状态为mating_history
的动物将不会在该表中记录,因为它们只是幼崽并且不会参与交配。
答案 0 :(得分:0)
尝试:
SELECT group_name, strain_name,death_cause, count(*) as Total
FROM ANIMALS a
JOIN GROUPS g ON a.group_id = g.id
JOIN STRAIN s ON a.strain_id = s.id
GROUP BY group_name, strain_name,death_cause
答案 1 :(得分:0)
您可以在加入表格之前进行聚合:
SELECT group_name, strain_name, death_cause, total
FROM (
SELECT group_id,
strain_id,
death_cause,
COUNT(*) AS total
FROM animals
GROUP BY group_id, strain_id, death_cause
) a
INNER JOIN groups g ON ( g.group_id = a.group_id )
INNER JOIN strain s ON ( s.strain_id = a.strain_id );