假设我有两个现有的桌子,“狗”和“猫”:
dog_name | owner
---------+------
Sparky | Bob
Rover | Bob
Snoopy | Chuck
Odie | Jon
cat_name | owner
---------+------
Garfield | Jon
Muffy | Sam
Stupid | Bob
如何使用此输出编写查询?
owner | num_dogs | num_cats
------+----------+---------
Bob | 2 | 1
Chuck | 1 | 0
Sam | 0 | 1
Jon | 1 | 1
答案 0 :(得分:10)
select owner, sum(num_dogs), sum(num_cats) from
(select owner, 1 as num_dogs, 0 as num_cats from dogs
union
select owner, 0 as num_dogs, 1 as num_cats from cats)
group by owner
答案 1 :(得分:5)
我更喜欢这个:
select owner
, count(dog_name) dogs
, count(cat_name) cats
from cats FULL OUTER JOIN dogs ON (cats.owner = dogs.owner)
答案 2 :(得分:2)
在SQL Server 2005的T-SQL中(如果没有,则用内联子查询替换CTE):
WITH ownership AS (
SELECT owner, COUNT(dog_name) AS num_dogs, 0 AS num_cats -- counts all non-NULL dog_name
FROM dogs
GROUP BY owner
UNION
SELECT owner, 0 AS num_dogs, COUNT(cat_name) as num_cats -- counts all non-NULL cat_name
FROM cats
GROUP BY owner
)
SELECT ownership.owner
,SUM(ownership.num_dogs) AS num_dogs
,SUM(ownership.num_cats) as num_cats
FROM ownership
GROUP BY ownership.owner
答案 3 :(得分:1)
我从Cade Roux的优秀答案开始,但更改了WITH ... AS()以使用表变量,因为我最终使用类似查询的结果来获得更多的聚合函数。
-- Table variable declaration
DECLARE @RainingCatsDogs TABLE
(
Owner nvarchar(255),
num_cats int,
num_dogs int
)
-- Populate the table variable with data from the union of the two SELECT statements
INSERT INTO @RainingCatsDogs
-- Get the count of doggies
SELECT
owner, COUNT(dog_name) AS num_dogs, 0 AS num_cats
FROM
dogs
GROUP BY
owner
-- join the results from the two SELECT statements
UNION
-- Get the count of kittehs
SELECT
owner, 0 AS num_dogs, COUNT(cat_name) as num_cats
FROM
cats
GROUP BY
owner
-- From the table variable, you can calculate the summed results
SELECT
owner,
SUM(num_dogs),
SUM(num_cats)
FROM
@RainingCatsDogs
答案 4 :(得分:1)
如果您的数据库可以处理它,我会加倍努力FerranB's solution,并写一个令人讨厌的NATURAL FULL JOIN
解决方案。我的意思是,您最后一次有机会这样做是什么时候?
SELECT owner, COUNT(dog_name), COUNT(cat_name)
FROM cats
NATURAL FULL JOIN dogs
GROUP BY owner