我正在处理一个与以下内容非常相似的查询：
CREATE TABLE #test (a char(1), b char(1))
INSERT INTO #test(a,b) VALUES
('A',NULL),
('A','B'),
('B',NULL),
('B',NULL)
SELECT DISTINCT a,b FROM #test
DROP TABLE #test
结果，不出所料，
a b
-------
A NULL
A B
B NULL
我实际上希望看到的输出是：
a b
-------
A B
B NULL
也就是说，如果一列在某些记录中具有值但在其他记录中没有值，我想为该列丢弃带有 NULL 的行。但是，如果一列的所有记录都具有 NULL 值，我想保留该 NULL。
在单个查询中执行此操作的最简单/最优雅的方法是什么？
我有一种感觉，如果我在星期五下午没有筋疲力尽，这将很简单。
试试这个：
select distinct * from test
where b is not null or a in (
select a from test
group by a
having max(b) is null)
你可以在这里得到小提琴。
请注意，如果 中只能有一个非空值b，则可以简化为：
select a, max(b) from test
group by a
;WITH CTE
AS
(
SELECT DISTINCT * FROM #test
)
SELECT a,b
FROM CTE
ORDER BY CASE WHEN b IS NULL THEN 9999 ELSE b END ;
Try this:
create table test(
x char(1),
y char(1)
);
insert into test(x,y) values
('a',null),
('a','b'),
('b', null),
('b', null)
Query:
with has_all_y_null as
(
select x
from test
group by x
having sum(case when y is null then 1 end) = count(x)
)
select distinct x,y from test
where
(
-- if a column has a value in some records but not in others,
x not in (select x from has_all_y_null)
-- I want to throw out the row with NULL
and y is not null
)
or
-- However, if a column has a NULL value for all records,
-- I want to preserve that NULL
(x in (select x from has_all_y_null))
order by x,y
Output:
X Y
A B
B NULL
Live test: http://sqlfiddle.com/#!3/259d6/16
EDIT
Seeing Mosty's answer, I simplified my code:
with has_all_y_null as
(
select x
from test
group by x
-- having sum(case when y is null then 1 end) = count(x)
-- should have thought of this instead of the code above. Mosty's logic is good:
having max(y) is null
)
select distinct x,y from test
where
y is not null
or
(x in (select x from has_all_y_null))
order by x,y
I just prefer CTE approach, it has a more self-documenting logic :-)
You can also put documentation on non-CTE approach, if you are conscious of doing so:
select distinct * from test
where b is not null or a in
( -- has all b null
select a from test
group by a
having max(b) is null)
SELECT DISTINCT t.a, t.b
FROM #test t
WHERE b IS NOT NULL
OR NOT EXISTS (SELECT 1 FROM #test u WHERE t.a = u.a AND u.b IS NOT NULL)
ORDER BY t.a, t.b
这是一个非常奇怪的要求。我想知道你怎么需要它。
SELECT DISTINCT a, b
FROM test t
WHERE NOT ( b IS NULL
AND EXISTS
( SELECT *
FROM test ta
WHERE ta.a = t.a
AND ta.b IS NOT NULL
)
)
AND NOT ( a IS NULL
AND EXISTS
( SELECT *
FROM test tb
WHERE tb.b = t.b
AND tb.a IS NOT NULL
)
)
Well, I don't particularly like this solution, but it seems the most appropriate to me. Note that your description of what you want sounds exactly like what you get with a LEFT JOIN, so:
SELECT DISTINCT a.a, b.b
FROM #test a
LEFT JOIN #test b ON a.a = b.a
AND b.b IS NOT NULL
SELECT a,b FROM #test t where b is not null
union
SELECT a,b FROM #test t where b is null
and not exists(select 1 from #test where a=t.a and b is not null)
Result:
a b
---- ----
A B
B NULL
I'll just put here a mix of two answers that solved my issue, because my View was more complex
--IdCompe int,
--Nome varchar(30),
--IdVanBanco int,
--IdVan int
--FlagAtivo bit,
--FlagPrincipal bit
select IdCompe
, Nome
, max(IdVanBanco)
, max(IdVan)
, CAST(MAX(CAST(FlagAtivo as INT)) AS BIT) FlagAtivo
, CAST(MAX(CAST(FlagPrincipal as INT)) AS BIT) FlagPrincipal
from VwVanBanco
where IdVan = {IdVan} or IdVan is null
group by IdCompe, Nome order by IdCompe asc
Thanks to mosty mostacho and kenwarner