我希望有人能指出我正确的方向。我有两个 任意表：
TableA TableB
---------- ----------
(cola-int) (colb-int) (colc-char) (cola-int) (colb-int) (colc-char)
1 10 Data1 1 NULL Match1
2 20 Data2 2 NULL Match2
1 10 Data3 2 30 Match3
2 30 Data4
2 30 Data5
查询：
select a.cola, a.colb, a.colc, b.cola, b.colb, b.colc
from TableA a
inner join TableB b on b.cola = a.cola
where a.colb = b.colb or b.colb is null
产生以下结果：
cola colb colc cola colb colc
1 10 Data1 1 NULL Match1
1 10 Data3 1 NULL Match1
2 20 Data2 2 NULL Match2
2 30 Data4 2 NULL Match2
2 30 Data5 2 NULL Match2
2 30 Data4 2 30 Match3
2 30 Data5 2 30 Match3
我怎样才能返回Data3和Data5的匹配3和
不是Data4和Data5的Match2，因为它们不匹配
在colb？我希望我的结果集看起来像这样：
cola colb colc cola colb colc
1 10 Data1 1 NULL Match1
1 10 Data3 1 NULL Match1
2 20 Data2 2 NULL Match2
2 30 Data4 2 30 Match3
2 30 Data5 2 30 Match3
我尝试使用外连接无效并尝试使用独占或
在具有以下查询的位置：
select a.cola, a.colb, a.colc, b.cola, b.colb, b.colc
from TableA a
left outer join TableB b on b.cola = a.cola
where (case when (a.colb = b.colb) then 1 else 0 end) +
(case when (b.colb is null) then 1 else 0 end) = 1
产生以下结果：
cola colb colc cola colb colc
1 10 Data1 1 NULL Match1
2 20 Data2 2 NULL Match2
1 10 Data3 1 NULL Match1
2 30 Data4 2 NULL Match2
2 30 Data4 2 30 Match3
2 30 Data5 2 NULL Match2
2 30 Data5 2 30 Match3
仍然为Data4和Data5提供Match2。
谢谢！
答案 0 :(得分：0)
在查看案例陈述版本（我认为在上面的评论中回答我的问题）之后对此进行了尝试。
您可以使用窗口函数或相关子查询来解决此问题：
窗口功能（比较a.cola的b.colb与max（col.b），看看我们是否要“保留”或“删除”记录。 ）
SELECT
acola,acolb,acolc,bcola,bcolb,bcolc
FROM
(
SELECT
a.cola as acola,
a.colb as acolb,
a.colc as acolc,
b.cola as bcola,
b.colb as bcolb,
b.colc as bcolc,
CASE
WHEN
b.colb = max(b.colb) OVER (PARTITION BY a.colc) OR
b.colb IS NULL AND count(*) OVER (PARTITION BY a.colc) = 1
THEN 'KEEP'
ELSE 'DROP'
END AS KeepOrDrop
FROM TableA a
INNER JOIN TableB b ON b.cola = a.cola
) as c
WHERE c.KeepOrDrop = 'Keep'
相关子查询（我们在WHERE子句中的相关子查询中找到max b.colb）
SELECT
a.cola as acola,
a.colb as acolb,
a.colc as acolc,
b.cola as bcola,
b.colb as bcolb,
b.colc as bcolc,
FROM TableA a
INNER JOIN TableB b ON b.cola = a.cola
WHERE coalesce(b.colb,0) IN (SELECT max(coalesce(colb,0)) FROM TableB WHERE cola = a.cola)
答案 1 :(得分：0)
我想出了这个：
/* Test data. */
DECLARE @TA TABLE (cola int, colb int, colc char(10))
DECLARE @TB TABLE (cola int, colb int, colc char(10))
INSERT INTO @TA VALUES (1, 10, 'Data1')
INSERT INTO @TA VALUES (2, 20, 'Data2')
INSERT INTO @TA VALUES (1, 10, 'Data3')
INSERT INTO @TA VALUES (2, 30, 'Data4')
INSERT INTO @TA VALUES (2, 30, 'Data5')
INSERT INTO @TB VALUES (1, NULL, 'Match1')
INSERT INTO @TB VALUES (2, NULL, 'Match2')
INSERT INTO @TB VALUES (2, 30, 'Match3')
/* End Test data. */
/* First get all that match on both fields [cola] and [colb]. */
; WITH BothMatch As (
SELECT
A.cola As acola
, A.colb As acolb
, A.colc As acolc
, B.cola As bcola
, B.colb As bcolb
, B.colc As bcolc
FROM
@TA As A
INNER JOIN @TB As B
ON A.cola = B.cola
AND A.colb = B.colb
)
SELECT
*
FROM
BothMatch
UNION
/* Then match all by [cola] that do not already exist in BothMatch. */
SELECT
*
FROM
@TA As A
INNER JOIN @TB As B
ON A.cola = B.cola
AND B.colb IS NULL
WHERE
NOT EXISTS
(
SELECT
BothMatch.acola
FROM
BothMatch
WHERE
BothMatch.acola = A.cola
AND BothMatch.acolb = A.colb
)
ORDER BY
acola
, acolb
结果：
acola acolb acolc bcola bcolb bcolc
1 10 Data1 1 NULL Match1
1 10 Data3 1 NULL Match1
2 20 Data2 2 NULL Match2
2 30 Data4 2 30 Match3
2 30 Data5 2 30 Match3
答案 2 :(得分：0)
这比Uporabnik003's answer更短的方式来实现这一目标：
/* Test data - code by Uporabnik003 */
declare @ta table (cola int, colb int, colc char(10))
declare @tb table (cola int, colb int, colc char(10))
insert into @ta values (1, 10, 'Data1')
insert into @ta values (2, 20, 'Data2')
insert into @ta values (1, 10, 'Data3')
insert into @ta values (2, 30, 'Data4')
insert into @ta values (2, 30, 'Data5')
insert into @tb values (1, NULL, 'Match1')
insert into @tb values (2, NULL, 'Match2')
insert into @tb values (2, 30, 'Match3')
/* End Test data. */
;with cteAllResults as (
select a_cola =a.cola
,a_colb =a.colb
,a_colc =a.colc
,b_cola =b.cola
,b_colb =b.colb
,b_colc =b.colc
-- The "best" row of each joining group will get RowNo=1
,RowNo =row_number() over ( partition by a.cola,a.colb,a.colc
-- Depending on your real data, you may want
-- to play around with the order by clause
-- to identify the "best" row differently.
order by b.colb desc
)
from @ta a
inner join @tb b on a.cola = b.cola
-- This additional join clause is equivalent to your
-- `where a.colb = b.colb or b.colb is null`
and a.colb = isnull(b.colb,a.colb)
)
select *
from cteAllResults
where RowNo=1
它产生与问题所需输出完全相同的结果。
但是，更喜欢＆＃34;更短＆＃34;代码更高效的代码。对于初学者来说，行数受到关于空格和代码布局的注释和（半）任意决定的影响很大。
更重要的是，Uporabnik003的解决方案在大型数据集上的表现可能比这更好。这个解决方案依赖于首先生成一个太大的行集，然后将其削减;他们的解决方案从两个较小的集合中构建正确的行集。
如果我在表格上有适当的主键，我会使用Uporabnik003的解决方案，但用where not exists替换left join子句，测试加入失败（{{1 }} 在右手侧）。