鉴于以下内容：
declare @a table
(
pkid int,
value int
)
declare @b table
(
otherID int,
value int
)
insert into @a values (1, 1000)
insert into @a values (1, 1001)
insert into @a values (2, 1000)
insert into @a values (2, 1001)
insert into @a values (2, 1002)
insert into @b values (-1, 1000)
insert into @b values (-1, 1001)
insert into @b values (-1, 1002)
如何查询@a中与@b完全匹配的所有值？
不会返回 {@a.pkid = 1, @b.otherID = -1}（3个值中只有2个匹配）
{@a.pkid = 2, @b.otherID = -1}（3个值中的3个匹配）
重构表可以是一种选择。
编辑：我在James和Tom H.的答案中取得了成功。
当我在@b中添加另一个案例时，它们会有点短暂。
insert into @b values (-2, 1000)
假设这应该返回另外两行（{@a.pkid = 1, @b.otherID = -2}和{@a.pkid = 2, @b.otherID = -2}，它不起作用。但是，对于我的项目，这不是问题。
答案 0 :(得分：7)
效率更高（使用TOP 1代替COUNT），并与(-2, 1000)一起使用：
SELECT *
FROM (
SELECT ab.pkid, ab.otherID,
(
SELECT TOP 1 COALESCE(ai.value, bi.value)
FROM (
SELECT *
FROM @a aii
WHERE aii.pkid = ab.pkid
) ai
FULL OUTER JOIN
(
SELECT *
FROM @b bii
WHERE bii.otherID = ab.otherID
) bi
ON ai.value = bi.value
WHERE ai.pkid IS NULL OR bi.otherID IS NULL
) unmatch
FROM
(
SELECT DISTINCT pkid, otherid
FROM @a a , @b b
) ab
) q
WHERE unmatch IS NOT NULL
答案 1 :(得分：6)
可能不是最便宜的方式：
SELECT a.pkId,b.otherId FROM
(SELECT a.pkId,CHECKSUM_AGG(DISTINCT a.value) as 'ValueHash' FROM @a a GROUP BY a.pkId) a
INNER JOIN (SELECT b.otherId,CHECKSUM_AGG(DISTINCT b.value) as 'ValueHash' FROM @b b GROUP BY b.otherId) b
ON a.ValueHash = b.ValueHash
你可以看到，基本上我正在为每个表创建一个新的结果集，表示每个表中每个Id的值集合的一个值，并且只在它们匹配的地方加入。
答案 2 :(得分：2)
以下查询为您提供了所需的结果：
select A.pkid, B.otherId
from @a A, @b B
where A.value = B.value
group by A.pkid, B.otherId
having count(B.value) = (
select count(*) from @b BB where B.otherId = BB.otherId)
答案 3 :(得分：1)
适用于您的示例，我认为它适用于所有情况，但我没有彻底测试过：
SELECT
SQ1.pkid
FROM
(
SELECT
a.pkid, COUNT(*) AS cnt
FROM
@a AS a
GROUP BY
a.pkid
) SQ1
INNER JOIN
(
SELECT
a1.pkid, b1.otherID, COUNT(*) AS cnt
FROM
@a AS a1
INNER JOIN @b AS b1 ON b1.value = a1.value
GROUP BY
a1.pkid, b1.otherID
) SQ2 ON
SQ2.pkid = SQ1.pkid AND
SQ2.cnt = SQ1.cnt
INNER JOIN
(
SELECT
b2.otherID, COUNT(*) AS cnt
FROM
@b AS b2
GROUP BY
b2.otherID
) SQ3 ON
SQ3.otherID = SQ2.otherID AND
SQ3.cnt = SQ1.cnt
答案 4 :(得分：1)
-- Note, only works as long as no duplicate values are allowed in either table
DECLARE @validcomparisons TABLE (
pkid INT,
otherid INT,
num INT
)
INSERT INTO @validcomparisons (pkid, otherid, num)
SELECT a.pkid, b.otherid, A.cnt
FROM (select pkid, count(*) as cnt FROM @a group by pkid) a
INNER JOIN (select otherid, count(*) as cnt from @b group by otherid) b
ON b.cnt = a.cnt
DECLARE @comparison TABLE (
pkid INT,
otherid INT,
same INT)
insert into @comparison(pkid, otherid, same)
SELECT a.pkid, b.otherid, count(*)
FROM @a a
INNER JOIN @b b
ON a.value = b.value
GROUP BY a.pkid, b.otherid
SELECT COMP.PKID, COMP.OTHERID
FROM @comparison comp
INNER JOIN @validcomparisons val
ON comp.pkid = val.pkid
AND comp.otherid = val.otherid
AND comp.same = val.num
答案 5 :(得分：1)
我添加了一些额外的测试用例。您可以通过更改在聚合中使用不同关键字的方式来更改重复处理。基本上，我得到了一些匹配，并将它与每个@a和@b中所需匹配的数量进行比较。
declare @a table
(
pkid int,
value int
)
declare @b table
(
otherID int,
value int
)
insert into @a values (1, 1000)
insert into @a values (1, 1001)
insert into @a values (2, 1000)
insert into @a values (2, 1001)
insert into @a values (2, 1002)
insert into @a values (3, 1000)
insert into @a values (3, 1001)
insert into @a values (3, 1001)
insert into @a values (4, 1000)
insert into @a values (4, 1000)
insert into @a values (4, 1001)
insert into @b values (-1, 1000)
insert into @b values (-1, 1001)
insert into @b values (-1, 1002)
insert into @b values (-2, 1001)
insert into @b values (-2, 1002)
insert into @b values (-3, 1000)
insert into @b values (-3, 1001)
insert into @b values (-3, 1001)
SELECT Matches.pkid, Matches.otherId
FROM
(
SELECT a.pkid, b.otherId, n = COUNT(*)
FROM @a a
INNER JOIN @b b
ON a.Value = b.Value
GROUP BY a.pkid, b.otherId
) AS Matches
INNER JOIN
(
SELECT
pkid,
n = COUNT(DISTINCT value)
FROM @a
GROUP BY pkid
) AS ACount
ON Matches.pkid = ACount.pkid
INNER JOIN
(
SELECT
otherId,
n = COUNT(DISTINCT value)
FROM @b
GROUP BY otherId
) AS BCount
ON Matches.otherId = BCount.otherId
WHERE Matches.n = ACount.n AND Matches.n = BCount.n
答案 6 :(得分：1)
如何查询@a中与@b完全匹配的所有值？
我担心这个定义不太清楚。从你的另一个例子看，你想要所有a.pkid，b.otherID对，给定b.otherID的每个b.value也是给定a.pkid的a.value。
换句话说，您希望@a中的pkids具有至少 b中其他ID的所有值。 @a中的额外值似乎没问题。同样，这是基于您的附加示例的推理，并且假设（1，-2）和（2，-2）将是有效结果。在这两种情况下，给定pkid的a.value值大于给定otherID的b.value值。
所以，考虑到这一点：
select
matches.pkid
,matches.otherID
from
(
select
a.pkid
,b.otherID
,count(1) as cnt
from @a a
inner join @b b
on b.value = a.value
group by
a.pkid
,b.otherID
) as matches
inner join
(
select
otherID
,count(1) as cnt
from @b
group by otherID
) as b_counts
on b_counts.otherID = matches.otherID
where matches.cnt = b_counts.cnt
答案 7 :(得分：0)
这样做的几种方法，但一个简单的方法是创建一个联合视图
create view qryMyUinion as
select * from table1
union all
select * from table2
小心使用union all，而不是简单的union，因为它会省略重复
然后这样做
select count( * ), [field list here]
from qryMyUnion
group by [field list here]
having count( * ) > 1
Union和Having语句往往是标准SQL中最容易被忽视的部分，但它们可以解决许多棘手的问题，否则需要程序代码
答案 8 :(得分：0)
如果您尝试仅返回完整的记录集，则可以尝试此操作。我肯定会建议使用有意义的别名，但是......
Cervo是对的，我们需要额外检查以确保a与b完全匹配，而不是b的超集。在这一点上，这更像是一种笨拙的解决方案，所以这只适用于其他解决方案中的分析功能不起作用的情况。
select
a.pkid,
a.value
from
@a a
where
a.pkid in
(
select
pkid
from
(
select
c.pkid,
c.otherid,
count(*) matching_count
from
(
select
a.pkid,
a.value,
b.otherid
from
@a a inner join @b b
on a.value = b.value
) c
group by
c.pkid,
c.otherid
) d
inner join
(
select
b.otherid,
count(*) b_record_count
from
@b b
group by
b.otherid
) e
on d.otherid = e.otherid
and d.matching_count = e.b_record_count
inner join
(
select
a.pkid match_pkid,
count(*) a_record_count
from
@a a
group by
a.pkid
) f
on d.pkid = f.match_pkid
and d.matching_count = f.a_record_count
)
答案 9 :(得分：0)
进一步重复这一点：
select a.*
from @a a
inner join @b b on a.value = b.value
这将返回@a中与@b
匹配的所有值答案 10 :(得分：0)
1）我假设你没有重复的身份
2）获取具有相同数值的键
3）键值等于等值数的行是目标
我希望这是你搜索的内容（你不搜索性能不是吗？）
declare @a table( pkid int, value int)
declare @b table( otherID int, value int)
insert into @a values (1, 1000)
insert into @a values (1, 1001)
insert into @a values (2, 1000)
insert into @a values (2, 1001)
insert into @a values (2, 1002)
insert into @a values (3, 1000)
insert into @a values (3, 1001)
insert into @a values (4, 1000)
insert into @a values (4, 1001)
insert into @b values (-1, 1000)
insert into @b values (-1, 1001)
insert into @b values (-1, 1002)
insert into @b values (-2, 1001)
insert into @b values (-2, 1002)
insert into @b values (-3, 1000)
insert into @b values (-3, 1001)
select cntok.cntid1 as cntid1, cntok.cntid2 as cntid2
from
(select cnt.cnt, cnt.cntid1, cnt.cntid2 from
(select acnt.cnt as cnt, acnt.cntid as cntid1, bcnt.cntid as cntid2 from
(select count(pkid) as cnt, pkid as cntid from @a group by pkid)
as acnt
full join
(select count(otherID) as cnt, otherID as cntid from @b group by otherID)
as bcnt
on acnt.cnt = bcnt.cnt)
as cnt
where cntid1 is not null and cntid2 is not null)
as cntok
inner join
(select count(1) as cnt, cnta.cntid1 as cntid1, cnta.cntid2 as cntid2
from
(select cnt, cntid1, cntid2, a.value as value1
from
(select cnt.cnt, cnt.cntid1, cnt.cntid2 from
(select acnt.cnt as cnt, acnt.cntid as cntid1, bcnt.cntid as cntid2 from
(select count(pkid) as cnt, pkid as cntid from @a group by pkid)
as acnt
full join
(select count(otherID) as cnt, otherID as cntid from @b group by otherID)
as bcnt
on acnt.cnt = bcnt.cnt)
as cnt
where cntid1 is not null and cntid2 is not null)
as cntok
inner join @a as a on a.pkid = cntok.cntid1)
as cnta
inner join
(select cnt, cntid1, cntid2, b.value as value2
from
(select cnt.cnt, cnt.cntid1, cnt.cntid2 from
(select acnt.cnt as cnt, acnt.cntid as cntid1, bcnt.cntid as cntid2 from
(select count(pkid) as cnt, pkid as cntid from @a group by pkid)
as acnt
full join
(select count(otherID) as cnt, otherID as cntid from @b group by otherID)
as bcnt
on acnt.cnt = bcnt.cnt)
as cnt
where cntid1 is not null and cntid2 is not null)
as cntok
inner join @b as b on b.otherid = cntok.cntid2)
as cntb
on cnta.cntid1 = cntb.cntid1 and cnta.cntid2 = cntb.cntid2 and cnta.value1 = cntb.value2
group by cnta.cntid1, cnta.cntid2)
as cntequals
on cntok.cnt = cntequals.cnt and cntok.cntid1 = cntequals.cntid1 and cntok.cntid2 = cntequals.cntid2
答案 11 :(得分：-1)
正如CQ所说，只需要一个简单的内部联接。
Select * -- all columns but only from #a
from #a
inner join #b
on #a.value = #b.value -- only return matching rows
where #a.pkid = 2