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T-SQL:如何从一个表中获取其值与另一个表的值完全匹配的行?

tsql sql-server

Austin Salonen gmlacrosse · 技术社区 · 16 年前

鉴于以下情况:

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?

{@a.pkid = 1, @b.otherID = -1} 不会返回(3个值中只有2个匹配)

{@a.pkid = 2, @b.otherID = -1} 将返回(3个值中有3个匹配)

重构表可以是一种选择。

编辑: 詹姆斯和汤姆·H的回答让我很满意。

当我在@b中添加另一个案例时,它们有点不足。

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。然而,对于我的项目来说,这不是问题。

12 回复 | 直到 10 年前

Quassnoi 16 年前

这更有效(它使用 TOP 1 而不是 COUNT ),并与 (-2, 1000) :

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

James 16 年前

可能不是最便宜的方法:

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

您可以看到,基本上我正在为每个表创建一个新的结果集,代表每个表中每个Id的值集的一个值,并仅在匹配的地方连接。

boes 16 年前

以下查询将为您提供请求的结果:

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)

Tom H zenazn 16 年前

适用于您的示例,我认为它适用于所有情况,但我还没有彻底测试它:

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

Cervo 16 年前

--注意,只有在两个表中都不允许有重复值的情况下才有效
声明@有效比较表(
pkid INT,
其他ID INT,
编号INT
)

插入@validcomparisons(pkid、otherid、num)
选择a.pkid、b.otherid、a.cnt
FROM(选择pkid,按pkid计数(*)为cnt FROM@a组)a
INNER JOIN(选择otherid,按otherid从@b组中计数(*)作为cnt)b
ON b.cnt=交流

声明@对照表(
pkid INT,
其他ID INT,
相同的INT)

插入@comparison(pkid、otherid、same)
选择a.pkid、b.otherid、count(*)
来自@a
内部加入@b b
ON a.value=b.value
分组方式a.pkid,b.otherid

选择COMP。PKID,COMP。其他ID
来自@比较comp
加入@validcomparisons val
ON comp.pkid=val.pkid
AND comp.therid=val.therid
AND comp.same=val.num

user19143 16 年前

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,我正在计算匹配数,并将其与每个@a和@b中所需的匹配数进行比较。

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

user19164 16 年前

如何查询@a中与@b完全匹配的所有值?

恐怕这个定义不太清楚。从您的附加示例中可以看出,您想要所有a.pkid、b.otherID对,其中给定b.otherIID的每个b.value也是给定a.pkid的a.value。

换言之,你希望@a中的pkids具有 不少于 b中otherIDs的所有值。@a中的额外值似乎是可以的。同样,这是基于您的附加示例以及(1,-2)和(2,-2)是有效结果的假设的推理。在这两种情况下,给定pkid的a.value值为超过给定otherID的b.value值。

因此,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

mslaven mslaven 16 年前

要进一步迭代该点:

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

Dave DuPlantis 16 年前

如果你只想返回完整的记录集,你可以试试这个。不过,我绝对建议使用有意义的别名。..

Cervo是对的,我们需要进行额外的检查,以确保a是b的精确匹配,而不是b的超集。在这一点上,这是一个更难处理的解决方案,因此这只有在其他解决方案中的分析函数不起作用的情况下才是合理的。

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
    )

C B dkretz 7 年前

1) 我假设你没有重复的id

2) 获取具有相同数值的密钥

3) 键值数量等于相等值数量的行是目标

我希望这就是你搜索的(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

C B dkretz 7 年前

有几种方法可以做到这一点,但一个简单的方法是创建一个联合视图,如下所示

create view qryMyUinion as
select * from table1 
union all
select * from table2

注意使用union-all,而不是简单的union,因为这会省略重复项

那就这么做

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需要过程代码的棘手问题

-1

Dave Jackson 16 年前

正如CQ所说,你只需要一个简单的内部连接。

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