Question

鉴于这两个表/集包含不同的项目组，~~如何找到set1中set2中多个组中的哪些组跨越set1？~~怎么样？我发现set2中的哪些群组无法由A (1,2,5)中的单个群组覆盖？

e.g。对于下表，s1是唯一跨越s2（ 1 ， 2 ，3）和B（2）的群组，3,4-，的 5 ）。 C和s2不是答案，因为两者都包含在一个组SQL中。

我更愿意使用Sql Server 2008 R2（set1 set2 +---------+----------+ +---------+----------+ | group | item | | group | item | `````````````````````+ `````````````````````+ | A | 1 | | s1 | 1 | | A | 2 | | s1 | 2 | | A | 5 | | s1 | 3 | | B | 4 | | s2 | 2 | | B | 5 | | s2 | 3 | | C | 3 | | s2 | 4 | | C | 5 | | s2 | 5 | +---------+----------+ +---------+----------+可用）。

感谢。

create table #set1 (grp varchar(5),item int)
create table #set2 (grp varchar(5),item int)

insert into #set1 select 'a',1 union select 'a',2 union select 'a',5 union select 'b',4 union select 'b',5 union select 'c',3 union select 'c',5
insert into #set2 select 's1',1 union select 's1',2 union select 's1',3 union select 's2',2 union select 's2',3 union select 's2',4 union select 's2',5

select * from #set1
select * from #set2

--drop table #set1
--drop table #set2

使用此sqlfiddle尝试：http://sqlfiddle.com/#!6/fac8a/3

或者使用下面的脚本生成临时表以试用答案：

func application(application: UIApplication, didFinishLaunchingWithOptions launchOptions: [NSObject: AnyObject]?) -> Bool {

    Flurry.startSession("XXXXXXXX")

    if let userEmail = email{
        // user logged with our login
        println("User logged via ff.")
        Flurry.setUserID(userEmail as String)
    }else if let userID = id{
        // user logged with Facebook
        println("User logged with facebook.")
        Flurry.setUserID(userID as String)
    }else{
        // user not logged
    }

}

Answer 1

从set1中选择set2中没有任何群组的群组，set1中set2中的所有内容都存在于select s1.grp from set1 s1 where not exists( select * from set2 s2 where not exists( select item from set1 s11 where s11.grp = s1.grp except select item from set2 s22 where s22.grp = s2.grp)) group by s1.grp中

self.HeaderView

Answer 2

确定。这很难看，但应该有效。我小提琴尝试了。我认为可以通过开窗来完成，但我必须考虑一下。

现在这是丑陋的。

WITH d1 AS (
SELECT set1.grp
     , COUNT(*) cnt
  FROM set1
 GROUP BY set1.grp
), d2 AS (  
SELECT set1.grp grp1
     , set2.grp grp2
     , COUNT(set1.item) cnt
  FROM set1
 INNER JOIN set2
    ON set1.item = set2.item
 GROUP BY set1.grp
     , set2.grp
 )
SELECT grp
  FROM d1
EXCEPT  
SELECT d1.grp 
  FROM d1
 INNER JOIN d2
    ON d2.grp1 = d1.grp
   AND d2.cnt = d1.cnt

Answer 3

您可以通过以下查询找到解决方案：

SELECT A.GROUP AS G1, A.ITEM AS T1, B.GROUP, B.ITEM
FROM SET1 A RIGHT JOIN SET2 B ON A.ITEM=B.ITEM
WHERE A.GROUP IS NULL

Answer 4

你能看看这个

吗？

SELECT DISTINCT a.Group1, a.Item, b.CNT
FROM SET1 a
INNER JOIN
(SELECT GroupA, COUNT(*) CNT
 FROM
 (
    SELECT DISTINCT a.Group1 GroupA, b.Group1 GroupB
    FROM SET1 a
        INNER JOIN SET2 b ON a.Item = b.Item
 ) a GROUP BY GroupA
) b ON a.Group1 = b.GroupA
 WHERE b.CNT > 1

Answer 5

感谢您的评论。我相信以下编辑的查询将起作用：

Select distinct grp1, initialRows, max(MatchedRows) from 
  (
  select a.grp as grp1, b.grp as grp2
  , count(distinct case when b.item is not null then a.item end) as MatchedRows
  , d.InitialRows
  from set1 a
  left join set2 b
  on a.item = b.item
  left join 
    (select grp, count(distinct Item) as InitialRows from set1
     group by grp) d
  on a.grp = d.grp
  group by a.grp, b.grp, InitialRows
) c
group by grp1, InitialRows
having max(MatchedRows) < InitialRows

Answer 6

我认为这样做会有所帮助。子查询通过计算匹配并将匹配计数与set1组计数进行比较，为每个set1组返回set2组，这些组与set1中的所有项匹配。

select s.grp from #set1 s
group by s.grp
having not exists ( 
    select s2.grp from #set2 s2 inner join #set1 s1 on s2.item = s1.item
    where s1.grp = s.grp
    group by s2.grp
    having count(s.item) = count(s2.item)   
    )

Answer 7

基本上和Robert Co一样我没有从他的答案中得到这个 - 独立提出这个

    select set1.group  
      from set1
except
    select set1count.group 
      from ( select set1.group            , count(*) as [count]  
               from set1 
           ) as set1count 
      join ( select set1.group as [group1], count(*) as [count]  
               from set1 
               join set2 
                 on set2.item = set1.item 
              group by set1.group, set2.group -- this is the magic
           ) as set1count 
        on set1count.group = set2count.[group1] -- note no set2.group match
       and set1count.count = set12count.count -- the items in set1 are in at least on set2 group

SQL与group by相交

7 个答案: