对于这种情况，做一个自加入的存在吗？

Question

我有一个表阿尔法2域组标识，成员:

GroupId | Member;
A1----------A;
A1----------B; 
A1----------C;
A2----------A;
A2----------B;
A3----------A;
A3----------D;
A3----------E;

目标： 给出的输入-A，B，C-我们来查询表，以找到，如果一组标识存在于这个确切组成员。所以，这是什么我的计划要做到：

1. 查询表为所有GroupIds其计数为3(由于我inpt是A，B，C.I，其3)
2. 这将给我A1、A3。现在，查询这个集确切匹配件值..这会给我A1。

我计划写一个存储的过程和将实现的目标。但是，我的问题可以这是实现在一个单一的查询...一个单一的自参加，也许。

澄清：该组(A、B、C)特A1。如果得到输入的(A，B，C，D)查询不应返回A1。

Answer 1

答案给予迄今为止假定会员领域是独特的任何给定的组标识.在工作我已经做了这个并不是这种情况。并且，如果集团中有你在寻找什么,外加一些额外的，你需要exlucde，小组。

SELECT
   [Alpha].GroupID
FROM
   [Alpha]
GROUP BY
   [Alpha].GroupID
HAVING
       SUM(CASE WHEN [alpha].Member IN ('A','B','C') THEN 1 ELSE 0 END) = 3
   AND MIN(CASE WHEN [alpha].Member IN ('A','B','C') THEN 1 ELSE 0 END) = 1

你还可以代替第一个加入到一个表格保持的成员正在寻找...

SELECT
   [Alpha].GroupID
FROM
   [Alpha]
LEFT JOIN
   [Search]
       ON [Search].Member
GROUP BY
   [Alpha].GroupID
HAVING
       SUM(CASE WHEN [alpha].Member = [search].Member THEN 1 ELSE 0 END) = (SELECT COUNT(*) FROM [search])
   AND MIN(CASE WHEN [alpha].Member = [search].Member THEN 1 ELSE 0 END) = 1

Answer 2

SELECT GroupID
  FROM ALPHA
 WHERE Member IN ('A', 'B', 'C')
 GROUP BY GroupID
HAVING COUNT(*) = 3

这依赖于你的写作出的名单中的成员在条款，并设定的数量(不同)项中的成员名单在有条款。如果你不能产生的SQL正是如此，那么你必须要更加努力地工作。

---

如指出的，在早期评论，这也依赖于口译，你想要团体在所有三个A、B、C(可能还有其他一些人)成员团。一种方法，不一定是最好的方式，得到'组包含三个人，即A、B、C'，是使用：

SELECT GroupID
  FROM ALPHA A1
 WHERE Member IN ('A', 'B', 'C')
   AND 3 = (SELECT COUNT(*) FROM ALPHA A2 WHERE A2.GroupID = A1.GroupID)
 GROUP BY GroupID
HAVING COUNT(*) = 3

这明确地检查的总人数在本小组是3和成员都是A、B和C(假定有一个独特的约束，在阿尔法(组标识，会员)使一件不能被列两次为属于同一集团)。

Answer 3

SELECT DISTINCT aa.GroupId
FROM Alpha aa
JOIN Alpha ab ON (aa.GroupId = ab.GroupId)
JOIN Alpha ac ON (aa.GroupId = ac.GroupId)
LEFT OUTER JOIN Alpha ax ON (aa.GroupId = ax.GroupId AND ax.Member NOT IN ('A', 'B', 'C')
WHERE aa.Member = 'A' AND ab.Member = 'B' AND ac.Member = 'C'
 AND ax.GroupId IS NULL;

还有涉及解决方案 GROUP BY 但是我找到的 JOIN 解决方案往往具有更好的业绩。我通常的工作MySQL，我理解MS SQL Server是更好的分组查询。所以尝试这两个解决方案，看看什么最为品牌的关系型数据库使用。

Answer 4

试试这个：

declare @YourTable table (GroupID char(2),Member char(1))

insert into @YourTable values ('A1','A')
insert into @YourTable values ('A1','B')
insert into @YourTable values ('A1','C')
insert into @YourTable values ('A2','A')
insert into @YourTable values ('A2','B')
insert into @YourTable values ('A3','A')
insert into @YourTable values ('A3','D')
insert into @YourTable values ('A3','E')
insert into @YourTable values ('A5','A')
insert into @YourTable values ('A5','B')
insert into @YourTable values ('A5','C')
insert into @YourTable values ('A5','D')
SELECT t1.GroupID
    FROM @YourTable t1
        LEFT OUTER JOIN @YourTable t2 ON t1.GroupID=t2.GroupID AND t2.Member NOT IN ('A', 'B', 'C') 
    WHERE t1.Member IN ('A', 'B', 'C') 
        AND t2.GroupID IS NULL
    GROUP BY t1.GroupID
    HAVING COUNT(*) = 3

输出：

GroupID
-------
A1

(1 row(s) affected)

这里是一个完整的解决方案：

之前你用我的功能，就需要设置"助手"表，你只需要做这一个时间每数据库：

CREATE TABLE Numbers
(Number int  NOT NULL,
    CONSTRAINT PK_Numbers PRIMARY KEY CLUSTERED (Number ASC)WITH (PAD_INDEX  = OFF, STATISTICS_NORECOMPUTE  = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS  = ON, ALLOW_PAGE_LOCKS  = ON) ON [PRIMARY]
) ON [PRIMARY]
DECLARE @x int
SET @x=0
WHILE @x<8000
BEGIN
    SET @x=@x+1
    INSERT INTO Numbers VALUES (@x)
END

使用这种功能分割你的串，不会循环和非常快：

CREATE FUNCTION [dbo].[FN_ListToTable]
(
     @SplitOn              char(1)              --REQUIRED, the character to split the @List string on
    ,@List                 varchar(8000)        --REQUIRED, the list to split apart
)
RETURNS
@ParsedList table
(
    ListValue varchar(500)
)
AS
BEGIN

/**
Takes the given @List string and splits it apart based on the given @SplitOn character.
A table is returned, one row per split item, with a column name "ListValue".
This function workes for fixed or variable lenght items.
Empty and null items will not be included in the results set.


Returns a table, one row per item in the list, with a column name "ListValue"

EXAMPLE:
----------
SELECT * FROM dbo.FN_ListToTable(',','1,12,123,1234,54321,6,A,*,|||,,,,B')

    returns:
        ListValue  
        -----------
        1
        12
        123
        1234
        54321
        6
        A
        *
        |||
        B

        (10 row(s) affected)

**/



----------------
--SINGLE QUERY-- --this will not return empty rows
----------------
INSERT INTO @ParsedList
        (ListValue)
    SELECT
        ListValue
        FROM (SELECT
                  LTRIM(RTRIM(SUBSTRING(List2, number+1, CHARINDEX(@SplitOn, List2, number+1)-number - 1))) AS ListValue
                  FROM (
                           SELECT @SplitOn + @List + @SplitOn AS List2
                       ) AS dt
                      INNER JOIN Numbers n ON n.Number < LEN(dt.List2)
                  WHERE SUBSTRING(List2, number, 1) = @SplitOn
             ) dt2
        WHERE ListValue IS NOT NULL AND ListValue!=''



RETURN

END --Function FN_ListToTable

你现在可以使用那个这样的功能以查询任何名单：

DECLARE @List varchar(100)
SET @List='A,B,C'
declare @YourTable table (GroupID char(2),Member char(1))

insert into @YourTable values ('A1','A')
insert into @YourTable values ('A1','B')
insert into @YourTable values ('A1','C')
insert into @YourTable values ('A2','A')
insert into @YourTable values ('A2','B')
insert into @YourTable values ('A3','A')
insert into @YourTable values ('A3','D')
insert into @YourTable values ('A3','E')
insert into @YourTable values ('A5','A')
insert into @YourTable values ('A5','B')
insert into @YourTable values ('A5','C')
insert into @YourTable values ('A5','D')

SELECT t1.GroupID
    FROM @YourTable t1
        LEFT OUTER JOIN @YourTable t2 ON t1.GroupID=t2.GroupID AND t2.Member NOT IN (SELECT ListValue FROM dbo.FN_ListToTable(',',@List))
    WHERE t1.Member IN (SELECT ListValue FROM dbo.FN_ListToTable(',',@List))
        AND t2.GroupID IS NULL
    GROUP BY t1.GroupID
    HAVING COUNT(*) = (SELECT COUNT(*) FROM dbo.FN_ListToTable(',',@List)) 

输出：

GroupID
-------
A1

Answer 5

select*from ALPHA 在件( 选择成员，从阿尔法 小组成员 具有最(*)=3)

Answer 6

试试这个:

SELECT GroupId
  FROM ALPHA
 GROUP BY GroupId
HAVING SUM(CASE WHEN Member='A' THEN 1.0
                WHEN Member='B' THEN 2.0
                WHEN Member='C' THEN 4.0
                ELSE 7.31415
          END) = 7.0

Answer 7

我的建议是解析，delimited string into a temp表，然后再尝试这样的事情。

create table #temp(member varchar(10))

create table #groups
(
groupID varchar(2),
member char(1)
)

--#temp holds the members from your delimited string.
--#groups holds your relationships.



select distinct groupID
from #groups
where 
 (select count(*) from #groups i, #temp t
  where i.member = t.member and i.groupID = #groups.groupID) = 
(select count(*) from #temp)