Question

采取下表示例：

Name  |  List |  FlagByList
---------------------------
  A   |   1   |   Y
  A   |   2   |   Y
  B   |   1   |   Y
  B   |   2   |   N
  C   |   -   |   -
  C   |   -   |   -

我想返回所有列表中包含“Y”的名称和任何列表中不存在的名称。

Answer 1

简单的聚合查询可以做到这一点。

SELECT Name
FROM Table
GROUP BY Name
HAVING COUNT(1) = COUNT(CASE FlagByList WHEN 'Y' THEN 1 END) --counts all rows with Y as Value
       OR COUNT(1) = COUNT(CASE WHEN FlagByList  IS NULL THEN 1 END);  --counts all rows with NULL as value

Answer 2

使用解码

 select name from example
    group by Name 
  having  sum(decode( FlagByList, 'Y',1, 0)) = count(*)  
            OR sum(decode(List, NULL, 0, 1)) = count(*)

Answer 3

请使用以下代码。它与SQL Server 2012一起工作正常。

DECLARE @Table TABLE (Name char(2),List  char(2) , FlagByList char(2))
INSERT @Table
(Name,List,FlagByList)
VALUES
('A','1','Y'),
('A','2','Y'),
('B','1','Y'),
('B','2','N'),
('C','-','-'),
('C','-','-')

SELECT DISTINCT(Name) FROM @Table WHERE FlagByList ='Y'
UNION
SELECT DISTINCT(Name) FROM @Table WHERE FlagByList ='-'
EXCEPT
SELECT DISTINCT(Name) FROM @Table WHERE FlagByList ='N'

使用聚合函数的SQL查询

3 个答案: