SQL查询组

Question

SQL查询问题

我有一个像

select proposal_id, service_id,account_type 
from table1 

结果查询是这样的：

proposal_id service_id account_type 
    1   1001  INTERVAL 
    1   1002  INTERVAL 
    2   1003  NON INTERVAL 
    2   1004  NON INTERVAL 
    3   1005  NON INTERVAL 
    3   1006  INTERVAL 

我想编写一个查询：对于每个PROPOSAL_ID，如果所有的服务都有INTERVAL然后得到'INTERVAL'，如果所有的非INTERVAL得到'非INTERVAL'，如果两者都得到'Both'

For上面的例子中，它应该返回

proposal_id account_type 
    1   INTERVAL 
    2   NON-INTERVAL 
    3   BOTH 

Answer 1

数据：

declare @table table (id int, sid int, acc nvarchar(20)) 
insert @table VALUES (1,1001,'INTERVAL'),(1,1002,'INTERVAL'),(2,1003,'NON INTERVAL'),(2,1004,'NON INTERVAL'), 
(3,1005,'NON INTERVAL'),(3,1006,'INTERVAL') 

查询：

select x.Id 
    , CASE counter 
      WHEN 1 THEN x.Account_Type 
      ELSE 'BOTH' 
     END AS Account_Type 
from (
    select Id, Count(DISTINCT(acc)) AS counter, MAX(acc) As Account_Type 
    from @table 
    GROUP BY Id 
) x 

结果

Id   Account_Type 
----------- -------------------- 
1   INTERVAL 
2   NON INTERVAL 
3   BOTH 

Answer 2

SELECT 
     b.proposal_id 
    ,CASE 
     WHEN s1.proposal_id IS NOT NULL AND s2.proposal_id IS NOT NULL THEN 'BOTH' 
     WHEN s1.proposal_id IS NOT NULL THEN 'INTERVAL' 
     WHEN s2.proposal_id IS NOT NULL THEN 'NON-INTERVAL' 
     ELSE 'UNKNOWN' 
    END [account_type] 
FROM table1 b 
    LEFT JOIN(
    SELECT proposal_id,account_type FROM table1 WHERE account_type = 'INTERVAL' 
    ) s1 
     ON b.proposal_id = s1.proposal_id 
    LEFT JOIN (
    SELECT proposal_id,account_type FROM table1 WHERE account_type = 'NON-INTERVAL' 
    )s2 
     ON b.proposal_id = s2.proposal_id 

Answer 3

你可以使用count distinct来确定的，如果它是既然后用CASE来确定的显示哪些

SELECT DISTINCT proposal.proposal_id, 
     CASE cou 
      WHEN 1 THEN type ELSE 'Both' END as TYPE 
FROM proposal 
INNER JOIN (SELECT proposal_id, count(distinct type) cou 
      FROM proposal GROUP BY proposal_id) inn 
     ON proposal.id = inn.id 

Answer 4

select proposal_id, 
case when count(distinct account_type) > 1 then 'BOTH' 
    else max(account_type) 
    end 
from table1 
group by proposal_id 

你有fiddler here。