Question

我有两张桌子

TABLE_1：

id  pa_name
 1     A
 2     B
 3     C

TABLE_2：

id  breakfast   lunch   dinner  day day_des
1      Yes       Yes      No     1    des1
1       No        No      Yes    2    des2
1      Yes        No      Yes    3    des3
2      Yes       Yes      Yes    1    des11
2      Yes        No       No    2    des22
2       No       Yes       No    3    des33

我想要那样的结果

id  pa_name breakfast lunch dinner day_des
 1     A       2       1      2     des1
 2     B       2       2      1     des11

我正在使用SQL Server 2008

Answer 1

这将以逗号分隔格式

为您提供值

;WITH CTE AS
(
    SELECT t1.id, t1.pa_name,
           SUM(CASE WHEN breakfast = 'YES' then 1 end) as breakfast,
           SUM(CASE WHEN lunch = 'YES' THEN 1 END) AS lunch,
           SUM(CASE WHEN dinner= 'YES' THEN 1 END) AS dinner       
    FROM Table_1 AS t1 
    JOIN Table_2 AS t2 ON t1.id = t2.id
    GROUP BY t1.id, t1.pa_name
)
SELECT *,
SUBSTRING(
        (SELECT ', ' + day_des
        FROM TABLE_2 T2 
        WHERE CTE.id=id 
        FOR XML PATH('')),2,200000) day_des
FROM CTE

的 SQL FIDDLE

Answer 2

将JOIN与GROUP BY结合使用：

select t1.id, t1.pa_name,
       SUM(case when breakfast = 'YES' then 1 end) as breakfast,
       SUM(case when lunch = 'YES' then 1 end) as lunch,
       SUM(case when dinner= 'YES' then 1 end) as dinner,
       MIN(day_des)
from Table_1 as t1 join Table_2 as t2 on t1.id = t2.id
group by t1.id, t1.pa_name

但是我对最近的编辑有点担心，如果有不同的值，day_des应该如何被选中？我的选择是MIN值。

sql server 2008为行计数选择语句

2 个答案: