1

表数据:

NAME     RS
A        10
A        20
A        30
B        15 
B        5
B        10
C        70
C        30 

预期输出:

A   -- column name
10
20
30
60  -- total of A

B
15
5
10 
30  -- total of B

C
70
30
100 -- total of C

到目前为止,我已经尝试过:

  1. UNION ALL
  2. 选择所有数据并使用前端进行操作。

没有其他最简单的解决方案UNION ALL吗?

4

4 回答 4

1

尝试这个

select  unnest(string_to_array( name ||','||rs||','||sum, ',')) AS elem 
from (
     select name
           ,string_agg(rs::text,',')  rs
           ,sum(rs) 
     from tbl 
     group by name
)t 

如果要Total Of结果中打印,请使用

select  unnest(string_to_array( name ||','||rs||','||sum, ',')) AS elem 
from (
     select name
           ,string_agg(rs::text,',')  rs
           ,'Total Of '||name||' '|| sum(rs) sum
     from tbl 
     group by name
)t 

> SQLFIDDLE

于 2015-01-21T11:51:24.753 回答
1

要使用纯 SQL 在单个查询中执行此操作:

SELECT unnest(arr)
FROM  (
   SELECT ARRAY[name] || array_agg(rs::text) || sum(rs)::text AS arr
   FROM   tbl
   GROUP  BY name
   ) sub
ORDER  BY arr[1];

类似于@WingedPanther发布的内容,但任何名称都干净且安全。

回报:

A
10
20
30
60
B
15
15
C
70
30
100

或者:

WITH cte AS (
   SELECT name, sum(RS)::text AS sum_rs
   FROM   tbl
   GROUP  BY 1
   )
SELECT unicol
FROM  ( 
   SELECT name AS unicol, name AS order1, 1 AS order2 FROM cte
   UNION  ALL
   SELECT sum_rs, name, 2 FROM cte
   ) sub
ORDER  BY order1, order2;

回报:

A
60
B
15
C
100

SQL小提琴。

或者只是从 CTE 发送结果,其余的交给您的客户。

于 2015-01-21T11:35:25.547 回答
0

不是 100% 确定你在追求什么,但听起来你只是想要这个:

Select name, Sum(RS)
from sometable
group by name
于 2015-01-21T11:00:48.590 回答
-2

尝试这个:

DECLARE @t TABLE(Name NVARCHAR(MAX), RS INT)

INSERT INTO @t VALUES
('A', 10),
('A', 20),
('A', 30),
('B', 15),
('C', 70),
('C', 30)

与联合

SELECT Name, RS, 0 AS OrderBit FROM @t
UNION ALL
SELECT Name, SUM(RS) AS Total, 1 AS OrderBit FROM @t
GROUP BY Name

ORDER BY Name, OrderBit

没有联合:

SELECT Name, SUM(RS) AS RS FROM @t 
GROUP BY GROUPING SETS((Name), (Name, RS))
于 2015-01-21T10:59:02.540 回答