Question

我的输入如下。

+----+-------+---------+----------+----+----+
| A  | B     |    C    |  D       | E  | F  |
+----+-------+---------+----------+----+----+
| 3  | 32447 |  168531 | 2/2/2017 | 10 | 20 |
| 4  | 32447 |  168531 | 2/3/2017 | 10 | 20 |
| 14 | 32447 |  168531 | 2/6/2017 | 10 | 20 |
| 14 | 32447 |  168531 | 2/6/2017 | 10 | 20 |
| 12 | 32447 |  168531 | 2/7/2017 | 10 | 20 |
| 12 | 32447 |  168531 | 2/7/2017 | 10 | 20 |
| 12 | 32447 |  168531 | 2/7/2017 | 10 | 20 |
| 46 | 32447 |  168531 | 2/8/2017 | 10 | 20 |
| 46 | 32447 |  168531 | 2/8/2017 | 10 | 20 |
| 46 | 32447 |  168531 | 2/8/2017 | 10 | 20 |
| 85 | 32447 |  168531 | 2/9/2017 | 10 | 20 |
| 85 | 32447 |  168531 | 2/9/2017 | 10 | 20 |
| 85 | 32447 |  168531 | 2/9/2017 | 10 | 20 |
+----+-------+---------+----------+----+----+

我的预期输出如

+----+-------+--------+----------+----+----+
| A  | B     | C      | D        | E  | F  |
+----+-------+--------+----------+----+----+
| 3  | 32447 | 168531 | 2/2/2017 | 10 | 20 |
| 4  | 32447 | 168531 | 2/3/2017 | 10 | 20 |
| 14 | 32447 | 168531 | 2/6/2017 | 20 | 40 |
| 12 | 32447 | 168531 | 2/7/2017 | 30 | 60 |
| 46 | 32447 | 168531 | 2/8/2017 | 30 | 60 |
| 85 | 32447 | 168531 | 2/9/2017 | 30 | 60 |
+----+-------+--------+----------+----+----+

我正在使用像

这样的查询

select b,c,d,sum(e),sum(f)
group by b,c,d

如何获得没有MAX概念的列值？感谢。

Answer 1

select max(a),b,c,d,sum(e),sum(f)
group by b,c,d

Answer 2

将SUM函数与GROUP BY子句一起使用。

<强>查询

select a, b, c, d, sum(e) e, sum(f) f
from your_table_name
group by a, b, c, d;

Answer 3

select a,b,c,d,sum(e)e,sum(f)f FROM Table_Name
group by a,b,c,d

Answer 4

试试这个..希望这对你有用

select a,b,c,d,sum(e),sum(f) FROM yourtable out_TB
WHERE A=(SELECT TOP 1 in_TA.A FROM yourtable in_TA WHERE in_TA.A=out_TB.A ORDER BY in_TA.A DESC)
group by a,b,c,d

Answer 5

使用distinct：

SELECT DISTINCT r.* FROM
(
   ...your query here...
 ) r
ORDER BY 1
/

如何获得总和的价值？

5 个答案: