Question

我有下表：

AMNT1   |  COLUMN1  |  COLUMN2  |  COLUMN3    | GROUP1
--------|-----------|-----------|-------------|--------
1.00    | COL1_ROW1 | COL2_ROW1 |   COL3_ROW1 | AAA
9.00    | COL1_ROW2 | COL2_ROW2 |   COL2_ROW2 | AAA
2.00    | COL1_ROW3 | COL2_ROW3 |   COL3_ROW3 | BBB
3.00    | COL1_ROW4 | COL2_ROW4 |   COL3_ROW4 | CCC

我想对GROUP1分组的AMNT1求和：

SELECT GROUP1, SUM(AMNT1) FROM ND_TEST GROUP BY GROUP1;

GROUP1 | SUM(AMNT1)
-------|-----------
AAA    | 10.00
BBB    | 2.00
CCC    | 3.00

另外，我想从一个行中选择COLUMN1，COLUMN2和COLUMN3。所以我的输出应该是这样的：

GROUP1 | SUM(AMNT1)|  COLUMN1  |  COLUMN2  |   COLUMN3  |
-------|-----------|-----------|-----------|------------|
AAA    | 10.00     | COL1_ROW1 | COL2_ROW1 |  COL3_ROW1 |
BBB    | 2.00      | COL1_ROW3 | COL2_ROW3 |  COL3_ROW3 |   
CCC    | 3.00      | COL1_ROW4 | COL2_ROW4 |  COL3_ROW4 |

如果在分区上使用总和，则每个组将得到重复...如果使用聚合函数，则不会从同一行中得到结果... 你有主意吗？

谢谢！

Answer 1

select group1, sum_amnt1, column1, column2, column3
from   (
         select group1, sum(amnt1) over (partition by group1) as sum_amnt1, 
                column1, column2, column3,
                row_number() over (partition by group1 order by null) as rn
         from   your_table
       )
where  rn = 1

order by null函数中的

row_number()与您的澄清（在注释中）相对应，即来自每个组的任何行都可以（您不在乎哪个）

Answer 2

您可以使用窗口功能：

select nt.*
from (select nt.*, sum(AMNT1) over (partition by GROUP1) as sum,
             row_number() over (partition by GROUP1 order by AMNT1) as seq
      from ND_TEST as nt
     ) nt
where seq = 1;

我需要按一列分组并从一个数据集中显示更多列

2 个答案: