Question

我尝试选择max（date_added）支付方面的行，只选择col_a和col_b中的重复值，其中 log_id 是唯一的。我已经看过使用DISTINCT ON并使用窗口函数的解决方案，但是我无法正确获取语法以获得我想要的结果。

从此表中：

+--------+-------+-------+-------+-------+-------+-------------------------+
| log_id | col_a | col_b | col_c | col_d | col_e |       date_added        |
+--------+-------+-------+-------+-------+-------+-------------------------+
|      1 | ACME  | West  |    14 |    27 | A     | 2016-01-01 12:45:00.453 |
|      2 | ACME  | West  |    17 |    31 | A     | 2016-01-02 07:33:24.551 |
|      3 | ACME  | East  |    13 |    27 | B     | 2016-01-01 11:43:21.223 |
|      4 | ACME  | East  |    17 |    28 | A     | 2016-01-03 09:15:50.113 |
+--------+-------+-------+-------+-------+-------+-------------------------+

我该如何归还：

+--------+-------+-------+-------+-------+-------+-------------------------+
| log_id | col_a | col_b | col_c | col_d | col_e |       date_added        |
+--------+-------+-------+-------+-------+-------+-------------------------+
|      2 | ACME  | West  |    17 |    31 | A     | 2016-01-02 07:33:24.551 |
|      4 | ACME  | East  |    17 |    28 | A     | 2016-01-03 09:15:50.113 |
+--------+-------+-------+-------+-------+-------+-------------------------+

Answer 1

一种方法是使用窗口函数......类似......

set my variable to 4

这样做是按照添加日期的降序为每个col_A和col_B分组（将每个A，b组的行号重置为1）分配行号，然后仅返回行号为1的记录（拥有最长日期）

或在窗口函数之前...我们在每个col_a和Col_B组中获取一组最大日期数据，然后连接回基本集以限制数据。

with cte as (SELECTlog_id,col_a,col_b,col_c,col_d, col_e, date_added,
                    row_number() over (partition by col_A, col_B order by date_added desc) rn
             FROM tableName )
SELECT log_id,col_a,col_b,col_c,col_d,col_e,date_added 
FROM cte 
WHERE rn = 1;

Answer 2

使用distinct on很简单：

select distinct on (col_a, col_b)
  *
from
  yourtable
order by
  col_a, col_b, -- columns from `distinct on` should be in the `order by` clause
  -- and here is a max value of date_added for every distinct (col_a, col_b) pairs 
  -- because of descending order
  date_added desc;

根据最大日期时间戳选择列

2 个答案: