用前一行值oracle sql填充表中的行

Question

+----+---------+-------+--------+---------+--------+
| id | counter | name  | duties | remarks | monies |
+----+---------+-------+--------+---------+--------+
|  1 |       1 | jake  |        |         |        |
|  2 |       0 |       | clean  | misc    | 12k    |
|  3 |       1 | james |        |         |        |
|  4 |       0 |       | clean  | misc    | 12k    |
|  5 |       0 |       | soap   | misc    | 12k    |
|  6 |       0 |       | shower | misc    | 12k    |
|  7 |       1 | john  |        |         |        |
|  8 |       0 |       | dry    | misc    | 12k    |
|  9 |       0 |       | scrub  | misc    | 12k    |
+----+---------+-------+--------+---------+--------+

上面的数据是按人分组的，然后列出了他们的职责：

我需要一种使用 oracle sql 使数据看起来像这样的方法：

+----+---------+-------+--------+---------+--------+
| id | counter | name  | duties | remarks | monies |
+----+---------+-------+--------+---------+--------+
|  1 |       1 | jake  |        |         |        |
|  2 |       0 | jake  | clean  | misc    | 12k    |
|  3 |       1 | james |        |         |        |
|  4 |       0 | james | clean  | misc    | 12k    |
|  5 |       0 | james | soap   | misc    | 12k    |
|  6 |       0 | james | shower | misc    | 12k    |
|  7 |       1 | john  |        |         |        |
|  8 |       0 | john  | dry    | misc    | 12k    |
|  9 |       0 | john  | scrub  | misc    | 12k    |
+----+---------+-------+--------+---------+--------+

遍历每一行有点麻烦...不要介意它的 plsql - sql 是否更喜欢

尝试了一些东西......但在每个人的第二行之后它又回到了 null

Answer 1

一种方法是lag(ignore nulls):

select coalesce(name, lag(name ignore nulls) over (order by id)),
       . . .
from t;

这是从name 列中获取先前的非NULL 值，其中“previous”基于id 列的顺序。