Question

我有一个名为quantity的表：

+----------+----------+
| date     | quantity |
+----------+----------+
| 30/11/17 | 90       |
+----------+----------+
| 01/12/17 |          |
+----------+----------+
| 02/12/17 |          |
+----------+----------+
| 03/12/17 | 1622     |
+----------+----------+
| 04/12/17 |          |
+----------+----------+
| 05/12/17 | 9092     |
+----------+----------+
| 06/12/17 |          |
+----------+----------+
| 07/12/17 |          |
+----------+----------+
| 08/12/17 | 2132     |
+----------+----------+
| 09/12/17 |          |
+----------+----------+
| 10/12/17 | 2889     |
+----------+----------+

我想选择它，以便我可以使用之前的非空值填充空白：

+----------+----------+
| date     | quantity |
+----------+----------+
| 30/11/17 | 90       |
+----------+----------+
| 01/12/17 | 90       |
+----------+----------+
| 02/12/17 | 90       |
+----------+----------+
| 03/12/17 | 1622     |
+----------+----------+
| 04/12/17 | 1622     |
+----------+----------+
| 05/12/17 | 9092     |
+----------+----------+
| 06/12/17 | 9092     |
+----------+----------+
| 07/12/17 | 9092     |
+----------+----------+
| 08/12/17 | 2132     |
+----------+----------+
| 09/12/17 | 2132     |
+----------+----------+
| 10/12/17 | 2889     |
+----------+----------+

我在i686-pc-linux-gnu上使用PostgreSQL 8.0.2，由GCC gcc（GCC）3.4.2 20041017（Red Hat 3.4.2-6.fc3）编译，Redshift 1.0.1499

我怎么能实现这个目标？

谢谢！

Answer 1

等等 last_value(quantity ignore nulls) over (order by date rows unbounded preceding)

它是一个窗口函数，返回指定窗口中的最后一个值

Answer 2

您可以使用以下内容：

select date,(select t.quantity from [tablename] t where t.quantity is not null and t.date <= t1.date
order by t.date desc limit 1) from [tablename] t1;

t.quantity is not null：确保您不会在结果集中获取空值。

t.date <= t1.date：确保选择最后一个已知值。

使用Redshift将空值替换为最后一个非空值

2 个答案: