使用Redshift将空值替换为最后一个非空值

Question

我有一个称为数量的表：

+----------+----------+
| 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     |
+----------+----------+

我想要选择它，以便可以使用先前的非空值来填充空白：

+----------+----------+
| 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），Redshift 1.0.1499编译

我怎样才能做到这一点？

谢谢！

Answer 1

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

这是一个窗口函数，它返回指定窗口中的最后一个值