Select 两个 SQLite 表之间的 n 个最新行

Question

我在 SQLite 数据库中有两个 tables A 和 B。 Table A 看起来像这样：

|  id  |  name         |  description  |  created     |
|------|---------------|---------------|--------------|
|  1   |  "Something"  |  "Something"  |  1466266963  |
|  2   |  "Something"  |  "Something"  |  1466266965  |
|  3   |  "Something"  |  "Something"  |  1466266967  |
|  4   |  "Something"  |  "Something"  |  1466266969  |
|  5   |  "Something"  |  "Something"  |  1466266971  |
|  6   |  "Something"  |  "Something"  |  1466266975  |

而 table B 是这样的：

|  id  |  title        |  content      |  created     |
|------|---------------|---------------|--------------|
|  1   |  "Something"  |  "Something"  |  1466266951  |
|  2   |  "Something"  |  "Something"  |  1466266953  |
|  3   |  "Something"  |  "Something"  |  1466266954  |
|  4   |  "Something"  |  "Something"  |  1466266956  |
|  5   |  "Something"  |  "Something"  |  1466266957  |
|  6   |  "Something"  |  "Something"  |  1466266978  |

而且我想获得两个 table 中最老的 4 个。

所以在这个例子中我会得到类似的东西：

|  id  |  title        |  name         |  content      |  description  |  created     |
|------|---------------|---------------|---------------|---------------|--------------|
|  6   |  "Something"  |               |  "Something"  |               |  1466266975  |
|  6   |               |  "Something"  |               |  "Something"  |  1466266975  |
|  5   |               |  "Something"  |               |  "Something"  |  1466266975  |
|  4   |               |  "Something"  |               |  "Something"  |  1466266975  |

我试过 SELECT A.*, B.* FROM A, B ORDER BY created DESC LIMIT 4 但我得到的结果是空的（不过我的 table 不是空的）。

我也查看了 JOIN，但没有成功。

有什么想法吗？谢谢。

Answer 1

使用union all

select * from(
select * from a
union all
select * from b
) t
order by created DESC LIMIT 4

Select 两个 SQLite 表之间的 n 个最新行

Select n latest rows between two SQLite tables

sqlite

select

multiple-tables