我不希望任何形式的JOIN
在这里.我正在使用PHP构建两个表的RSS提要,我想从两个表中选择所有行,保持行分开但按公共created
列排序.
例如,如果我有一个表foo
:
id downloads views created
-----------------------------------------------
1 12 23 2011-07-22 00:10:16
2 51 900 2011-07-22 10:11:45
3 8 80 2011-07-23 04:12:18
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还有一张桌子bar
:
id title body created
-----------------------------------------------
1 foo ogblog 2011-07-21 10:54:07
3 bar zip 2011-07-24 10:54:07
4 zip bar 2011-07-25 10:54:07
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我想从公共created
列排序的两个表中选择所有数据,因此示例结果集将是(忽略,bar.id
因为它不需要):
id title body downloads views created | table
-------------------------------------------------------------------------------
NULL bar zip NULL NULL 2011-07-24 10:54:07 | bar
NULL foo ogblog NULL NULL 2011-07-21 10:54:07 | bar
1 NULL NULL 12 23 2011-07-22 00:10:16 | foo
2 NULL NULL 51 900 2011-07-22 10:11:45 | foo
3 NULL NULL 8 80 2011-07-23 04:12:18 | foo
NULL zip bar NULL NULL 2011-07-25 10:54:07 | bar
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table
不需要该列; 我添加它以使事情更容易理解.
希望很明显我想做什么; 而不是JOIN
在从两个表的列生成行的位置,我希望获得具有公共列布局的所有行数据,其中表中不存在的任何列都NULL
放入其中.
如果您需要澄清,请告诉我.
Bre*_*len 32
使用虚拟列来计算不同的结构,使用联合来连接它们,使用父选择来处理排序:
SELECT * FROM (
(SELECT foo.id, NULL AS title, NULL AS body, foo.downloads, foo.views, foo.created FROM foo)
UNION ALL
(SELECT NULL AS id, bar.title, bar.body, NULL AS downloads, NULL AS views, bar.created FROM bar)
) results
ORDER BY created ASC
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Ike*_*ker 15
扩展@Brendan Bullen的建议,这里有一个使用UNION ALL
它的例子应该适合你:
SELECT id as id, NULL as title, NULL as body, downloads as downloads,
views as views, created as created, 'foo' as table_name
FROM foo
UNION ALL
SELECT NULL as id, title as title, body as body, NULL as downloads,
NULL as views, created as created, 'bar' as table_name
FROM bar
ORDER BY created ASC
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