lau*_*kok 5 mysql sql inner-join self-join left-join
例如,我在表格中有这些数据
id name parent parent_id
1 add self 100
2 manage null 100
3 add 10 200
4 manage null 200
5 add 20 300
6 manage null 300
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我怎么能left join或inner join这个表本身所以我得到这个结果?
id name parent
2 manage self
4 manage 10
6 manage 20
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正如我所知,我只想用关键字'manage'查询行,但我希望列parent的数据在add的行中作为结果中manage的行.
可能吗?
编辑:
我的实际表的简化版- system,
system_id parent_id type function_name name main_parent make_accessible sort
31 30 left main Main NULL 0 1
32 31 left page_main_add Add self 0 1
33 31 left page_main_manage Manage NULL 0 2
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我的实际查询已经非常混乱......
SELECT
a.system_id,
a.main_parent,
b.name,
b.make_accessible,
b.sort
FROM system AS a
INNER JOIN -- self --
(
SELECT system_id, name, make_accessible, sort
FROM system AS s2
LEFT JOIN -- search --
(
SELECT system_id AS parent_id
FROM system AS s1
WHERE s1.function_name = 'page'
) AS s1
ON s1.parent_id = s2.parent_id
WHERE s2.parent_id = s1.parent_id
AND s2.system_id != s1.parent_id
ORDER BY s2.sort ASC
) b
ON b.system_id = a.parent_id
WHERE a.function_name LIKE '%manage%'
ORDER BY b.sort ASC
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我得到的结果,
system_id main_parent name make_accessible sort
33 NULL Main 0 1
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但我在此之后,
system_id main_parent name make_accessible sort
33 self Main 0 1
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您只需要两次引用该表:
select t1.id, t1.name, t2.id, t2.name
from TableA t1
inner join TableA t2
on t1.parent_id = t2.Id
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如果要在列表中查看根,请将内部替换为左连接.
更新:
我误解了你的问题.在我看来,你总是有两行,管理一行并添加一行.要从manage获得"添加":
select system.*, (select parent
from system s2
where s2.parent_id = system.parent_id
and s2.name = 'add')
AS parent
from system
where name = 'manage'
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或者,您可以将表拆分为两个派生表,并通过parent_id连接它们:
select *
from system
inner join
(
select * from system where name = 'add'
) s2
on system.parent_id = s2.parent_id
where system.name = 'manage'
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这将允许您使用s2中的所有列.