我如何加入一个 sql table 并第一次出现在另一个 table 中，保持所有从第一次出现 table
How do I join all in one sql table and with first occurence from another table, keeping all from first table
以下 SQLite（android 房间）查询：
（更新 2021.01.03 07:43 GMT：我增强了 Q）
select o.*, gi.*
from orders as o
left join galleryitems as gi on gi.siteId = o.siteId
order by orderId asc
使用这些给定的表格：
orders: unique key is orderid
o.orderId | siteId | col1 | col2
----------+-------------+-----------+----------
456 | 1001 | o.456c1r1 | o.456c2r1
457 | 1002 | o.457c1r1 | o.457c2r1
458 | 1003 | o.458c1r1 | o.458c2r1
459 | 1004 | o.459c1r1 | o.459c2r1
460 | 1005 | o.460c1r1 | o.460c2r1
461 | 1006 | o.461c1r1 | o.461c2r1
462 | 1007 | o.462c1r1 | o.462c2r1
galleryitems: unique key is galleryId, reference to ordertable is gi.siteId
gi.galleryID | gi.col1 | gi.col2 | gi.siteId
-------------+-------------+-------------+----------
0001 | gi1001.c1r1 | gi1001.c2r1 | 1001
0002 | gi1001.c1r2 | gi1001.c2r2 | 1001
0003 | gi1002.c1r1 | gi1002.c1r1 | 1002
0004 | gi1003.c1r1 | gi1003.c1r1 | 1005
0005 | gi1003.c1r2 | gi1003.c1r2 | 1005
给出这个结果：
o.orderId | o.siteId | col1 | col2 | gi.galleryID | gi.col1 | gi.col2 | gi.siteId
----------+----------+-----------+-----------+--------------+-------------+-------------+----------
456 | 1001 | o.456c1r1 | o.456c2r1 | 0001 | gi1001.c1r1 | gi1001.c2r1 | 1001 <-- "duplicates"
456 | 1001 | o.456c1r1 | o.456c2r1 | 0002 | gi1001.c1r2 | gi1001.c2r2 | 1001 <-- "duplicates"
457 | 1002 | o.457c1r1 | o.457c2r1 | 0003 | gi1002.c1r1 | gi1002.c1r1 | 1002
458 | 1003 | o.458c1r1 | o.458c2r1 | <null> | <null> | <null> | <null>
459 | 1004 | o.459c1r1 | o.459c2r1 | <null> | <null> | <null> | <null>
460 | 1005 | o.460c1r1 | o.460c2r1 | 0004 | gi1003.c1r1 | gi1003.c1r1 | 1005 <-- "duplicates"
460 | 1005 | o.460c1r1 | o.460c2r1 | 0005 | gi1003.c1r2 | gi1003.c1r2 | 1005 <-- "duplicates"
461 | 1006 | o.461c1r1 | o.461c2r1 | <null> | <null> | <null> | <null>
462 | 1007 | o.462c1r1 | o.462c2r1 | <null> | <null> | <null> | <null>
正如您在下面看到的，我希望只有在 gi 出现时才会出现数据，但只有第一次出现时才会出现，所以它们不会“重复”（如果 [=39 只需填写客栈数据=] 被发现（第一次出现是可以的，或者甚至被计算在内，如果在加入期间可能的话），否则 gi 字段应该是 <null> ）：
o.orderId | o.siteId | col1 | col2 | gi.galleryID | gi.col1 | gi.col2 | gi.siteId
----------+----------+-----------+-----------+--------------+-------------+-------------+----------
456 | 1001 | o.456c1r1 | o.456c2r1 | 0001 | gi1001.c1r1 | gi1001.c2r1 | 1001 <-- no "duplicates"
457 | 1002 | o.457c1r1 | o.457c2r1 | 0003 | gi1002.c1r1 | gi1002.c1r1 | 1002
458 | 1003 | o.458c1r1 | o.458c2r1 | <null> | <null> | <null> | <null>
459 | 1004 | o.459c1r1 | o.459c2r1 | <null> | <null> | <null> | <null>
460 | 1005 | o.460c1r1 | o.460c2r1 | 0004 | gi1003.c1r1 | gi1003.c1r1 | 1005 <-- no "duplicates"
461 | 1006 | o.461c1r1 | o.461c2r1 | <null> | <null> | <null> | <null>
462 | 1007 | o.462c1r1 | o.462c2r1 | <null> | <null> | <null> | <null>
我看到有很多建议，但它们无法应对 SQLite 有限的库。
有什么建议吗？
使用 androidx 房间。
您可以通过聚合来获得最小值为 galleryitems.row1 的行，方法是使用 SQLite 的一个非标准 SQL 但它是documented:
select min(gi.row1) as row1, gi.row2, gi.siteId, o.*
from orders as o left join galleryitems as gi
on gi.siteId = o.siteId
group by o.siteId
order by orderId asc
或使用galleryitems的rowid来定义第一次出现:
select gi.*, o.*
from orders as o left join galleryitems as gi
on gi.siteId = o.siteId
group by o.siteId
having min(gi.rowid) or min(gi.rowid) is null
order by orderId asc
参见demo。
@forpas 将我纠正到宽阔的道路上，我什至可以加强它。谢谢！
select count(gi.siteId) as cc, gi.*, o.*
from orders as o
left join galleryitems as gi on gi.siteId = o.siteId
group by o.siteId
这个结果现在变成了这个结果：（注意 !! group by o.siteId 而不是 gi.siteId）
cc |o.orderId | o.siteId | col1 | col2 | gi.galleryID | gi.col1 | gi.col2 | gi.siteId
---+----------+----------+-----------+-----------+--------------+-------------+-------------+----------
2 |456 | 1001 | o.456c1r1 | o.456c2r1 | 0001 | gi1001.c1r1 | gi1001.c2r1 | 1001 <-- no "duplicates"
1 |457 | 1002 | o.457c1r1 | o.457c2r1 | 0003 | gi1002.c1r1 | gi1002.c1r1 | 1002
0 |458 | 1003 | o.458c1r1 | o.458c2r1 | <null> | <null> | <null> | <null>
0 |459 | 1004 | o.459c1r1 | o.459c2r1 | <null> | <null> | <null> | <null>
2 |460 | 1005 | o.460c1r1 | o.460c2r1 | 0004 | gi1003.c1r1 | gi1003.c1r1 | 1005 <-- no "duplicates"
0 |461 | 1006 | o.461c1r1 | o.461c2r1 | <null> | <null> | <null> | <null>
0 |462 | 1007 | o.462c1r1 | o.462c2r1 | <null> | <null> | <null> | <null>
以下 SQLite（android 房间）查询：
（更新 2021.01.03 07:43 GMT：我增强了 Q）
select o.*, gi.*
from orders as o
left join galleryitems as gi on gi.siteId = o.siteId
order by orderId asc
使用这些给定的表格：
orders: unique key is orderid
o.orderId | siteId | col1 | col2
----------+-------------+-----------+----------
456 | 1001 | o.456c1r1 | o.456c2r1
457 | 1002 | o.457c1r1 | o.457c2r1
458 | 1003 | o.458c1r1 | o.458c2r1
459 | 1004 | o.459c1r1 | o.459c2r1
460 | 1005 | o.460c1r1 | o.460c2r1
461 | 1006 | o.461c1r1 | o.461c2r1
462 | 1007 | o.462c1r1 | o.462c2r1
galleryitems: unique key is galleryId, reference to ordertable is gi.siteId
gi.galleryID | gi.col1 | gi.col2 | gi.siteId
-------------+-------------+-------------+----------
0001 | gi1001.c1r1 | gi1001.c2r1 | 1001
0002 | gi1001.c1r2 | gi1001.c2r2 | 1001
0003 | gi1002.c1r1 | gi1002.c1r1 | 1002
0004 | gi1003.c1r1 | gi1003.c1r1 | 1005
0005 | gi1003.c1r2 | gi1003.c1r2 | 1005
给出这个结果：
o.orderId | o.siteId | col1 | col2 | gi.galleryID | gi.col1 | gi.col2 | gi.siteId
----------+----------+-----------+-----------+--------------+-------------+-------------+----------
456 | 1001 | o.456c1r1 | o.456c2r1 | 0001 | gi1001.c1r1 | gi1001.c2r1 | 1001 <-- "duplicates"
456 | 1001 | o.456c1r1 | o.456c2r1 | 0002 | gi1001.c1r2 | gi1001.c2r2 | 1001 <-- "duplicates"
457 | 1002 | o.457c1r1 | o.457c2r1 | 0003 | gi1002.c1r1 | gi1002.c1r1 | 1002
458 | 1003 | o.458c1r1 | o.458c2r1 | <null> | <null> | <null> | <null>
459 | 1004 | o.459c1r1 | o.459c2r1 | <null> | <null> | <null> | <null>
460 | 1005 | o.460c1r1 | o.460c2r1 | 0004 | gi1003.c1r1 | gi1003.c1r1 | 1005 <-- "duplicates"
460 | 1005 | o.460c1r1 | o.460c2r1 | 0005 | gi1003.c1r2 | gi1003.c1r2 | 1005 <-- "duplicates"
461 | 1006 | o.461c1r1 | o.461c2r1 | <null> | <null> | <null> | <null>
462 | 1007 | o.462c1r1 | o.462c2r1 | <null> | <null> | <null> | <null>
正如您在下面看到的，我希望只有在 gi 出现时才会出现数据，但只有第一次出现时才会出现，所以它们不会“重复”（如果 [=39 只需填写客栈数据=] 被发现（第一次出现是可以的，或者甚至被计算在内，如果在加入期间可能的话），否则 gi 字段应该是 <null> ）：
o.orderId | o.siteId | col1 | col2 | gi.galleryID | gi.col1 | gi.col2 | gi.siteId
----------+----------+-----------+-----------+--------------+-------------+-------------+----------
456 | 1001 | o.456c1r1 | o.456c2r1 | 0001 | gi1001.c1r1 | gi1001.c2r1 | 1001 <-- no "duplicates"
457 | 1002 | o.457c1r1 | o.457c2r1 | 0003 | gi1002.c1r1 | gi1002.c1r1 | 1002
458 | 1003 | o.458c1r1 | o.458c2r1 | <null> | <null> | <null> | <null>
459 | 1004 | o.459c1r1 | o.459c2r1 | <null> | <null> | <null> | <null>
460 | 1005 | o.460c1r1 | o.460c2r1 | 0004 | gi1003.c1r1 | gi1003.c1r1 | 1005 <-- no "duplicates"
461 | 1006 | o.461c1r1 | o.461c2r1 | <null> | <null> | <null> | <null>
462 | 1007 | o.462c1r1 | o.462c2r1 | <null> | <null> | <null> | <null>
我看到有很多建议，但它们无法应对 SQLite 有限的库。
有什么建议吗？
使用 androidx 房间。
您可以通过聚合来获得最小值为 galleryitems.row1 的行，方法是使用 SQLite 的一个非标准 SQL 但它是documented:
select min(gi.row1) as row1, gi.row2, gi.siteId, o.*
from orders as o left join galleryitems as gi
on gi.siteId = o.siteId
group by o.siteId
order by orderId asc
或使用galleryitems的rowid来定义第一次出现:
select gi.*, o.*
from orders as o left join galleryitems as gi
on gi.siteId = o.siteId
group by o.siteId
having min(gi.rowid) or min(gi.rowid) is null
order by orderId asc
参见demo。
@forpas 将我纠正到宽阔的道路上，我什至可以加强它。谢谢！
select count(gi.siteId) as cc, gi.*, o.*
from orders as o
left join galleryitems as gi on gi.siteId = o.siteId
group by o.siteId
这个结果现在变成了这个结果：（注意 !! group by o.siteId 而不是 gi.siteId）
cc |o.orderId | o.siteId | col1 | col2 | gi.galleryID | gi.col1 | gi.col2 | gi.siteId
---+----------+----------+-----------+-----------+--------------+-------------+-------------+----------
2 |456 | 1001 | o.456c1r1 | o.456c2r1 | 0001 | gi1001.c1r1 | gi1001.c2r1 | 1001 <-- no "duplicates"
1 |457 | 1002 | o.457c1r1 | o.457c2r1 | 0003 | gi1002.c1r1 | gi1002.c1r1 | 1002
0 |458 | 1003 | o.458c1r1 | o.458c2r1 | <null> | <null> | <null> | <null>
0 |459 | 1004 | o.459c1r1 | o.459c2r1 | <null> | <null> | <null> | <null>
2 |460 | 1005 | o.460c1r1 | o.460c2r1 | 0004 | gi1003.c1r1 | gi1003.c1r1 | 1005 <-- no "duplicates"
0 |461 | 1006 | o.461c1r1 | o.461c2r1 | <null> | <null> | <null> | <null>
0 |462 | 1007 | o.462c1r1 | o.462c2r1 | <null> | <null> | <null> | <null>