我坐在下面的问题上:我正在编写一个视图,将几个表连接到一个person表中。现在我试图加入partners表,但是我只需要历史上的最后一个有效的合作伙伴行:
伙伴桌:
id,
name,
married_at,
divorced_at,
died_at,
someone_id如你所见,这是关于你结婚的伴侣的。一个人一次只能有一个伴侣,但历史上有几个伴侣。因此,某人(someone_id)的最后一个合伙人可能是:
我只需要为某人找到最后一排搭档。
到目前为止我得到的是:
select *
from someone_table s
left join partners p on (p.someone_id = s.id and (p.divorced_at is null and p.died_at is null) )但这--很明显--只给了我还活着和已婚的伴侣。当然,这些伴侣是某人的最后一个伴侣,但所有其他“某人”谁是最后一个伴侣是离婚或死亡将不会在声明的结果。我如何得到其他的和只有一排为每一个人?
我还尝试了一个select语句作为表并使用rownum。
select *
from someone s,
(select * from partners p where p.someone_id = s.id and ROWNUM = 1 order by p.married_at)但是这条语句总是失败,出现了“无效标识符s.id”错误。
注意:表格结构是固定的,不能更改。DBMS是甲骨文。
提前感谢
编辑:样本数据
partners_table
╔════╦═════════╦════════════╦═════════════╦════════════╦════════════╗
║ id ║ name ║ married_at ║ divorced_at ║ died_at ║ someone_id ║
╠════╬═════════╬════════════╬═════════════╬════════════╬════════════╣
║ 1 ║ partner ║ 01.01.2000 ║ ║ ║ 12 ║
║ 2 ║ honey1 ║ 15.01.2000 ║ 15.01.2001 ║ ║ 15 ║
║ 3 ║ honey2 ║ 16.02.2001 ║ ║ ║ 15 ║
║ 4 ║ beauty ║ 23.03.2005 ║ ║ 25.03.2005 ║ 16 ║
║ 5 ║ lady1 ║ 11.11.2000 ║ 11.12.2000 ║ ║ 20 ║
║ 6 ║ lady2 ║ 12.12.2000 ║ 01.01.2001 ║ ║ 20 ║
║ 7 ║ lady3 ║ 02.02.2001 ║ ║ 04.02.2004 ║ 20 ║
║ 8 ║ lady4 ║ 05.05.2005 ║ ║ ║ 20 ║
║ 9 ║ mate ║ 23.06.2003 ║ 12.12.2009 ║ ║ 25 ║
╚════╩═════════╩════════════╩═════════════╩════════════╩════════════╝最后的历史行是:
╔════╦═════════╦════════════╦═════════════╦════════════╦════════════╗
║ id ║ name ║ married_at ║ divorced_at ║ died_at ║ someone_id ║
╠════╬═════════╬════════════╬═════════════╬════════════╬════════════╣
║ 1 ║ partner ║ 01.01.2000 ║ ║ ║ 12 ║
║ 3 ║ honey2 ║ 16.02.2001 ║ ║ ║ 15 ║
║ 4 ║ beauty ║ 23.03.2005 ║ ║ 25.03.2005 ║ 16 ║
║ 8 ║ lady4 ║ 05.05.2005 ║ ║ ║ 20 ║
║ 9 ║ mate ║ 23.06.2003 ║ 12.12.2009 ║ ║ 25 ║
╚════╩═════════╩════════════╩═════════════╩════════════╩════════════╝发布于 2015-05-12 11:12:16
这应该能做你想做的事:
with partners (id, name, married_at, divorced_at, died_at, someone_id) as (select 1, 'partner', to_date('01/01/2000', 'dd/mm/yyyy'), null, null, 12 from dual union all
select 2, 'honey1', to_date('15/01/2000', 'dd/mm/yyyy'), to_date('15/01/2001', 'dd/mm/yyyy'), null, 15 from dual union all
select 3, 'honey2', to_date('16/02/2001', 'dd/mm/yyyy'), null, null, 15 from dual union all
select 4, 'beauty', to_date('23/03/2005', 'dd/mm/yyyy'), null, to_date('25/03/2005', 'dd/mm/yyyy'), 16 from dual union all
select 5, 'lady1', to_date('11/11/2000', 'dd/mm/yyyy'), to_date('11/12/2000', 'dd/mm/yyyy'), null, 20 from dual union all
select 6, 'lady2', to_date('12/12/2000', 'dd/mm/yyyy'), to_date('01/01/2001', 'dd/mm/yyyy'), null, 20 from dual union all
select 7, 'lady3', to_date('02/02/2001', 'dd/mm/yyyy'), null, to_date('04/02/2004', 'dd/mm/yyyy'), 20 from dual union all
select 8, 'lady4', to_date('05/05/2005', 'dd/mm/yyyy'), null, null, 20 from dual union all
select 9, 'mate', to_date('23/06/2003', 'dd/mm/yyyy'), to_date('12/12/2009', 'dd/mm/yyyy'), null, 25 from dual)
select id,
name,
married_at,
divorced_at,
died_at,
someone_id
from (select id,
name,
married_at,
divorced_at,
died_at,
someone_id,
row_number() over (partition by someone_id order by married_at desc) rn
from partners)
where rn = 1;
ID NAME MARRIED_AT DIVORCED_AT DIED_AT SOMEONE_ID
---------- ------- ---------- ----------- ---------- ----------
1 partner 01/01/2000 12
3 honey2 16/02/2001 15
4 beauty 23/03/2005 25/03/2005 16
8 lady4 05/05/2005 20
9 mate 23/06/2003 12/12/2009 25发布于 2015-05-12 10:43:08
如果我理解你的问题(我相信我理解你的问题),你应该这样做:
SELECT *
FROM someone_table s
left join (
SELECT *
FROM (
SELECT *
FROM partners p
WHERE p.someone_id = s.id
ORDER BY GREATEST(died_at, divorced_at, married_at)
) x
WHERE ROWNUM = 1
) y注意:我不是甲骨文,我大部分的工作都是在使用服务器,但是according to this post greatest应该在oracle数据库上工作。
发布于 2015-05-12 11:40:43
方法1:
SELECT
*
FROM
partners
WHERE
someone_id = $someone_id
AND
married_at = (SELECT MAX(married_at) FROM partners WHERE someone_id = $someone_id GROUP BY someone_id);方法2:
SELECT
p.*
FROM
partners p
INNER JOIN
(
SELECT
someone_id, MAX(married_at) as lastmarried_at
FROM
partners
GROUP BY
someone_id
) m
ON m.someone_id = p.someone_id AND m.lastmarried_at = p.married_at
where p.someone_id in ($someone_id1, $someone_id2);注意:用实际值替换$someone_id
https://stackoverflow.com/questions/30188181
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