我需要SELECT
一行中issue_date = maturity_date
的另一行id
,且相同的amount_usd。
我尝试过自我加入,但我没有得到正确的结果。
这是我的表格的简化版本:
ID ISSUE_DATE MATURITY_DATE AMOUNT_USD
1 2010-01-01 00:00:00.000 2015-12-01 00:00:00.000 5000
1 2010-01-01 00:00:00.000 2001-09-19 00:00:00.000 700
2 2014-04-09 00:00:00.000 2019-04-09 00:00:00.000 400
1 2015-12-01 00:00:00.000 2016-12-31 00:00:00.000 5000
5 2015-02-24 00:00:00.000 2015-02-24 00:00:00.000 8000
4 2012-11-29 00:00:00.000 2015-11-29 00:00:00.000 10000
3 2015-01-21 00:00:00.000 2018-01-21 00:00:00.000 17500
2 2015-02-02 00:00:00.000 2015-12-05 00:00:00.000 12000
1 2015-01-12 00:00:00.000 2018-01-12 00:00:00.000 18000
2 2015-12-05 00:00:00.000 2016-01-10 00:00:00.000 12000
Result should be:
ID ISSUE_DATE MATURITY_DATE AMOUNT_USD
1 2015-12-01 00:00:00.000 2016-12-31 00:00:00.000 5000
2 2015-12-05 00:00:00.000 2016-01-10 00:00:00.000 12000
提前致谢!
答案 0 :(得分:0)
请执行以下操作:http://sqlfiddle.com/#!6/c0a02/1
select a.id, a.issue_date, a.maturity_date, a.amount_usd
from tbl a
inner join tbl b
on a.id = b.id
and a.maturity_date = b.issue_date
-- added to prevent same maturity date and issue date
where a.maturity_date <> a.issue_date
输出:
| id | issue_date | maturity_date | amount_usd |
|----|----------------------------|----------------------------|------------|
| 1 | January, 01 2010 00:00:00 | December, 01 2015 00:00:00 | 5000 |
| 2 | February, 02 2015 00:00:00 | December, 05 2015 00:00:00 | 12000 |