问题
I have the following data:
id date mia
1 1/1/2017 3
1 1/2/2017 1
1 1/3/2017 2
2 1/4/2017 1
2 1/5/2017 4
2 1/6/2017 6
.
.
.
.
and so on.
If I give input as id=1 I should fetch record of 2017-02-01 and if input id=2 then record of 2017-05-01 i.e record of previous month of the highest date.
回答1:
You could use:
SELECT *
FROM (SELECT *, ROW_NUMBER() OVER(PARTITION BY id ORDER BY mia DESC) AS rn
FROM table) sub
WHERE rn = 2;
回答2:
you may not define a column named as date, instead i use date_.
For former versions you may refer to @lad2025 's answer, if you're on oracle12c, you may query with the following :
select min(date_) min_date
from
(
select date_
from mytable
where id = &i_id
group by id, date_
order by date_
fetch first 2 rows only
);
来源：https://stackoverflow.com/questions/47535648/find-second-highest-record-from-oracle-db