Reputation: 3484
I have table with the data in following format.
User ID| Date Paid | Amount Paid
- - - - - - - - - - - - - - - - -
1 | 2019-04-01 | 120
1 | 2019-03-01 | 120
1 | 2019-05-01 | 130
2 | 2019-03-01 | 100
2 | 2019-04-01 | 110
3 | 2019-05-01 | 100
I need export this table into a format where all of the records for a single user should be in a single row, and date paid and amount records should start from left with the oldest record first.
User ID | 1st Date Paid | 1st Amount Paid | 2nd Date Paid | 2nd Amount Paid | 3rd Date Paid | 3rd Amount Paid | . . . . .
- - - - - - - - - - - - - - - - - -- - - - - - -- -- - - -- - - - - - - -
1 | 2019-03-01 | 120 | 2019-04-01 | 120 | 2019-05-01 | 130
2 | 2019-03-01 | 100 | 2019-40-01 | 110 |
3 | 2019-05-01 | 100 |
There could be up to a maximum of 10 records per user, so the export should contain the date paid and amount paid columns 10 times. If a user doesn't contain 10 records, then those columns will be left empty.
What is the best way to achieve this result?
Upvotes: 0
Views: 30
Reputation: 1269493
You can use conditional aggregation:
select user_id,
max(case when seqnum = 1 then date_paid end) as date_paid1,
max(case when seqnum = 1 then amount_paid end) as amount_paid1,
max(case when seqnum = 2 then date_paid end) as date_paid2,
max(case when seqnum = 2 then amount_paid end) as amount_paid2,
max(case when seqnum = 3 then date_paid end) as date_paid3,
max(case when seqnum = 3 then amount_paid end) as amount_paid3
from (select t.*,
row_number() over (partition by user_id order by date_paid) as seqnum
from t
) t
group by user_id;
Upvotes: 1