I have a table named PAYMENT
. Within this table I have a user ID, an account number, a ZIP code and a date. I would like to find all records for all users that have more than one payment per day with the same account number.
UPDATE: Additionally, there should be a filter than only counts the records whose ZIP code is different.
This is how the table looks like:
| user_id | account_no | zip | date | | 1 | 123 | 55555 | 12-DEC-09 | | 1 | 123 | 66666 | 12-DEC-09 | | 1 | 123 | 55555 | 13-DEC-09 | | 2 | 456 | 77777 | 14-DEC-09 | | 2 | 456 | 77777 | 14-DEC-09 | | 2 | 789 | 77777 | 14-DEC-09 | | 2 | 789 | 77777 | 14-DEC-09 |
The result should look similar to this:
| user_id | count | | 1 | 2 |
How would you express this in a SQL query? I was thinking self join but for some reason my count is wrong.
Use the HAVING clause and GROUP By the fields that make the row unique
The below will find
all users that have more than one payment per day with the same account number
SELECT
user_id ,
COUNT(*) count
FROM
PAYMENT
GROUP BY
account,
user_id ,
date
HAVING
COUNT(*) > 1
Update If you want to only include those that have a distinct ZIP you can get a distinct set first and then perform you HAVING/GROUP BY
SELECT
user_id,
account_no ,
date,
COUNT(*)
FROM
(SELECT DISTINCT
user_id,
account_no ,
zip,
date
FROM
payment
)
payment
GROUP BY
user_id,
account_no ,
date
HAVING COUNT(*) > 1