Posted By: Anonymous
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
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