SQL query for finding records where count > 1

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.

Answer

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