CODEWITHSUNDEEP
sqlcountgroup-byhaving
Benjamin Muschko
Benjamin Muschko

Reputation: 33436

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.

Upvotes: 262

Views: 1030242

Answers (4)

Conrad Frix
Conrad Frix

Reputation: 52645

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

Upvotes: 482

user4019456
user4019456

Reputation: 681

Try this query:

SELECT column_name
  FROM table_name
 GROUP BY column_name
HAVING COUNT(column_name) = 1;

Upvotes: 68

iryndin
iryndin

Reputation: 570

create table payment(
    user_id int(11),
    account int(11) not null,
    zip int(11) not null,
    dt date not null
);

insert into payment values
(1,123,55555,'2009-12-12'),
(1,123,66666,'2009-12-12'),
(1,123,77777,'2009-12-13'),
(2,456,77777,'2009-12-14'),
(2,456,77777,'2009-12-14'),
(2,789,77777,'2009-12-14'),
(2,789,77777,'2009-12-14');

select foo.user_id, foo.cnt from
(select user_id,count(account) as cnt, dt from payment group by account, dt) foo
where foo.cnt > 1;

Upvotes: 6

onedaywhen
onedaywhen

Reputation: 57023

I wouldn't recommend the HAVING keyword for newbies, it is essentially for legacy purposes.

I am not clear on what is the key for this table (is it fully normalized, I wonder?), consequently I find it difficult to follow your specification:

I would like to find all records for all users that have more than one payment per day with the same account number... Additionally, there should be a filter than only counts the records whose ZIP code is different.

So I've taken a literal interpretation.

The following is more verbose but could be easier to understand and therefore maintain (I've used a CTE for the table PAYMENT_TALLIES but it could be a VIEW:

WITH PAYMENT_TALLIES (user_id, zip, tally)
     AS
     (
      SELECT user_id, zip, COUNT(*) AS tally
        FROM PAYMENT
       GROUP 
          BY user_id, zip
     )
SELECT DISTINCT *
  FROM PAYMENT AS P
 WHERE EXISTS (
               SELECT * 
                 FROM PAYMENT_TALLIES AS PT
                WHERE P.user_id = PT.user_id
                      AND PT.tally > 1
              );

Upvotes: 8

Related Questions