Opposite of DISTINCT without Group By

[sw9]

I have a table with many columns, including: group_id, contact_id

It has 20,000 rows in it. I cannot have entries where group_id and contact_id are the same more than once, so for example:

 

group_id  |    contact_id

100                606

100                606

 

Are bad. But

 

group_id  |  contact_id

100              606

68                606

 

are fine. I need to find instances where they are duplicate entries. If I run this query:

SELECT DISTINCT group_id, contact_id
FROM table1
GROUP BY group_id, contact_id

then I believe it's showing me how many are distinct, and that leaves me to believe that there are about 400 cases of duplicates. I need to see all the ones that are duplicates. How do I write this query?

[ignace]

SELECT * FROM table GROUP BY group_id, contact_id HAVING count(*) > 1

[sw9]

thanks for that suggestion. there had to be a dupe of the same table to really get what was needed. i ended up using this query:

 

SELECT gc.* FROM group_contact gc

      INNER JOIN ( SELECT id, group_id, contact_id

                          FROM group_contact

                          GROUP BY group_id, contact_id HAVING count(*) > 1 ) dup_gc

                  ON ( gc.group_id = dup_gc.group_id

                          AND gc.contact_id = dup_gc.contact_id

                          AND gc.id <> dup_gc.id )