how to select all duplicates efficiently

I want to select all rows that have a value that al开发者_开发百科ready exists in the table. I did not find a better solution than

select * 
from provisioning_requests tt 
where code in (select code 
               from provisioning_requests tt2 
               where tt2.id <> tt.id)

This seems a bit naive. Does anybody have a better solution?

select * 
from provisioning_requests t1
 join (select code from provisioning_requests group by code having count(*)>1) t2
 ON t1.code = t2.code

OR

select * 
from provisioning_requests
 WHERE code in (select code from provisioning_requests group by code having count(*)>1)

An Auto join do the job

select tt.* 
from provisioning_requests tt 
    INNER JOIN provisioning_requests tt2 
        ON tt.code = tt2.code
        AND tt2.id <> tt.id

select t.*
from(
    select *, count(1) over(partition by code) as cnt
    from test
) as t
where t.cnt > 1

You can use operator exists, it produces better performance:

select * 
from provisioning_requests tt 
where exists
(
    select 1
    from provisioning_requests tt2
    where tt2.id <> tt.id and tt2.code = tt.code
)

How about using the distinct keyword?

SELECT col1, col2, col3, ..., DISTINCT(code) from provisioning_requests;

Perhaps using self join and also an index over Code column would make it perform better.

select pr1.* from provisioning_requests pr1
join provisioning_requests pr2 on pr1.code = pr2.code and pr1.id <> pr2.id

you can monitor the row Codes by

select code 
from provisioning_requests tt 
group by code
having count(code) > 1

how about :

SELECT *,COUNT(*) FROM provisioning_requests HAVING COUNT(*)>1