MySQL select records for duplicates using multiple columns

If you want to count duplicates among multiple columns, use group by:

select ColumnA, ColumnB, ColumnC, count(*) as NumDuplicates
from table
group by ColumnA, ColumnB, ColumnC

If you only want the values that are duplicated, then the count is bigger than 1. You get this using the having clause:

select ColumnA, ColumnB, ColumnC, count(*) as NumDuplicates
from table
group by ColumnA, ColumnB, ColumnC
having NumDuplicates > 1

If you actually want all the duplicate rows returns, then join the last query back to the original data:

select t.*
from table t join
     (select ColumnA, ColumnB, ColumnC, count(*) as NumDuplicates
      from table
      group by ColumnA, ColumnB, ColumnC
      having NumDuplicates > 1
     ) tsum
     on t.ColumnA = tsum.ColumnA and t.ColumnB = tsum.ColumnB and t.ColumnC = tsum.ColumnC

This will work, assuming none of the column values are NULL. If so, then try:

     on (t.ColumnA = tsum.ColumnA or t.ColumnA is null and tsum.ColumnA is null) and
        (t.ColumnB = tsum.ColumnB or t.ColumnB is null and tsum.ColumnB is null) and
        (t.ColumnC = tsum.ColumnC or t.ColumnC is null and tsum.ColumnC is null)

EDIT:

If you have NULL values, you can also use the NULL-safe operator:

     on t.ColumnA <=> tsum.ColumnA and
        t.ColumnB <=> tsum.ColumnB and
        t.ColumnC <=> tsum.ColumnC 

why don't you try using union or creating temporary table. but personally, i do recommend using union than that of creating temporary table cause it would take you a longer time doing that. try doing this:

  select field1, field2 from(
   select '' as field2, field1, count(field1) as cnt FROM list GROUP BY field2 HAVING cnt > 1
    union
    select ''as field1, field2, cound(field2) as cnt from list group by field1 having cnt > 1
  )

hope this make sense.:)