ML wrote:
> Try this:
>
> select sum(derived.count_mid) as sum_count_mid
> from (
> select count(distinct mid) as count_mid from mytab
> union
> select count(distinct mid) from mytab_old
> union
> select count(distinct mid) from mytab_hist
> ) derived
>
>
> ML
>
> ---
> http://milambda.blogspot.com/

On Sun, 21 Jan 2007 23:54:52 -0800, Jami <jami.khan@yahoo.com> wrote:

>Dear All!
>
>i have following query
>
>select count(distinct mid) from mytab
>union
>select count(distinct mid) from mytab_old
>union
>select count(distinct mid) from mytab_hist
>
>this return me three out puts
>
>100045
>11245
>334562
>
>but i a single result sum of al these how will it b possible
>
>regards,