Thank you both for both for your replies.
2009/7/16 Maximilian Tyrtania <maximilian.tyrtania@onlinehome.de>:
> am 16.07.2009 11:57 Uhr schrieb A B unter gentosaker@gmail.com:
>
>> Hi.
>> I need to do a
>>
>> select distinct userid from usertable,<more tables> where <conditions>;
>>
>> and then count the rows in the result
>>
>> How do I do that?
>
> Try:
>
> select count(distinct(userid)) from usertable,<more tables> where
> <conditions>;
>
> Best wishes from Berlin,
>
> Maximilian Tyrtania
>
>
>