Subject | Re: [firebird-support] SELECT COUNT(sub_select.*) as COUNT, sub_select.user_name ... |
---|---|
Author | Helen Borrie |
Post date | 2004-01-15T10:21:41Z |
At 09:55 AM 15/01/2004 +0000, you wrote:
(select count(ss.user_name) from USER ss) as "COUNT",
u.user_name,
u.password
FROM USER u;
/hb
>Content of my "USER" table:SELECT
>
>user_name password name surname
>_________ ______ __________ ________
>
>g aa aa
>h bb bb
>i cc cc
>j dd dd
>k ee ee
>l ff ff
>
>Result of SQL must be :
>
>COUNT user_name password
>_____ ______ __________
>6 g aa
>6 h bb
>6 i cc
>6 j dd
>6 k ee
>6 l ff
(select count(ss.user_name) from USER ss) as "COUNT",
u.user_name,
u.password
FROM USER u;
/hb