Hi\I think you can achieve this using a single query like this: SELECT * FROM
approvals WHERE adminname != 'admin2'and photo_uid in (select photo_uid from
approvals where adminname='admin2'), of course, maybe it's not the best
solution, just for your information.在2009-11-22，"Ashley M. Kirchner"
<kira...@gmail.com> 写道：
>I'm stuck again ... and perhaps it's just not designed right, but I
>inherited this setup, so if anyone has suggestions on how to make it better,
>I'm all ears. This is all one table ...
>
>describe approvals;
>+-----------+--------------------------+------+-----+---------+----------------+
>| Field | Type | Null | Key | Default |
>Extra |
>+-----------+--------------------------+------+-----+---------+----------------+
>| id | int(6) | NO | PRI | NULL |
>auto_increment |
>| photo_uid | int(7) unsigned zerofill | NO | | NULL
>| |
>| adminname | varchar(100) | NO | | NULL
>| |
>| status | int(1) | NO | | NULL
>| |
>+-----------+--------------------------+------+-----+---------+----------------+
>
>
>SELECT * FROM approvals;
>+----+-----------+-----------+--------+
>| id | photo_uid | adminname | status |
>+----+-----------+-----------+--------+
>| 6 | 0000028 | admin1 | 1 |
>| 13 | 0000016 | admin1 | 0 |
>| 49 | 0000016 | admin2 | 1 |
>| 16 | 0000018 | admin1 | 1 |
>| 50 | 0000018 | admin2 | 0 |
>+----+-----------+-----------+--------+
>
>
>The goal, here is to capture everything that does not have 'admin2' in the
>adminname column, however duplicated records in the photo_uid column should
>also be excluded.
>
>SELECT * FROM approvals WHERE adminname != 'admin2';
>+----+-----------+-----------+--------+
>| id | photo_uid | adminname | status |
>+----+-----------+-----------+--------+
>| 6 | 0000028 | admin1 | 1 |
>| 13 | 0000016 | admin1 | 0 |
>| 16 | 0000018 | admin1 | 1 |
>+----+-----------+-----------+--------+
>
>
>However, I also need to exclude photo_uid '0000016' and '0000018' because
>both of them are already tagged by 'admin2'. The only record that needs to
>be returned here is '0000028' ...
>
>I just don't know how to do that, or if it's even possible to do in a single
>query. I can get it done in two, but I'm hoping for a single query here ...
