Am I missing something? Will this not do the trick:
SELECT DISTINCT member_id FROM table WHERE specialty_id IN(6,33); Graham > -----Original Message----- > From: Johan Höök [mailto:[EMAIL PROTECTED] > Sent: 30 November 2004 19:56 > To: Mike Zornek > Cc: [EMAIL PROTECTED] > Subject: Re: Select member when it meets two requirements > > > Hi Mike, > you should be able to do: > > SELECT DISTINCT t.member_id > FROM table t > INNER JOIN table t2 ON t2.member_id = t.member_id AND t2.speciality_id = 2 > WHERE t.speciality_id = 6 > > /Johan > > Mike Zornek wrote: > > I'm very much a noob when it comes to MySQL .. Historically > I've only used > > it for storage. I need help. > > > > I have a table: > > > > > +--------------------+-----------------------+------+-----+------- > --+------- > > ---------+ > > | Field | Type | Null | Key | > Default | Extra > > | > > > +--------------------+-----------------------+------+-----+------- > --+------- > > ---------+ > > | memberspecialty_id | int(10) unsigned | | PRI | NULL | > > auto_increment | > > | member_id | smallint(10) unsigned | | MUL | 0 | > > | > > | specialty_id | tinyint(3) unsigned | | MUL | 0 | > > | > > > +--------------------+-----------------------+------+-----+------- > --+------- > > ---------+ > > > > How would I select all distinct member_id that have a > specialty_id of 6 and > > 33? > > > > Thanks! > > > > ~ Mike > > ----- > > Mike Zornek > > Web Designer, Media Developer, Programmer and Geek > > Personal site: <http://MikeZornek.com> > > > > > > > > -- > Johan Höök, Pythagoras Engineering Group > - MailTo:[EMAIL PROTECTED] > - http://www.pythagoras.se > Pythagoras AB, Stormbyv. 2-4, SE-163 55 SPÅNGA, Sweden > Phone: +46 8 760 00 10 Fax: +46 8 761 22 77 > > > -- > MySQL General Mailing List > For list archives: http://lists.mysql.com/mysql > To unsubscribe: > http://lists.mysql.com/[EMAIL PROTECTED] > > -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe: http://lists.mysql.com/[EMAIL PROTECTED]
