select field1, field2, field3 from venues, volunteers_2009 where volunteers_2009.venue_id = venues.id
Galen
2008-10-13 22:14:00
A:
A:
General SQL LEFT Outer join syntax.
SELECT volunteers.id, volunteers.lname, volunteers.fname, volunteers.venue_id, venues.venue_name FROM Volunteers_2009 AS Volunteers LEFT OUTER JOIN Venues ON (Volunteers.venue_id = Venues.id)
Hapkido
2008-10-13 22:22:06
thank you. $query = "SELECT volunteers_2009.id, volunteers_2009.lname, volunteers_2009.fname, volunteers_2009.venue_id, venues.venue_name FROM volunteers_2009 AS volunteers_2009 LEFT OUTER JOIN venues ON (volunteers_2009.venue_id = venues.id) ORDER by $order $sort";
Brad
2008-10-13 23:48:13
A:
I cannot add any comment, but the syntax select field1, field2, field3 from venues, volunteers_2009 where volunteers_2009.venue_id = venues.id will only gave you volunteers with a venue; not those without a venue.
Hapkido
2008-10-13 22:24:31
I want to go through the table and display the venue_name, if there is none, then display "No Venue Assigned". The problem I am having is to display the appropriate venue_name for each volunteer.Right now I can just display venue names, but they do not display next to the appropriate volunteer.
Brad
2008-10-13 22:38:00