Thread: Selecting rows as if they were columns?
Gang,
I've got what I think should be a pretty simple problem- I just can't
figure out how I'd do it. Heck, I can't even figure out how I'd search
for an answer to this problem.
I've got 2 tables: member and member_attr
Member is just a standard entry for a member. Very simple:
table member {
id integer,
name character varying(30)
}
Member_attr is just a table of name/value pairs, such as:
table memberextra {
memberid integer,
param1 character varying(512),
param2 character varying(512),
}
where param1 is the name, param2 is the value.
Right now, I can:
select member.id, member.name, member_attr.param1, member_attr.param2
from member, member_attr where member.id = member_attr.memberid;
...this gives me n rows per member where n = # of attributes for that
member, like this:
id | name | param1 | param2
=======================================
1 | dave | home | /home/dave
1 | dave | testing | 1
2 | john | home | /home/john
...is there some way to make this return like this:
id | name | home | testing
======================================
1 | dave | /home/dave | 1
2 | john | /home/john |
...where it sort of "expands" the rows out into columns? Am I just
going about this the wrong way? Thoughts on the subject would be
greatly appreciated.
thanks!
/kurt
In article <3F855F38.9070803@hotdogrecords.com>,
Kurt Overberg <kurt@hotdogrecords.com> writes:
> Gang,
> I've got what I think should be a pretty simple problem- I just can't
> figure out how I'd do it. Heck, I can't even figure out how I'd
> search for an answer to this problem.
> I've got 2 tables: member and member_attr
> Member is just a standard entry for a member. Very simple:
> table member {
> id integer,
> name character varying(30)
> }
> Member_attr is just a table of name/value pairs, such as:
> table memberextra {
> memberid integer,
> param1 character varying(512),
> param2 character varying(512),
> }
> where param1 is the name, param2 is the value.
> Right now, I can:
> select member.id, member.name, member_attr.param1, member_attr.param2
> from member, member_attr where member.id = member_attr.memberid;
> ...this gives me n rows per member where n = # of attributes for that
> member, like this:
> id | name | param1 | param2
> =======================================
> 1 | dave | home | /home/dave
> 1 | dave | testing | 1
> 2 | john | home | /home/john
> ...is there some way to make this return like this:
> id | name | home | testing
> ======================================
> 1 | dave | /home/dave | 1
> 2 | john | /home/john |
> ...where it sort of "expands" the rows out into columns? Am I just
> going about this the wrong way? Thoughts on the subject would be
> greatly appreciated.
This should work fine as long as you know all possible values of
param1 in advance:
SELECT m.id, m.name, m1.param2, m2.param2
FROM member m
LEFT JOIN memberextra m1 ON m1.memberid = m.id AND m1.param1 = 'home'
LEFT JOIN memberextra m2 ON m2.memberid = m.id AND m2.param1 = 'testing'
Kurt, > I've got what I think should be a pretty simple problem- I just can't > figure out how I'd do it. Heck, I can't even figure out how I'd search > for an answer to this problem. This is an old common SQL problem ... the "crosstab" table. There are 3 standard ways to solve it, depending on the exact circumstances of your case. It's probably best if you buy Joe Celko's "SQL for Smarties, 2nd Ed.", which has a 50-page chapter on the topic and explores the methods in detail. -- Josh Berkus Aglio Database Solutions San Francisco
On Thursday 09 October 2003 18:14, Josh Berkus wrote: > Kurt, > > > I've got what I think should be a pretty simple problem- I just can't > > figure out how I'd do it. Heck, I can't even figure out how I'd search > > for an answer to this problem. > > This is an old common SQL problem ... the "crosstab" table. There are 3 > standard ways to solve it, depending on the exact circumstances of your > case. It's probably best if you buy Joe Celko's "SQL for Smarties, 2nd > Ed.", which has a 50-page chapter on the topic and explores the methods in > detail. Or see the contrib/tablefunc functions which can do this sort of thing for you. -- Richard Huxton Archonet Ltd