On Sep 21, 2007, at 2:17 PM, Madison Kelly wrote:
... Or something like that. :)
Sorry for so many questions! I have another "how do I create this
query?" question, if it's okay.
I've got three tables; 'foo', 'bar' and 'baz'.
In 'foo' I've got 'foo_id' which is a PK. I've also got a bunch of
other info, but in essence this is the "parent" table that all others
reference in some way.
In 'bar' I've got 'bar_id' which is also a PK. I also have
'bar_foo_id' which is a FK pointing to 'foo' -> 'foo_id', to show what
'foo' row it (primarily) belongs to.
Lastly, I've got a table called 'baz' which has 'baz_id'. In it,
there
are just two columns;
- 'baz_foo_id' which is a FK pointing to 'foo' -> 'foo_id'.
- 'baz_bar_id' which is a FK pointing to 'bar' -> 'bar_id'.
This last table, 'baz' is used as a way for saying 'bar *also*
belongs
to a given 'foo' row,
So now my question;
I want to create a query that will allow me to say "show me all
'foo'
rows and tell me if a specific 'baz_id' belongs to it". Normally, I
would do this:
SELECT foo_id FROM foo;
(for each returned row)
{
# Where '$foo_id' is the current 'foo_id' and '$bar_id' is
# the specific/static 'bar_id' we are checking.
SELECT COUNT(*) FROM baz z, bar r WHERE baz_foo_id=$foo_id AND
baz_bar_id=$bar_id;
( if count > 0 ) { # TRUE } else { # FALSE }
}
This is pretty inefficient, obviously. How could I create a query
that
returned a TRUE/FALSE column that checks if there is a 'baz' record
for
a specified 'bar_id' in all 'foo_id's in one query?
I hope this isn't too muddy. I think part of my problem is I am
having
trouble even visualizing my question...
Thanks as always!
Madi
*Not tested* If this isn't quite it, then it should at least get you
close
SELECT foo.foo_id, CASE WHEN baz.foo_id IS NULL THEN FALSE ELSE TRUE
as has_baz
FROM foo
LEFT JOIN baz ON (foo.foo_id=baz.baz_foo_id AND baz.baz_bar_id=$bar_id)
Erik Jones
Software Developer | Emma®
erik@xxxxxxxxxx
800.595.4401 or 615.292.5888
615.292.0777 (fax)
Emma helps organizations everywhere communicate & market in style.
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