try something like this:
location_one = db.item_location.with_alias('location_one')
location_two = db.item_location.with_alias('location_two')
rows = db((db.item_location.item == db.item.id) & (db.item.id != itemid) & (
db.item_location.location_one == location_one.id) &
(db.item_location.location_two
== location_two.id) &
(location_one.geom.st_equals(point_x_y)) & (location_two.geom.st_equals
(point_n_m))).select(db.item.ALL)
Alex
Am Dienstag, 27. August 2013 19:43:08 UTC+2 schrieb Apple Mason:
>
> I have this many to many relationship example:
>
> db.define_table('location',
> Field('geom', 'geometry()'))
>
> db.define_table('item',
> Field('name'))
>
> db.define_table('item_location',
> Field('item', db.item),
> Field('location_one', db.location),
> Field('location_two', db.location))
>
>
>
> The goal is to find all items that are NOT the given item.id, but matches
> locations.
> An example query would be:
>
> Given: item.id==1 and two points POINT(x,y) and POINT(n,m),
> Result: "get all items that are not item.id==1, but has
> location_one==POINT(x,y) and location_two==POINT(n,m)"
>
> I am able to get it matching one of the points, but not the other with
> this:
>
>
> point_x_y = "POINT(1,2)"
> point_n_m = "POINT(3,4)"
> itemid = 1
>
> t = db( (db.item.id==db.item_location.item) &
> ( (db.location.id==db.item_location.location_one) & (db.
> location.geom.st_equals(point_x_y))))
>
> result = t( db.item.id != itemid ).select()
>
>
> This will successfully match all items that do not have an id=1, and has
> location_one as point_x_y.
>
> The problem is I do not know how to match location_two with point_n_m. I
> tried this, but it doesn't make sense (it also returns in nothing):
>
> t = db( (db.item.id==db.item_location.item) &
> ( (db.location.id==db.item_location.location_one) & (db.
> location.geom.st_equals(point_x_y)))
> ( (db.location.id==db.item_location.location_two) & (db.
> location.geom.st_equals(point_n_m))))
>
> Any help would be great!
>
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