Thanks, so the use of field_id is important! I tried your first test which
can be written with the proper field_id *m.id.* This Works:
> def test1():
id_company = 1
m, cm = db.meeting, db.co_meet
q = cm.ref_company == id_company
q = q & (m.id == cm.ref_meeting)
grid = SQLFORM
Sorry. What I wrote is wrong.
Try this ways:
def test1():
id_company = 1
m, cm = db.meeting, db.co_meet
cm.ref_company.default = id_company
cm.ref_company.readable = cm.ref_company.writable = False
cm.ref_meeting.requires = IS_IN_DB(db, m.id ,'%(title)s')
q = cm.ref_co
m, cm = db.meeting, db.co_meet
q = m.id == cm.ref_meeting
q = q & (cm.ref_company == 1)
grid = SQLFORM.grid(q, field_id=m.id, fields=[m.id, m.title])
On Sun, May 31, 2015 at 4:48 AM, Ben Lawrence wrote:
> Both of your answers give a grid of co_meet.
>
> What would the query be i
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