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 =
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 =
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 benlawr...@gmail.com wrote:
Both of your answers give a grid of co_meet.
What
Both of your answers give a grid of co_meet.
What would the query be in SQLFORM.grid such that it would be a grid of
db.meeting for one company (not db.co_meet)?
On Thursday, May 7, 2015 at 8:26:47 PM UTC-7, 黄祥 wrote:
had you tried it?
another work around is u can use smartgrid constraints
had you tried it?
another work around is u can use smartgrid constraints
e.g.
def test():
table = db.co_meet
query = db.co_meet.ref_company == 1 # whatever value that refer to
table company
grid = SQLFORM.smartgrid(table, constraints = dict(co_meet=query) )
return locals()
best
Thanks! But wouldn't that just give me a grid of 'co_meet' and not
'meeting'?
On Monday, May 4, 2015 at 9:20:01 AM UTC-7, 黄祥 wrote:
perhaps, you can do
e.g.
def test():
query = db.co_meet.ref_company == 1 # whatever value that refer to
table company
grid=SQLFORM.grid(query)
perhaps, you can do
e.g.
def test():
query = db.co_meet.ref_company == 1 # whatever value that refer to
table company
grid=SQLFORM.grid(query)
return locals()
best regards,
stifan
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