[web2py] Re: smartgrid: linked_tables=dict(parent=['child'], child=[''])
How did you insert icons for edit in your form
I am using SQLFORM.smartgrid and I want my grid to be more bootstrap like.
Also I am having problems with displaying my child table fields (They are
not displayed :))
Any suggestions?
Here is my code
workers = db(db.worker.w_organisation == 10).select(db.worker.id,
db.worker.w_organisation, db.worker.w_first_name,
db.worker.w_last_name,db.worker.w_nick_name,db.worker.w_email,db.worker.w_status,db.worker.w_note).as_list()
#Define the query object. Here we are pulling all contacts having date of
birth less than 18 Nov 1990
query = ((db.worker.w_organisation == 10) )#
(db.worker.w_status==db.status.s_id_s))
#Define the fields to show on grid. Note: (you need to specify id field in
fields section in 1.99.2
fields = (db.worker.id,db.worker.w_first_name,
db.worker.w_last_name,db.worker.w_nick_name,db.worker.w_email,db.status.s_code,db.worker.w_note)
#Define headers as tuples/dictionaries
headers = {'worker.id' : 'Id',
'worker.w_first_name' : 'Ime',
'worker.w_last_name' : 'Priimek',
'worker.w_nick_name' : 'Vzdevek',
'worker.w_email' : 'E-posta',
'status.s_code': 'Status',
'worker.w_note' : 'Komentar' }
#Let's specify a default sort order on date_of_birth column in grid
default_sort_order=[db.worker.w_last_name]
form = SQLFORM.smartgrid(db.worker,fields = fields,headers=
headers,linked_tables=['status'])
And db.py
db.define_table('worker',
Field('id', type ='id'),
Field('w_organisation', type ='integer'),
Field('w_user', type ='integer'),
Field('w_status', 'reference status'),
Field('w_first_name',type='text'),
Field('w_last_name',type='text'),
Field('w_nick_name',type='text'),
Field('w_email',type='text'),
Field('w_note',type='text'),
migrate=settings.migrate
)
db.define_table('status',
Field('s_id_s', type ='id'),
Field('s_code', type ='text'),
Field('s_description', type ='text'),
migrate=settings.migrate
)
thank you
best regards
On Friday, September 21, 2012 5:58:05 PM UTC+2, Adi wrote:
I'm wondering what to do in this situation? I have self-referencing fields
in the child table, and due to that smartgrid display links, which
basically can't do anything.
I'm trying to eliminate them, but not sure what would be a proper way?
Thanks.
Simplified code sample:
db.define_table('campaign',
Field('tbl_uuid', length=64, default=lambda:str(uuid.
uuid4())),
Field('name','string', label=T('Campaing Name')),
format='%(name)s',
)
db.define_table('message',
Field('tbl_uuid', length=64, default=lambda:str(uuid.
uuid4())),
Field('name','string', label=T('Name')),
Field('campaign_id', 'reference campaign', label=T(
'Campaign')),
Field('action_yes_id', 'reference message)',
label=T('Action
Yes'),),
Field('action_no_id', 'reference message)',
label=T('Action
No')),
migrate=True)
grid=SQLFORM.smartgrid(Campaign, details=False, links_in_grid=True,
linked_tables=['message'],
#linked_tables=dict(campaign=['message'],
message=['']),
links=dict(campaign=[lambda row:(_get_messages(row))
]),
)
Enter code here...
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[web2py] Re: smartgrid: linked_tables=dict(parent=['child'], child=[''])
I just added the syntax you requested:
linked_tables=dict(parent=['child'], child=[])
Please help me test it.
On Saturday, 22 September 2012 18:55:57 UTC-5, Adi wrote:
If I may rephrase and clarify a question:
Is it possible to add control, through dictionary in (smartgrid)
linked_tables when will links display?
e.g: linked_tables=dict(parent=['child'], child=[''])
wishful outcome: display children links for parent table, but don't
display any related links for child table.
Thanks,
Adnan
On Friday, September 21, 2012 11:58:05 AM UTC-4, Adi wrote:
I'm wondering what to do in this situation? I have self-referencing
fields in the child table, and due to that smartgrid display links, which
basically can't do anything.
I'm trying to eliminate them, but not sure what would be a proper way?
Thanks.
Simplified code sample:
db.define_table('campaign',
Field('tbl_uuid', length=64, default=lambda:str(uuid.
uuid4())),
Field('name','string', label=T('Campaing Name')),
format='%(name)s',
)
db.define_table('message',
Field('tbl_uuid', length=64, default=lambda:str(uuid.
uuid4())),
Field('name','string', label=T('Name')),
Field('campaign_id', 'reference campaign', label=T(
'Campaign')),
Field('action_yes_id', 'reference message)',
label=T('Action
Yes'),),
Field('action_no_id', 'reference message)',
label=T('Action
No')),
migrate=True)
grid=SQLFORM.smartgrid(Campaign, details=False, links_in_grid=True,
linked_tables=['message'],
#linked_tables=dict(campaign=['message'],
message=['']),
links=dict(campaign=[lambda row:(_get_messages(row))
]),
)
Enter code here...
--
Re: [web2py] Re: smartgrid: linked_tables=dict(parent=['child'], child=[''])
Thanks Massimo!
I'm trying to test it, but got this error bellow:
File /Users/adnan/web2py-trunk/gluon/sqlhtml.py, line 2280, in smartgrid
linked_tables = linked_tables.get(tablename,[])
UnboundLocalError: local variable 'tablename' referenced before assignment
On Sun, Sep 23, 2012 at 5:31 PM, Massimo Di Pierro
massimo.dipie...@gmail.com wrote:
I just added the syntax you requested:
linked_tables=dict(parent=['**child'], child=[])
Please help me test it.
On Saturday, 22 September 2012 18:55:57 UTC-5, Adi wrote:
If I may rephrase and clarify a question:
Is it possible to add control, through dictionary in (smartgrid)
linked_tables when will links display?
e.g: linked_tables=dict(parent=['**child'], child=[''])
wishful outcome: display children links for parent table, but don't
display any related links for child table.
Thanks,
Adnan
On Friday, September 21, 2012 11:58:05 AM UTC-4, Adi wrote:
I'm wondering what to do in this situation? I have self-referencing
fields in the child table, and due to that smartgrid display links, which
basically can't do anything.
I'm trying to eliminate them, but not sure what would be a proper way?
Thanks.
Simplified code sample:
db.define_table('campaign',
Field('tbl_uuid', length=64, default=lambda:str(uuid
.uuid4(**))),
Field('name','string', label=T('Campaing Name')),
format='%(name)s',
)
db.define_table('message',
Field('tbl_uuid', length=64, default=lambda:str(uuid
.uuid4(**))),
Field('name','string', label=T('Name')),
Field('campaign_id', 'reference campaign', label=T(
'Campaign')),
Field('action_yes_id', 'reference message)', label=T
('Action Yes'),),
Field('action_no_id', 'reference message)',
label=T('Action
No')),
migrate=True)
grid=SQLFORM.smartgrid(Campaig**n, details=False, links_in_grid=True,
linked_tables=['message'],
#linked_tables=dict(campaign=[**'message'],
message=['']),
links=dict(campaign=[lambda row:(_get_messages(row))
]),
)
Enter code here...
--
--
Thanks,
Adnan
video: http://vimeo.com/24653283
--
Re: [web2py] Re: smartgrid: linked_tables=dict(parent=['child'], child=[''])
Can you please try again?
On Sunday, 23 September 2012 18:03:48 UTC-5, Adi wrote:
Thanks Massimo!
I'm trying to test it, but got this error bellow:
File /Users/adnan/web2py-trunk/gluon/sqlhtml.py, line 2280, in smartgrid
linked_tables = linked_tables.get(tablename,[])
UnboundLocalError: local variable 'tablename' referenced before assignment
On Sun, Sep 23, 2012 at 5:31 PM, Massimo Di Pierro
massimo@gmail.comjavascript:
wrote:
I just added the syntax you requested:
linked_tables=dict(parent=['**child'], child=[])
Please help me test it.
On Saturday, 22 September 2012 18:55:57 UTC-5, Adi wrote:
If I may rephrase and clarify a question:
Is it possible to add control, through dictionary in (smartgrid)
linked_tables when will links display?
e.g: linked_tables=dict(parent=['**child'], child=[''])
wishful outcome: display children links for parent table, but don't
display any related links for child table.
Thanks,
Adnan
On Friday, September 21, 2012 11:58:05 AM UTC-4, Adi wrote:
I'm wondering what to do in this situation? I have self-referencing
fields in the child table, and due to that smartgrid display links, which
basically can't do anything.
I'm trying to eliminate them, but not sure what would be a proper way?
Thanks.
Simplified code sample:
db.define_table('campaign',
Field('tbl_uuid', length=64, default=lambda:str(
uuid.uuid4(**))),
Field('name','string', label=T('Campaing Name')),
format='%(name)s',
)
db.define_table('message',
Field('tbl_uuid', length=64, default=lambda:str(
uuid.uuid4(**))),
Field('name','string', label=T('Name')),
Field('campaign_id', 'reference campaign', label=T(
'Campaign')),
Field('action_yes_id', 'reference message)', label=
T('Action Yes'),),
Field('action_no_id', 'reference message)', label=T
('Action No')),
migrate=True)
grid=SQLFORM.smartgrid(Campaig**n, details=False, links_in_grid=True,
linked_tables=['message'],
#linked_tables=dict(campaign=[**'message'],
message=['']),
links=dict(campaign=[lambda row:(_get_messages(row
))
]),
)
Enter code here...
--
--
Thanks,
Adnan
video: http://vimeo.com/24653283
--
Re: [web2py] Re: smartgrid: linked_tables=dict(parent=['child'], child=[''])
Works perfect. Thank you Massimo.
On Sun, Sep 23, 2012 at 8:16 PM, Massimo Di Pierro
massimo.dipie...@gmail.com wrote:
Can you please try again?
On Sunday, 23 September 2012 18:03:48 UTC-5, Adi wrote:
Thanks Massimo!
I'm trying to test it, but got this error bellow:
File /Users/adnan/web2py-trunk/**gluon/sqlhtml.py, line 2280, in
smartgrid
linked_tables = linked_tables.get(tablename,[]**)
UnboundLocalError: local variable 'tablename' referenced before assignment
On Sun, Sep 23, 2012 at 5:31 PM, Massimo Di Pierro massimo@gmail.com
wrote:
I just added the syntax you requested:
linked_tables=dict(parent=['**ch**ild'], child=[])
Please help me test it.
On Saturday, 22 September 2012 18:55:57 UTC-5, Adi wrote:
If I may rephrase and clarify a question:
Is it possible to add control, through dictionary in (smartgrid)
linked_tables when will links display?
e.g: linked_tables=dict(parent=['**ch**ild'], child=[''])
wishful outcome: display children links for parent table, but don't
display any related links for child table.
Thanks,
Adnan
On Friday, September 21, 2012 11:58:05 AM UTC-4, Adi wrote:
I'm wondering what to do in this situation? I have self-referencing
fields in the child table, and due to that smartgrid display links, which
basically can't do anything.
I'm trying to eliminate them, but not sure what would be a proper way?
Thanks.
Simplified code sample:
db.define_table('campaign',
Field('tbl_uuid', length=64, default=lambda:str(
uuid.uuid4())),
Field('name','string', label=T('Campaing Name')),
format='%(name)s',
)
db.define_table('message',
Field('tbl_uuid', length=64, default=lambda:str(
uuid.uuid4())),
Field('name','string', label=T('Name')),
Field('campaign_id', 'reference campaign', label=T
('Campaign')),
Field('action_yes_id', 'reference message)', label
=T('Action Yes'),),
Field('action_no_id', 'reference message)', label=
T('Action No')),
migrate=True)
grid=SQLFORM.smartgrid(Campaign, details=False, links_in_grid=True
,
linked_tables=['message'],
#linked_tables=dict(campaign=['message'],
message=['']),
links=dict(campaign=[lambda row:(_get_messages(row
))
]),
)
Enter code here...
--
--
[web2py] Re: smartgrid: linked_tables=dict(parent=['child'], child=[''])
If I may rephrase and clarify a question:
Is it possible to add control, through dictionary in (smartgrid)
linked_tables when will links display?
e.g: linked_tables=dict(parent=['child'], child=[''])
wishful outcome: display children links for parent table, but don't display
any related links for child table.
Thanks,
Adnan
On Friday, September 21, 2012 11:58:05 AM UTC-4, Adi wrote:
I'm wondering what to do in this situation? I have self-referencing fields
in the child table, and due to that smartgrid display links, which
basically can't do anything.
I'm trying to eliminate them, but not sure what would be a proper way?
Thanks.
Simplified code sample:
db.define_table('campaign',
Field('tbl_uuid', length=64, default=lambda:str(uuid.
uuid4())),
Field('name','string', label=T('Campaing Name')),
format='%(name)s',
)
db.define_table('message',
Field('tbl_uuid', length=64, default=lambda:str(uuid.
uuid4())),
Field('name','string', label=T('Name')),
Field('campaign_id', 'reference campaign', label=T(
'Campaign')),
Field('action_yes_id', 'reference message)',
label=T('Action
Yes'),),
Field('action_no_id', 'reference message)',
label=T('Action
No')),
migrate=True)
grid=SQLFORM.smartgrid(Campaign, details=False, links_in_grid=True,
linked_tables=['message'],
#linked_tables=dict(campaign=['message'],
message=['']),
links=dict(campaign=[lambda row:(_get_messages(row))
]),
)
Enter code here...
--
