Re: [web2py] create a link SQLFORM to another table with id.belongs
Hi Fabiano,
Thank you for your reply. What do you mean by *edit the registry of URL
arguments are changed. *
Here is my code:
def query_table2():
table2_records = []
table1_inv_record = request.args(0)
row = db(db.table1.id==table1_inv_record).select()
for line in row:
joined_records = line.inv_id
query=db.table1.id.belongs(joined_records)
grid=SQLFORM.grid(query, user_signature=False)
return dict(grid=grid)
view for query_table2.html:
{{=grid}}
On Friday, June 6, 2014 3:43:12 PM UTC-4, Fabiano Almeida wrote:
type 'exceptions.ValueError' invalid literal for int() with base 10:
'edit'
In some part of your code has converting string to numeric type. Probably
you should be picking up the URL argument and doing the conversion, but
when you edit the registry of URL arguments are changed.
In my example, I use session and try...except to resolve this.
2014-06-06 15:29 GMT-03:00 Fabiano Almeida fab...@techno7.com.br
javascript::
Show your code
2014-06-06 15:18 GMT-03:00 LoveWeb2py atayl...@gmail.com javascript::
Now I get this error when I try to click edit on the query database:
type 'exceptions.ValueError' invalid literal for int() with base 10:
'edit'
I've tried changing signature to false, but I don't think thats the
problem.
On Friday, June 6, 2014 1:17:39 PM UTC-4, LoveWeb2py wrote:
yes
On Fri, Jun 6, 2014 at 1:09 PM, Fabiano Almeida fab...@techno7.com.br
javascript: wrote:
Are you logged in your app?
On Friday, June 6, 2014 1:03:35 PM UTC-4, Fabiano Almeida wrote:
Try:
grid = SQLFORM.grid(db.table, user_signature=False)
Em sexta-feira, 6 de junho de 2014 13h10min34s UTC-3, LoveWeb2py
escreveu:
Hi Fabiano,
I actually was already using grid= SQLFORM.grid(db.new_table). That
is when I get the error.
If I do grid='' to test it and just return the table without
SQLFORM I can see it in the view, but when I apply SQLFORM I get the
error
'Rows' object has no attribute '_db'.
I'm guessing this is because of the way SQLFORM handles the grid?
Could there be something in my model messing it up?
On Friday, June 6, 2014 12:02:05 PM UTC-4, Fabiano Almeida wrote:
Hi,
You send var grid (see: return dict(grid=grid)). The first var
grid is a send var to your view, de second var grid is a local
var of your function.
Then,
in controller use:
grid = SQLFORM.grid(db.new_table)
return dict(grid=grid)
in the view use:
{{=grid}}
Fabiano.
2014-06-06 12:45 GMT-03:00 LoveWeb2py atayl...@gmail.com:
SQLFORM isn't working for me with
new_table = db(db.table1.id.belongs(record_ids)).select()
if I do {{=new_table}} in my view I can see the records which
belong to record_ids, but if I do:
SQLFORM.grid(new_table)
return dict(grid=grid)
I get an error 'Rows' object has no attribute '_db'
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Re: [web2py] create a link SQLFORM to another table with id.belongs
Hi,
Sorry for my poor english, by lazy, sometimes I use google translate
(portuguese - english), and don't check the translation.
table1_inv_record = request.args(0)
In this line you get de first args, but on click to edit record in grid,
the function it's called again and url change args, then first args in url
is 'edit', and this is literal, not numerical.
row = db(db.table1.id==table1_inv_record).select()
In edit mode (and changed url), get error because var table1_inv_record is
not numerical at this time. I think this is a line in your traceback error.
So I use try...except block to convert first arg in integer, if ok, use
session to store de right id sent in the original url.
alm = None # this is for prevent error case haven't args(0)
if request.args: alm = request.args(0)
try:
if alm:
alm = int(alm)
*session.almoxarifado*=alm# this line executes only if the
previous line no error
except ValueError:
pass
Objeto.almoxarifado_id.default = *session.almoxarifado*
2014-06-09 13:47 GMT-03:00 LoveWeb2py atayloru...@gmail.com:
Hi Fabiano,
Thank you for your reply. What do you mean by *edit the registry of URL
arguments are changed. *
Here is my code:
def query_table2():
table2_records = []
table1_inv_record = request.args(0)
row = db(db.table1.id==table1_inv_record).select()
for line in row:
joined_records = line.inv_id
query=db.table1.id.belongs(joined_records)
grid=SQLFORM.grid(query, user_signature=False)
return dict(grid=grid)
view for query_table2.html:
{{=grid}}
On Friday, June 6, 2014 3:43:12 PM UTC-4, Fabiano Almeida wrote:
type 'exceptions.ValueError' invalid literal for int() with base 10:
'edit'
In some part of your code has converting string to numeric type. Probably
you should be picking up the URL argument and doing the conversion, but
when you edit the registry of URL arguments are changed.
In my example, I use session and try...except to resolve this.
2014-06-06 15:29 GMT-03:00 Fabiano Almeida fab...@techno7.com.br:
Show your code
2014-06-06 15:18 GMT-03:00 LoveWeb2py atayl...@gmail.com:
Now I get this error when I try to click edit on the query database:
type 'exceptions.ValueError' invalid literal for int() with base 10:
'edit'
I've tried changing signature to false, but I don't think thats the
problem.
On Friday, June 6, 2014 1:17:39 PM UTC-4, LoveWeb2py wrote:
yes
On Fri, Jun 6, 2014 at 1:09 PM, Fabiano Almeida fab...@techno7.com.br
wrote:
Are you logged in your app?
On Friday, June 6, 2014 1:03:35 PM UTC-4, Fabiano Almeida wrote:
Try:
grid = SQLFORM.grid(db.table, user_signature=False)
Em sexta-feira, 6 de junho de 2014 13h10min34s UTC-3, LoveWeb2py
escreveu:
Hi Fabiano,
I actually was already using grid= SQLFORM.grid(db.new_table).
That is when I get the error.
If I do grid='' to test it and just return the table without
SQLFORM I can see it in the view, but when I apply SQLFORM I get the
error
'Rows' object has no attribute '_db'.
I'm guessing this is because of the way SQLFORM handles the grid?
Could there be something in my model messing it up?
On Friday, June 6, 2014 12:02:05 PM UTC-4, Fabiano Almeida wrote:
Hi,
You send var grid (see: return dict(grid=grid)). The first var
grid is a send var to your view, de second var grid is a local
var of your function.
Then,
in controller use:
grid = SQLFORM.grid(db.new_table)
return dict(grid=grid)
in the view use:
{{=grid}}
Fabiano.
2014-06-06 12:45 GMT-03:00 LoveWeb2py atayl...@gmail.com:
SQLFORM isn't working for me with
new_table = db(db.table1.id.belongs(record_ids)).select()
if I do {{=new_table}} in my view I can see the records which
belong to record_ids, but if I do:
SQLFORM.grid(new_table)
return dict(grid=grid)
I get an error 'Rows' object has no attribute '_db'
--
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- http://web2py.com/book (Documentation)
- http://github.com/web2py/web2py (Source code)
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Re: [web2py] create a link SQLFORM to another table with id.belongs
No problem at all and I appreciate you trying to respond to English.
Is Object your table name and Almoxarifado_id is your id field in Objeto?
Objeto.almoxarifado_id
On Monday, June 9, 2014 1:30:49 PM UTC-4, Fabiano Almeida wrote:
Hi,
Sorry for my poor english, by lazy, sometimes I use google translate
(portuguese - english), and don't check the translation.
table1_inv_record = request.args(0)
In this line you get de first args, but on click to edit record in grid,
the function it's called again and url change args, then first args in url
is 'edit', and this is literal, not numerical.
row = db(db.table1.id==table1_inv_record).select()
In edit mode (and changed url), get error because var table1_inv_record
is not numerical at this time. I think this is a line in your traceback
error.
So I use try...except block to convert first arg in integer, if ok, use
session to store de right id sent in the original url.
alm = None # this is for prevent error case haven't args(0)
if request.args: alm = request.args(0)
try:
if alm:
alm = int(alm)
*session.almoxarifado*=alm# this line executes only if the
previous line no error
except ValueError:
pass
Objeto.almoxarifado_id.default = *session.almoxarifado*
2014-06-09 13:47 GMT-03:00 LoveWeb2py atayl...@gmail.com javascript::
Hi Fabiano,
Thank you for your reply. What do you mean by *edit the registry of URL
arguments are changed. *
Here is my code:
def query_table2():
table2_records = []
table1_inv_record = request.args(0)
row = db(db.table1.id==table1_inv_record).select()
for line in row:
joined_records = line.inv_id
query=db.table1.id.belongs(joined_records)
grid=SQLFORM.grid(query, user_signature=False)
return dict(grid=grid)
view for query_table2.html:
{{=grid}}
On Friday, June 6, 2014 3:43:12 PM UTC-4, Fabiano Almeida wrote:
type 'exceptions.ValueError' invalid literal for int() with base 10:
'edit'
In some part of your code has converting string to numeric type.
Probably you should be picking up the URL argument and doing the
conversion, but when you edit the registry of URL arguments are changed.
In my example, I use session and try...except to resolve this.
2014-06-06 15:29 GMT-03:00 Fabiano Almeida fab...@techno7.com.br:
Show your code
2014-06-06 15:18 GMT-03:00 LoveWeb2py atayl...@gmail.com:
Now I get this error when I try to click edit on the query database:
type 'exceptions.ValueError' invalid literal for int() with base 10:
'edit'
I've tried changing signature to false, but I don't think thats the
problem.
On Friday, June 6, 2014 1:17:39 PM UTC-4, LoveWeb2py wrote:
yes
On Fri, Jun 6, 2014 at 1:09 PM, Fabiano Almeida
fab...@techno7.com.br wrote:
Are you logged in your app?
On Friday, June 6, 2014 1:03:35 PM UTC-4, Fabiano Almeida wrote:
Try:
grid = SQLFORM.grid(db.table, user_signature=False)
Em sexta-feira, 6 de junho de 2014 13h10min34s UTC-3, LoveWeb2py
escreveu:
Hi Fabiano,
I actually was already using grid= SQLFORM.grid(db.new_table).
That is when I get the error.
If I do grid='' to test it and just return the table without
SQLFORM I can see it in the view, but when I apply SQLFORM I get the
error
'Rows' object has no attribute '_db'.
I'm guessing this is because of the way SQLFORM handles the grid?
Could there be something in my model messing it up?
On Friday, June 6, 2014 12:02:05 PM UTC-4, Fabiano Almeida wrote:
Hi,
You send var grid (see: return dict(grid=grid)). The first var
grid is a send var to your view, de second var grid is a local
var of your function.
Then,
in controller use:
grid = SQLFORM.grid(db.new_table)
return dict(grid=grid)
in the view use:
{{=grid}}
Fabiano.
2014-06-06 12:45 GMT-03:00 LoveWeb2py atayl...@gmail.com:
SQLFORM isn't working for me with
new_table = db(db.table1.id.belongs(record_ids)).select()
if I do {{=new_table}} in my view I can see the records which
belong to record_ids, but if I do:
SQLFORM.grid(new_table)
return dict(grid=grid)
I get an error 'Rows' object has no attribute '_db'
--
Resources:
- http://web2py.com
- http://web2py.com/book (Documentation)
- http://github.com/web2py/web2py (Source code)
- https://code.google.com/p/web2py/issues/list (Report Issues)
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Re: [web2py] create a link SQLFORM to another table with id.belongs
Objeto it's the table
almoxarifado_id it's a FK (reference field)
2014-06-09 14:48 GMT-03:00 LoveWeb2py atayloru...@gmail.com:
No problem at all and I appreciate you trying to respond to English.
Is Object your table name and Almoxarifado_id is your id field in Objeto?
Objeto.almoxarifado_id
On Monday, June 9, 2014 1:30:49 PM UTC-4, Fabiano Almeida wrote:
Hi,
Sorry for my poor english, by lazy, sometimes I use google translate
(portuguese - english), and don't check the translation.
table1_inv_record = request.args(0)
In this line you get de first args, but on click to edit record in grid,
the function it's called again and url change args, then first args in url
is 'edit', and this is literal, not numerical.
row = db(db.table1.id==table1_inv_record).select()
In edit mode (and changed url), get error because var table1_inv_record
is not numerical at this time. I think this is a line in your traceback
error.
So I use try...except block to convert first arg in integer, if ok, use
session to store de right id sent in the original url.
alm = None # this is for prevent error case haven't args(0)
if request.args: alm = request.args(0)
try:
if alm:
alm = int(alm)
*session.almoxarifado*=alm# this line executes only if
the previous line no error
except ValueError:
pass
Objeto.almoxarifado_id.default = *session.almoxarifado*
2014-06-09 13:47 GMT-03:00 LoveWeb2py atayl...@gmail.com:
Hi Fabiano,
Thank you for your reply. What do you mean by *edit the registry of
URL arguments are changed. *
Here is my code:
def query_table2():
table2_records = []
table1_inv_record = request.args(0)
row = db(db.table1.id==table1_inv_record).select()
for line in row:
joined_records = line.inv_id
query=db.table1.id.belongs(joined_records)
grid=SQLFORM.grid(query, user_signature=False)
return dict(grid=grid)
view for query_table2.html:
{{=grid}}
On Friday, June 6, 2014 3:43:12 PM UTC-4, Fabiano Almeida wrote:
type 'exceptions.ValueError' invalid literal for int() with base 10:
'edit'
In some part of your code has converting string to numeric type.
Probably you should be picking up the URL argument and doing the
conversion, but when you edit the registry of URL arguments are changed.
In my example, I use session and try...except to resolve this.
2014-06-06 15:29 GMT-03:00 Fabiano Almeida fab...@techno7.com.br:
Show your code
2014-06-06 15:18 GMT-03:00 LoveWeb2py atayl...@gmail.com:
Now I get this error when I try to click edit on the query database:
type 'exceptions.ValueError' invalid literal for int() with base
10: 'edit'
I've tried changing signature to false, but I don't think thats the
problem.
On Friday, June 6, 2014 1:17:39 PM UTC-4, LoveWeb2py wrote:
yes
On Fri, Jun 6, 2014 at 1:09 PM, Fabiano Almeida
fab...@techno7.com.br wrote:
Are you logged in your app?
On Friday, June 6, 2014 1:03:35 PM UTC-4, Fabiano Almeida wrote:
Try:
grid = SQLFORM.grid(db.table, user_signature=False)
Em sexta-feira, 6 de junho de 2014 13h10min34s UTC-3, LoveWeb2py
escreveu:
Hi Fabiano,
I actually was already using grid= SQLFORM.grid(db.new_table).
That is when I get the error.
If I do grid='' to test it and just return the table without
SQLFORM I can see it in the view, but when I apply SQLFORM I get
the error
'Rows' object has no attribute '_db'.
I'm guessing this is because of the way SQLFORM handles the
grid? Could there be something in my model messing it up?
On Friday, June 6, 2014 12:02:05 PM UTC-4, Fabiano Almeida wrote:
Hi,
You send var grid (see: return dict(grid=grid)). The first var
grid is a send var to your view, de second var grid is a local
var of your function.
Then,
in controller use:
grid = SQLFORM.grid(db.new_table)
return dict(grid=grid)
in the view use:
{{=grid}}
Fabiano.
2014-06-06 12:45 GMT-03:00 LoveWeb2py atayl...@gmail.com:
SQLFORM isn't working for me with
new_table = db(db.table1.id.belongs(record_ids)).select()
if I do {{=new_table}} in my view I can see the records which
belong to record_ids, but if I do:
SQLFORM.grid(new_table)
return dict(grid=grid)
I get an error 'Rows' object has no attribute '_db'
--
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- http://web2py.com/book (Documentation)
- http://github.com/web2py/web2py (Source code)
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Re: [web2py] create a link SQLFORM to another table with id.belongs
Hi Fabiano,
I'm still trying to get this
For your Objeto.almoxarifado_id.default = session.almoxarifado
Shouldn't there be a db.Objeto.almoxarifado_id ? If I try to just put
table2.id I get an error saying table2 isn't defined.
Here is what I have so far
def query_table2():
table1_inv_record = None
table2_records = []
table1_inv_record = request.args(0)
row = db(db.table1.id==table1_inv_record).select()
for line in row:
joined_records = line.inv_id
if table1_inv_record:
inv1_record = int(table1_inv_record)
session.table1_id = inv1_record
query=db.table1.id.belongs(joined_records)
db.table1.id.default = session.table1.id
grid=SQLFORM.grid(query, user_signature=False)
return dict(grid=grid)
Still getting the same error though
My db.model is like this: db.define_table('table2',
Field('table1_inv_record', 'list:integer')
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Re: [web2py] create a link SQLFORM to another table with id.belongs
def query_table2():
table1_inv_record = None
table2_records = []
if request.args:
try:
session.table1_id = int(request.args(0))
except:
pass
row = db(db.table1.id==session.table1_id).select()
for line in row:
joined_records = line.inv_id
query=db.table1.id.belongs(joined_records)
db.table1.id.default = session.table1_id
grid=SQLFORM.grid(query, user_signature=False)
return dict(grid=grid)
2014-06-09 17:09 GMT-03:00 LoveWeb2py atayloru...@gmail.com:
Hi Fabiano,
I'm still trying to get this
For your Objeto.almoxarifado_id.default = session.almoxarifado
Shouldn't there be a db.Objeto.almoxarifado_id ? If I try to just put
table2.id I get an error saying table2 isn't defined.
Here is what I have so far
def query_table2():
table1_inv_record = None
table2_records = []
table1_inv_record = request.args(0)
row = db(db.table1.id==table1_inv_record).select()
for line in row:
joined_records = line.inv_id
if table1_inv_record:
inv1_record = int(table1_inv_record)
session.table1_id = inv1_record
query=db.table1.id.belongs(joined_records)
db.table1.id.default = session.table1.id
grid=SQLFORM.grid(query, user_signature=False)
return dict(grid=grid)
Still getting the same error though
My db.model is like this: db.define_table('table2',
Field('table1_inv_record', 'list:integer')
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Re: [web2py] create a link SQLFORM to another table with id.belongs
My db.py:
Almoxarifado = db.define_table('almoxarifado', Field('name'))
Objeto = db.define_table('objeto', Field('name'), Field('almoxarifado_id',
db.almoxarifado))
Objeto.almoxarifado_id.requires=IS_IN_DB(db, 'almoxarifado.id', '%(nome)s')
2014-06-09 17:09 GMT-03:00 LoveWeb2py atayloru...@gmail.com:
Hi Fabiano,
I'm still trying to get this
For your Objeto.almoxarifado_id.default = session.almoxarifado
Shouldn't there be a db.Objeto.almoxarifado_id ? If I try to just put
table2.id I get an error saying table2 isn't defined.
Here is what I have so far
def query_table2():
table1_inv_record = None
table2_records = []
table1_inv_record = request.args(0)
row = db(db.table1.id==table1_inv_record).select()
for line in row:
joined_records = line.inv_id
if table1_inv_record:
inv1_record = int(table1_inv_record)
session.table1_id = inv1_record
query=db.table1.id.belongs(joined_records)
db.table1.id.default = session.table1.id
grid=SQLFORM.grid(query, user_signature=False)
return dict(grid=grid)
Still getting the same error though
My db.model is like this: db.define_table('table2',
Field('table1_inv_record', 'list:integer')
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Re: [web2py] create a link SQLFORM to another table with id.belongs
It's working! Thank you so much! I actually didn't need the
db.table1.id.default = session.table1.id
I think this was the line that fixed it:
row = db(db.table1.id==session.table1_id).select()
Thank you, thank you, thank you!! Such a huge help. Have a virtual beer on
me :)
Once
On Monday, June 9, 2014 4:38:04 PM UTC-4, Fabiano Almeida wrote:
My db.py:
Almoxarifado = db.define_table('almoxarifado', Field('name'))
Objeto = db.define_table('objeto', Field('name'), Field('almoxarifado_id',
db.almoxarifado))
Objeto.almoxarifado_id.requires=IS_IN_DB(db, 'almoxarifado.id',
'%(nome)s')
2014-06-09 17:09 GMT-03:00 LoveWeb2py atayl...@gmail.com javascript::
Hi Fabiano,
I'm still trying to get this
For your Objeto.almoxarifado_id.default = session.almoxarifado
Shouldn't there be a db.Objeto.almoxarifado_id ? If I try to just put
table2.id I get an error saying table2 isn't defined.
Here is what I have so far
def query_table2():
table1_inv_record = None
table2_records = []
table1_inv_record = request.args(0)
row = db(db.table1.id==table1_inv_record).select()
for line in row:
joined_records = line.inv_id
if table1_inv_record:
inv1_record = int(table1_inv_record)
session.table1_id = inv1_record
query=db.table1.id.belongs(joined_records)
db.table1.id.default = session.table1.id
grid=SQLFORM.grid(query, user_signature=False)
return dict(grid=grid)
Still getting the same error though
My db.model is like this: db.define_table('table2',
Field('table1_inv_record', 'list:integer')
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Re: [web2py] create a link SQLFORM to another table with id.belongs
Hi, Good! db.table1.id.default = session.table1.id This line is used to insert new record automatically by SQLFORM. For the id field is unnecessary, it is useful when you want to set a default value, eg FK. All beers!! -- Resources: - http://web2py.com - http://web2py.com/book (Documentation) - http://github.com/web2py/web2py (Source code) - https://code.google.com/p/web2py/issues/list (Report Issues) --- You received this message because you are subscribed to the Google Groups web2py-users group. To unsubscribe from this group and stop receiving emails from it, send an email to web2py+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/d/optout.
Re: [web2py] create a link SQLFORM to another table with id.belongs
SQLFORM isn't working for me with
new_table = db(db.table1.id.belongs(record_ids)).select()
if I do {{=new_table}} in my view I can see the records which belong to
record_ids, but if I do:
SQLFORM.grid(new_table)
return dict(grid=grid)
I get an error 'Rows' object has no attribute '_db'
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Re: [web2py] create a link SQLFORM to another table with id.belongs
Hi,
You send var grid (see: return dict(grid=grid)). The first var grid is a
send var to your view, de second var grid is a local var of your function.
Then,
in controller use:
grid = SQLFORM.grid(db.new_table)
return dict(grid=grid)
in the view use:
{{=grid}}
Fabiano.
2014-06-06 12:45 GMT-03:00 LoveWeb2py atayloru...@gmail.com:
SQLFORM isn't working for me with
new_table = db(db.table1.id.belongs(record_ids)).select()
if I do {{=new_table}} in my view I can see the records which belong to
record_ids, but if I do:
SQLFORM.grid(new_table)
return dict(grid=grid)
I get an error 'Rows' object has no attribute '_db'
--
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Re: [web2py] create a link SQLFORM to another table with id.belongs
Hi Fabiano,
I actually was already using grid= SQLFORM.grid(db.new_table). That is when
I get the error.
If I do grid='' to test it and just return the table without SQLFORM I can
see it in the view, but when I apply SQLFORM I get the error 'Rows' object
has no attribute '_db'.
I'm guessing this is because of the way SQLFORM handles the grid? Could
there be something in my model messing it up?
On Friday, June 6, 2014 12:02:05 PM UTC-4, Fabiano Almeida wrote:
Hi,
You send var grid (see: return dict(grid=grid)). The first var grid is a
send var to your view, de second var grid is a local var of your function.
Then,
in controller use:
grid = SQLFORM.grid(db.new_table)
return dict(grid=grid)
in the view use:
{{=grid}}
Fabiano.
2014-06-06 12:45 GMT-03:00 LoveWeb2py atayl...@gmail.com javascript::
SQLFORM isn't working for me with
new_table = db(db.table1.id.belongs(record_ids)).select()
if I do {{=new_table}} in my view I can see the records which belong to
record_ids, but if I do:
SQLFORM.grid(new_table)
return dict(grid=grid)
I get an error 'Rows' object has no attribute '_db'
--
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Re: [web2py] create a link SQLFORM to another table with id.belongs
Try:
grid = SQLFORM.grid(db.table, user_signature=False)
Em sexta-feira, 6 de junho de 2014 13h10min34s UTC-3, LoveWeb2py escreveu:
Hi Fabiano,
I actually was already using grid= SQLFORM.grid(db.new_table). That is
when I get the error.
If I do grid='' to test it and just return the table without SQLFORM I can
see it in the view, but when I apply SQLFORM I get the error 'Rows' object
has no attribute '_db'.
I'm guessing this is because of the way SQLFORM handles the grid? Could
there be something in my model messing it up?
On Friday, June 6, 2014 12:02:05 PM UTC-4, Fabiano Almeida wrote:
Hi,
You send var grid (see: return dict(grid=grid)). The first var grid is a
send var to your view, de second var grid is a local var of your
function.
Then,
in controller use:
grid = SQLFORM.grid(db.new_table)
return dict(grid=grid)
in the view use:
{{=grid}}
Fabiano.
2014-06-06 12:45 GMT-03:00 LoveWeb2py atayl...@gmail.com:
SQLFORM isn't working for me with
new_table = db(db.table1.id.belongs(record_ids)).select()
if I do {{=new_table}} in my view I can see the records which belong to
record_ids, but if I do:
SQLFORM.grid(new_table)
return dict(grid=grid)
I get an error 'Rows' object has no attribute '_db'
--
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Re: [web2py] create a link SQLFORM to another table with id.belongs
How do you define tables?
2014-06-06 14:03 GMT-03:00 Fabiano Almeida fabi...@techno7.com.br:
Try:
grid = SQLFORM.grid(db.table, user_signature=False)
Em sexta-feira, 6 de junho de 2014 13h10min34s UTC-3, LoveWeb2py escreveu:
Hi Fabiano,
I actually was already using grid= SQLFORM.grid(db.new_table). That is
when I get the error.
If I do grid='' to test it and just return the table without SQLFORM I
can see it in the view, but when I apply SQLFORM I get the error 'Rows'
object has no attribute '_db'.
I'm guessing this is because of the way SQLFORM handles the grid? Could
there be something in my model messing it up?
On Friday, June 6, 2014 12:02:05 PM UTC-4, Fabiano Almeida wrote:
Hi,
You send var grid (see: return dict(grid=grid)). The first var grid is
a send var to your view, de second var grid is a local var of your
function.
Then,
in controller use:
grid = SQLFORM.grid(db.new_table)
return dict(grid=grid)
in the view use:
{{=grid}}
Fabiano.
2014-06-06 12:45 GMT-03:00 LoveWeb2py atayl...@gmail.com:
SQLFORM isn't working for me with
new_table = db(db.table1.id.belongs(record_ids)).select()
if I do {{=new_table}} in my view I can see the records which belong to
record_ids, but if I do:
SQLFORM.grid(new_table)
return dict(grid=grid)
I get an error 'Rows' object has no attribute '_db'
--
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Re: [web2py] create a link SQLFORM to another table with id.belongs
That was it! THANK YOU! Why does it work without a signature though?
On Friday, June 6, 2014 1:03:35 PM UTC-4, Fabiano Almeida wrote:
Try:
grid = SQLFORM.grid(db.table, user_signature=False)
Em sexta-feira, 6 de junho de 2014 13h10min34s UTC-3, LoveWeb2py escreveu:
Hi Fabiano,
I actually was already using grid= SQLFORM.grid(db.new_table). That is
when I get the error.
If I do grid='' to test it and just return the table without SQLFORM I
can see it in the view, but when I apply SQLFORM I get the error 'Rows'
object has no attribute '_db'.
I'm guessing this is because of the way SQLFORM handles the grid? Could
there be something in my model messing it up?
On Friday, June 6, 2014 12:02:05 PM UTC-4, Fabiano Almeida wrote:
Hi,
You send var grid (see: return dict(grid=grid)). The first var grid is
a send var to your view, de second var grid is a local var of your
function.
Then,
in controller use:
grid = SQLFORM.grid(db.new_table)
return dict(grid=grid)
in the view use:
{{=grid}}
Fabiano.
2014-06-06 12:45 GMT-03:00 LoveWeb2py atayl...@gmail.com:
SQLFORM isn't working for me with
new_table = db(db.table1.id.belongs(record_ids)).select()
if I do {{=new_table}} in my view I can see the records which belong to
record_ids, but if I do:
SQLFORM.grid(new_table)
return dict(grid=grid)
I get an error 'Rows' object has no attribute '_db'
--
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Re: [web2py] create a link SQLFORM to another table with id.belongs
Are you logged in your app?
2014-06-06 14:07 GMT-03:00 LoveWeb2py atayloru...@gmail.com:
That was it! THANK YOU! Why does it work without a signature though?
On Friday, June 6, 2014 1:03:35 PM UTC-4, Fabiano Almeida wrote:
Try:
grid = SQLFORM.grid(db.table, user_signature=False)
Em sexta-feira, 6 de junho de 2014 13h10min34s UTC-3, LoveWeb2py escreveu:
Hi Fabiano,
I actually was already using grid= SQLFORM.grid(db.new_table). That is
when I get the error.
If I do grid='' to test it and just return the table without SQLFORM I
can see it in the view, but when I apply SQLFORM I get the error 'Rows'
object has no attribute '_db'.
I'm guessing this is because of the way SQLFORM handles the grid? Could
there be something in my model messing it up?
On Friday, June 6, 2014 12:02:05 PM UTC-4, Fabiano Almeida wrote:
Hi,
You send var grid (see: return dict(grid=grid)). The first var grid is
a send var to your view, de second var grid is a local var of your
function.
Then,
in controller use:
grid = SQLFORM.grid(db.new_table)
return dict(grid=grid)
in the view use:
{{=grid}}
Fabiano.
2014-06-06 12:45 GMT-03:00 LoveWeb2py atayl...@gmail.com:
SQLFORM isn't working for me with
new_table = db(db.table1.id.belongs(record_ids)).select()
if I do {{=new_table}} in my view I can see the records which belong
to record_ids, but if I do:
SQLFORM.grid(new_table)
return dict(grid=grid)
I get an error 'Rows' object has no attribute '_db'
--
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Re: [web2py] create a link SQLFORM to another table with id.belongs
yes
On Fri, Jun 6, 2014 at 1:09 PM, Fabiano Almeida fabi...@techno7.com.br
wrote:
Are you logged in your app?
2014-06-06 14:07 GMT-03:00 LoveWeb2py atayloru...@gmail.com:
That was it! THANK YOU! Why does it work without a signature though?
On Friday, June 6, 2014 1:03:35 PM UTC-4, Fabiano Almeida wrote:
Try:
grid = SQLFORM.grid(db.table, user_signature=False)
Em sexta-feira, 6 de junho de 2014 13h10min34s UTC-3, LoveWeb2py
escreveu:
Hi Fabiano,
I actually was already using grid= SQLFORM.grid(db.new_table). That is
when I get the error.
If I do grid='' to test it and just return the table without SQLFORM I
can see it in the view, but when I apply SQLFORM I get the error 'Rows'
object has no attribute '_db'.
I'm guessing this is because of the way SQLFORM handles the grid? Could
there be something in my model messing it up?
On Friday, June 6, 2014 12:02:05 PM UTC-4, Fabiano Almeida wrote:
Hi,
You send var grid (see: return dict(grid=grid)). The first var grid
is a send var to your view, de second var grid is a local var of your
function.
Then,
in controller use:
grid = SQLFORM.grid(db.new_table)
return dict(grid=grid)
in the view use:
{{=grid}}
Fabiano.
2014-06-06 12:45 GMT-03:00 LoveWeb2py atayl...@gmail.com:
SQLFORM isn't working for me with
new_table = db(db.table1.id.belongs(record_ids)).select()
if I do {{=new_table}} in my view I can see the records which belong
to record_ids, but if I do:
SQLFORM.grid(new_table)
return dict(grid=grid)
I get an error 'Rows' object has no attribute '_db'
--
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Re: [web2py] create a link SQLFORM to another table with id.belongs
Now I get this error when I try to click edit on the query database:
type 'exceptions.ValueError' invalid literal for int() with base 10:
'edit'
I've tried changing signature to false, but I don't think thats the problem.
On Friday, June 6, 2014 1:17:39 PM UTC-4, LoveWeb2py wrote:
yes
On Fri, Jun 6, 2014 at 1:09 PM, Fabiano Almeida fabi...@techno7.com.br
wrote:
Are you logged in your app?
On Friday, June 6, 2014 1:03:35 PM UTC-4, Fabiano Almeida wrote:
Try:
grid = SQLFORM.grid(db.table, user_signature=False)
Em sexta-feira, 6 de junho de 2014 13h10min34s UTC-3, LoveWeb2py
escreveu:
Hi Fabiano,
I actually was already using grid= SQLFORM.grid(db.new_table). That is
when I get the error.
If I do grid='' to test it and just return the table without SQLFORM I
can see it in the view, but when I apply SQLFORM I get the error 'Rows'
object has no attribute '_db'.
I'm guessing this is because of the way SQLFORM handles the grid?
Could there be something in my model messing it up?
On Friday, June 6, 2014 12:02:05 PM UTC-4, Fabiano Almeida wrote:
Hi,
You send var grid (see: return dict(grid=grid)). The first var grid
is a send var to your view, de second var grid is a local var of
your function.
Then,
in controller use:
grid = SQLFORM.grid(db.new_table)
return dict(grid=grid)
in the view use:
{{=grid}}
Fabiano.
2014-06-06 12:45 GMT-03:00 LoveWeb2py atayl...@gmail.com:
SQLFORM isn't working for me with
new_table = db(db.table1.id.belongs(record_ids)).select()
if I do {{=new_table}} in my view I can see the records which belong
to record_ids, but if I do:
SQLFORM.grid(new_table)
return dict(grid=grid)
I get an error 'Rows' object has no attribute '_db'
--
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Re: [web2py] create a link SQLFORM to another table with id.belongs
Show your code
2014-06-06 15:18 GMT-03:00 LoveWeb2py atayloru...@gmail.com:
Now I get this error when I try to click edit on the query database:
type 'exceptions.ValueError' invalid literal for int() with base 10:
'edit'
I've tried changing signature to false, but I don't think thats the
problem.
On Friday, June 6, 2014 1:17:39 PM UTC-4, LoveWeb2py wrote:
yes
On Fri, Jun 6, 2014 at 1:09 PM, Fabiano Almeida fabi...@techno7.com.br
wrote:
Are you logged in your app?
On Friday, June 6, 2014 1:03:35 PM UTC-4, Fabiano Almeida wrote:
Try:
grid = SQLFORM.grid(db.table, user_signature=False)
Em sexta-feira, 6 de junho de 2014 13h10min34s UTC-3, LoveWeb2py
escreveu:
Hi Fabiano,
I actually was already using grid= SQLFORM.grid(db.new_table). That
is when I get the error.
If I do grid='' to test it and just return the table without SQLFORM
I can see it in the view, but when I apply SQLFORM I get the error 'Rows'
object has no attribute '_db'.
I'm guessing this is because of the way SQLFORM handles the grid?
Could there be something in my model messing it up?
On Friday, June 6, 2014 12:02:05 PM UTC-4, Fabiano Almeida wrote:
Hi,
You send var grid (see: return dict(grid=grid)). The first var grid
is a send var to your view, de second var grid is a local var of
your function.
Then,
in controller use:
grid = SQLFORM.grid(db.new_table)
return dict(grid=grid)
in the view use:
{{=grid}}
Fabiano.
2014-06-06 12:45 GMT-03:00 LoveWeb2py atayl...@gmail.com:
SQLFORM isn't working for me with
new_table = db(db.table1.id.belongs(record_ids)).select()
if I do {{=new_table}} in my view I can see the records which
belong to record_ids, but if I do:
SQLFORM.grid(new_table)
return dict(grid=grid)
I get an error 'Rows' object has no attribute '_db'
--
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- http://web2py.com/book (Documentation)
- http://github.com/web2py/web2py (Source code)
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Re: [web2py] create a link SQLFORM to another table with id.belongs
type 'exceptions.ValueError' invalid literal for int() with base 10:
'edit'
In some part of your code has converting string to numeric type. Probably
you should be picking up the URL argument and doing the conversion, but
when you edit the registry of URL arguments are changed.
In my example, I use session and try...except to resolve this.
2014-06-06 15:29 GMT-03:00 Fabiano Almeida fabi...@techno7.com.br:
Show your code
2014-06-06 15:18 GMT-03:00 LoveWeb2py atayloru...@gmail.com:
Now I get this error when I try to click edit on the query database:
type 'exceptions.ValueError' invalid literal for int() with base 10:
'edit'
I've tried changing signature to false, but I don't think thats the
problem.
On Friday, June 6, 2014 1:17:39 PM UTC-4, LoveWeb2py wrote:
yes
On Fri, Jun 6, 2014 at 1:09 PM, Fabiano Almeida fabi...@techno7.com.br
wrote:
Are you logged in your app?
On Friday, June 6, 2014 1:03:35 PM UTC-4, Fabiano Almeida wrote:
Try:
grid = SQLFORM.grid(db.table, user_signature=False)
Em sexta-feira, 6 de junho de 2014 13h10min34s UTC-3, LoveWeb2py
escreveu:
Hi Fabiano,
I actually was already using grid= SQLFORM.grid(db.new_table). That
is when I get the error.
If I do grid='' to test it and just return the table without SQLFORM
I can see it in the view, but when I apply SQLFORM I get the error
'Rows'
object has no attribute '_db'.
I'm guessing this is because of the way SQLFORM handles the grid?
Could there be something in my model messing it up?
On Friday, June 6, 2014 12:02:05 PM UTC-4, Fabiano Almeida wrote:
Hi,
You send var grid (see: return dict(grid=grid)). The first var grid
is a send var to your view, de second var grid is a local var of
your function.
Then,
in controller use:
grid = SQLFORM.grid(db.new_table)
return dict(grid=grid)
in the view use:
{{=grid}}
Fabiano.
2014-06-06 12:45 GMT-03:00 LoveWeb2py atayl...@gmail.com:
SQLFORM isn't working for me with
new_table = db(db.table1.id.belongs(record_ids)).select()
if I do {{=new_table}} in my view I can see the records which
belong to record_ids, but if I do:
SQLFORM.grid(new_table)
return dict(grid=grid)
I get an error 'Rows' object has no attribute '_db'
--
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- http://web2py.com
- http://web2py.com/book (Documentation)
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Re: [web2py] create a link SQLFORM to another table with id.belongs
I use this:
grid for table1
def index():
return dict(grid=SQLFORM.grid(Almoxarifado, links = [lambda row:
A('Listar Objetos',_href=URL('objeto','index',args=[row.id]))],
user_signature=False, csv=False))
grid for table2
def index():
if (not request.args) and (not request.vars):
redirect(URL('almoxarifado','index'))
alm = None
if request.args: alm = request.args(0)
try:
if alm:
alm = int(alm)
session.almoxarifado=alm
session.nomealmox = Almoxarifado[alm].nome
except ValueError:
pass
Objeto.almoxarifado_id.default = session.almoxarifado
form = SQLFORM.grid(Objeto.almoxarifado_id == session.almoxarifado,
fields=[Objeto.nome, Objeto.quantidade, Objeto.localizacao],
user_signature=False, csv=False)
return dict(form=form, nome=session.nomealmox)
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Re: [web2py] create a link SQLFORM to another table with id.belongs
This doesn't seem to work for me Fabiano,
When I click the button it just keeps table1 on the screen. There has got
to be an easier way (I hope)
On Thursday, June 5, 2014 7:41:51 PM UTC-4, Fabiano Almeida wrote:
I use this:
grid for table1
def index():
return dict(grid=SQLFORM.grid(Almoxarifado, links = [lambda row:
A('Listar Objetos',_href=URL('objeto','index',args=[row.id]))],
user_signature=False, csv=False))
grid for table2
def index():
if (not request.args) and (not request.vars):
redirect(URL('almoxarifado','index'))
alm = None
if request.args: alm = request.args(0)
try:
if alm:
alm = int(alm)
session.almoxarifado=alm
session.nomealmox = Almoxarifado[alm].nome
except ValueError:
pass
Objeto.almoxarifado_id.default = session.almoxarifado
form = SQLFORM.grid(Objeto.almoxarifado_id == session.almoxarifado,
fields=[Objeto.nome, Objeto.quantidade, Objeto.localizacao],
user_signature=False, csv=False)
return dict(form=form, nome=session.nomealmox)
--
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