def Grid_Home():
if 'view' in request.args:
redirect(URL('default', 'displayContents', args=request.args[-1]))
grid = SQLFORM.grid(db.somedatabase,csv=False,user_signature=False,
editable=False, create=False)
return dict(grid=grid)
The id of the requested record will be in request.args[-1], so that is
passed on to displayContents in the redirect.
Alternatively, if you want to avoid a redirect, you can instead disable the
standard View button (by setting details=False), and instead use the
"links" argument to create your own custom button/link, which could point
directly to the displayContents URL.
Anthony
On Saturday, June 18, 2016 at 8:47:42 AM UTC-4, Natalie Cluck wrote:
>
> How do I override what the View button does when clicked in SQLFORM.grid?
> This is what I have:
>
> def Grid_Home():
> grid = SQLFORM.grid(db.somedatabase,csv=False,user_signature=False,
> editable=False, create=False)
> return dict(grid=grid)
>
> def displayContents():
> # new page when View button is clicked
>
> I want to create a new grid on the displayContents page that uses my
> existing databases to display contents of a item listed in the "home grid."
>
--
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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