Hi Together
Now i got a little more Work down for this issue.
My Template recieves the follow dict.
test = FastDataGrid( fields=[
('Id','id'),
DataGrid.Column('user_name','user_name', 'user_name',
options=dict(sortable=True)),
('Email','email_address')],
template='myapp.widgets.backend.templates.datagrid')
return dict(active_section='home',
title='home',
navigations = navigations,
list_widget = test,
data=[user for user in User.select()])
I my template i have.
${list_widget.display(data)}
This shows me the FastDataGrid in the standart form. But, when i click
on the link edit ofcourse i get an error because i dont get anything
for /controller/edit/1. Is it possible to map in an easy way?
Maybe is it possible to extend the FastDataController with my own widgets?
Thanks so far.
Best Regards Ivo
---------- Forwarded message ----------
From: Ivo Looser <[EMAIL PROTECTED]>
Date: 17.05.2006 21:02
Subject: FastDataGrid as Widget
To: [email protected]
Hi Together
I have in my Template a Widget for Navigation:
<div id="navigation_container"
py:content="NavigationWidget(navigations=navigations,
active_section=active_section).display()" />
In my Controller i have:
return dict(active_section='home',
title='home',
navigations = navigations)
Everything works fine. But i whould like to extend with a
FastDataController. Just to pass in the dict the Controller or
somthing and use it in the same Template like:
${list_widget.display(data)}
Is this possible? How?
Thanks in Advance Ivo
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