Mike Coyle a écrit :
> Hi Folks,
>
> What is the preferred way to display a hyperlink within a DataGrid widget?
>
> Thanks,
> Mike
Hello I wrote some code for testing.
BUT :
- I used DataGrigPaginate in place of simple datagrid
- I used non SQL data (sort order is not working with non SQLObject )
Finaly I don't like this way and prefere to use hand made template in
state of widget
This is only my advice.
--- here is some code from my controler ---
from elementtree import ElementTree as ET
# generate the link inside the datagrid
class MakeLink:
def __init__(self, baseurl, id, title, action):
self.baseurl=baseurl
self.id=id
self.title=title
self.action=action
def __call__(self, obj):
url=controllers.url(self.baseurl,
dict(id=obj[self.id],action=self.action))
link = ET.Element('a', href=url, style='text-decoration:
underline' )
link.text = obj[self.title]
return link
# to get this kind of link :
http://centos41-srv:8080/person?action=edit&id=6
mylink=MakeLink('person', 'id', 'name', 'edit')
# just to get a element from a dictionary
def get(fieldname):
return lambda x: x.get(fieldname)
data_grid = PaginateDataGrid(
fields = [
PaginateDataGrid.Column(name='name', getter=mylink,
title='Name', options=dict(sortable=True)),
PaginateDataGrid.Column(name='age', getter=get('age'),
title='Age', options=dict(sortable=True, reverse_order=True)),
])
class Root(controllers.RootController):
....
# this is like paginate2 except it has sorting support
@turbogears.expose(template='.templates.paginate2')
@turbogears.paginate('persons', default_order='name')
def paginate4(self):
return dict(persons=self.persons(), list=data_grid)
def persons(self):
data = []
for i in range(100):
data.append(dict(id=i, name='name%d'%i, age=100-i))
return data
@turbogears.expose(template='.templates.person')
def person(self, id, action):
person=self.persons()[int(id)]
return dict(person=person)
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