Re: how do I specify optional params in url?

Al 31/03/13 19:00, En/na frocco ha escrit:
> Hello,
>
> Here is my current url.
>
> url(r'^narrow_category/(\d+)/(\w*)/(\d*)/(\d*)/(\d*)/$',
> 'catalog.views.narrow_category', name="narrow_category"),
>
> d+ is required
> w* can be optional
> last three d* can be optional
>
> This setting is not working.

You should be a bit more explicit about what "is not working" means.

I presume that you expect that an URL like "narrow_category/23" should
be matched against your url definition. No, it won't, since the slashes
are not optional. All those will match against your regular expression:

narrow_category/23/////
narrow_category/23/foo////
narrow_category/23/45////
narrow_category/23///23//

I think that the cleanest solution is defining a set of urls and,
somehow, provide default values for the missing parts:

def my_view(request, foo, bar="spam", baz=1234):
pass

url(r"^narrow/(\d+)/$", views.my_view, ...)
url(r"^narrow/(\d+)/(\w+)/$", views.my_view, ...)
url(r"^narrow/(\d+)/(\w+)/(\d+)/$", views.my_view, ...)

Note that the parts are no longer optional (+ instead of *).

It works like this:

/narrow/23/
matches the 1st rule
django calls: my_view(request, "23")
the view gets the arguments: foo="23", bar="spam", baz=1234

/narrow/23/baz/
2nd rule
my_view(request, "23", "baz")
foo="23", bar="baz", baz=1234

/narrow/23/baz/45/
3rd rule
my_view(request, "23", "baz", "45")
foo="23", bar="baz", baz="45"

/narrow/23/45/
2nd rule
my_view(request, "23", "45")
foo="23", bar="45", baz=1234


HTH

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