在 Django 通用视图中指定不同的模板名称

Question

I have the code in my urls.py for my generic views;

infodict = {
'queryset': Post.objects.all(),
'date_field': 'date',
'template_name': 'index.html',
'template_object_name': 'latest_post_list',
}

urlpatterns += patterns('django.views.generic.date_based',
(r'^gindex/$', 'archive_index', infodict),
)

So going to the address /gindex/ will use a generic view with the template of 'index.html'.

But since I will have more generic views in this urlpattern, how am I supposed to provide a different template name using the same infodict? I don't want to have to use lots of infodicts, and I can't use the default template name.

Please note this also applies to template object name within infodict.

Thanks for your help!

Edit : This is one of my first questions on stackoverflow and I am amazed with the thorough answers. I prefer using the dict constructor which I didn't know about. I find using the python documentation a bit harder as I can't find what i'm looking for usually!

Thanks again for all the answers and different approaches.

Answer 1

Use the dict() constructor:

infodict = {
    'queryset': Post.objects.all(),
    'date_field': 'date',
    'template_name': 'index.html',
    'template_object_name': 'latest_post_list',
}

urlpatterns = patterns('django.views.generic.date_based',
    url(r'^gindex/$', 'archive_index', dict(infodict, template_name='gindex.html')),
    url(r'^hindex/$', 'archive_index', dict(infodict, template_name='hindex.html')),
)

Answer 2

If you want to supply different template names to different views, the common practice is indeed to pass in a unique dictionary to each URL pattern. For example:

urlpatterns = patterns('',
    url(r'^home/$', 'my.views.home', {'template_name': 'home.html'}, name='home'),
    url(r'^about/$', 'my.views.about', {'template_name': 'about.html'}, name='about'),
)

This kind of pattern is common and acceptable.

Answer 3

Not as simple, but possibly useful if you have a whole lot of different patterns matching the same view:

base_dict={
...
#defaults go here
}
def make_dict(template_name,template_object_name):
    base_dict.update({
        'template_name':template_name,
        'template_object_name':template_object_name,
    })
    return base_dict

urlpatterns += patterns('django.views.generic.date_based',
(r'^gindex/$', 'archive_index', make_dict('index1.html','latest_poll_list')),
(r'^hindex/$', 'archive_index', make_dict('index2.html','oldest_poll_list')),
)

For a lot of similar generic views, this will condense your code calls a wee bit, at the expense of a little bit of transparency. If you've got many lines customizing the same few parameters, this might be easiest to read.

Finally, if all or most of your views require some, but not all, of the same information, never forget how useful a context processor is. It takes only a little more work to set up than the above solutions, but it extends much better, because it will guarantee that (unless you use the render_to_response shortcut without the RequestContext keyword) the defaults will always be available to your template no matter how your view or url conf changes.

Answer 4

you can define wrapper view functions to parametrize generic views. In your urls.py add pattern

url(r'^/(?P<tmpl_name>\w+)/$', 'my.views.datebasedproxy')

in your views.py add view function

def datebasedproxy(request, tmpl_name):
    return django.views.generic.date_based(request,otherparameters,
    template_name=tmpl_matrix[tmpl_name])

where tmpl_matrix is hypothetic list that matches template file name with parameter and otherparameters stands for other dictionary items required for date_based function