I have an app that makes use of \'django-registration\' a custom built model named \'idea\' and also a \'userprofile\'
I\'m having some trouble trying to understand how to create a dynamic choice field in django. I have a model set up something like:
I need to write a simple Django template tag t开发者_运维问答hat will display the user\'s IP Address?You need to take the information from the request (via the view function) in order to present it on
From the Django documentation... When you\'re only dealing with simple many-to-many relationships such as mixing and matching pizzas and toppings, a standard ManyToManyField is all you need. However,
I have a template showing a list of events. To prepare list of events I\'m using generic views, and set \'paginate_by\' parameter. Strangely when I load my page I see :
What is the best way to pass request information to a model class method? I\'m wondering whether I have my logic in the wrong place.Maybe I need to move it out of my model.
I am currently trying to break a list of people (aprox 20 to 30 items) into a table with 4 columns. Here is my current code.
I am trying to render the开发者_高级运维 dictionary content using Django template like this for example : result contain dictionary X
I have this in my view: string_location = myaddress2 geodata = [] for place, (lat, lng) in g.geocode(string_location,exactly_one=False):
I am trying to access the contents of an array using forloop.counter0 in django templates but I can\'t get it working.
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