i have a navigation element that is determined by values in a database.
no matter what view it is, i need to get these navigation objects out of the database.
where in the code can i tell it to set a template variable containing all the navigation objeccts without setting it in every view?
It sounds like a good use for a context processor.
Right way to do this is to use templatetag. Then you don't have to include it in every view, just in your templates like {% load navigation %} {% navigation %}
How to write one:
django docs on template tags (read overview and inclusion tags)
anoher resource
Related
I'm working on my first Django project and have my templates setup with a base that all the others extend. In that base I want to have some user-specific navigation which means loading some values from the database to build the contents of a drop down menu. However I don't want to have to do this inside each view. Coming from Symfony2/Twig I would normally do this using a sub-request where I tell the template to render a view and that will use it's own template. Using syntax like:
{% render 'Bundle:Controller:action' with {} %}
How would I accomplish this same thing with Django? I've read over the docs a couple of times but can't find any way to do this.
You have two approaches:
(better)
- add the code to base.html (the one you're always extending) and only override it when you need to.
or
(worse)
- in every template use {% include %} to include your menus.html template.
Update: re-reading your question: you could modify the request in context-processor so your base.html would then have this information.
Custom template tags are what you want.
I'm building a django application that has similar-looking display elements on several different pages.
For example, the projects.html page has a table that lists projects and some related information; the documents.html page has a similar-looking table
It seems like there ought to be a way to define a "my_kind_of_table" template and then insert it into different pages as necessary:
{% proj_list | create_my_kind_of_table:name,description,last_update %}
and then...
{% doc_list | create_my_kind_of_table:name,header,owner %}
I suspect that django can already do this, but I don't know what to search for. Any suggestions?
How about {% include ... %}?
To add context variables that all tables needs, then you have three solutions to pick: One is what I would call the "old fashioned" way, and is to have special function that is called by all views that needs to add context function. The second is to create a function decorator, that is used on the function that returns the response. The third way can be used if you use the new 1.3 class-based views, then you can create a mixin-class that your view-class inherits, and that adds these things in its own get method.
Custom template tags and filters will accomplish this: https://docs.djangoproject.com/en/dev/howto/custom-template-tags/
I store my sites navigation menu in database. I want to load all objects in the list once I launch my server, but I haven’t got idea how to do it. I really need to do it, because it will be problem in future to load data from database in views to display menu.
I tried to put loading code in settings.py, but there was an error, and in views after imports, but there was no effect.
I think you need Template context processors.
Here is a nice tutorial by James Bennett on that:
http://www.b-list.org/weblog/2006/jun/14/django-tips-template-context-processors/
Well, I'm very new to django, and maybe I'm wrong, but I think you are looking for something like caching. Read the docs, and decide whether is it fits you or not.
You can cache just a part of your view:
https://docs.djangoproject.com/en/1.3/topics/cache/#template-fragment-caching
surround your navigation bar like this:
{% cache 500 navbar %}
... put your navbar code
{% endcache %}
and ensure to have
{% load cache %}
at the top of your template or the base template.
I'm very new to django, about a week into it.
I'm making a site where users enter stuff, then other users can vote on whether they like the stuff or not. I know it's not so novel, but it's a good project to learn a bunch of tools.
I have a many-to-many table for storing who likes or dislikes what. Before I render the page, I pull out all the likes and dislikes for the current user, along with the stuff I'm going to show on the page.
When I render the page, I go through the list of stuff I'm going to show and print them out one at a time. I want to show the user which stuff they liked, and which they didn't.
So in my django template, I have an object called entry. I also have two lists of objects called likes and dislikes. Is there any way to determine if entry is a member of either list, inside my django template.
I think what I'm looking for is a filter where I can say something like
{% if entry|in:likes %}
or
{% if likes|contains:entry %}
I know I could add a method to my model and check for each entry individually, but that seems like it would be database intensive.
Is there a better way to think about this problem?
If you're using latest django version, then it's just
{% if entry in likes %}
Refer django docs
Go here. Very similar to what they're using on trunk. "Save this as smart_if.py in the templatetags folder of one of your apps. Then a simple {% load smart_if %} replaces the boring built-in Django {% if %} template with the new smart one."
If you're not running trunk one of the following should work:
Filter:
Replacement "if" tag, largely the basis for the new functionality in the upcoming 1.2 release:
I'm using Django to create a site that provides a separate web UI for sorts of producers and consumers. Both UIs (or "subsites") have different layouts, menus and graphics. However they access the same database and models, just from different aspects (producer vs. consumer...).It's all hosted under a single domain, the UI differentiation is done with URLs.
The problem comes when I want to integrate a CMS to this system, to take care of the menu structures and textual content. How should I go about handling the two different menus for the different UIs? I took a look at django-cms and django-page-cms, and they seem to maintain only a single menu hierarchy.
Any ideas?
One dirtyish solution would be to add e.g. a different prefix for each UI's menu items in the CMS, and hack the CMS code so that it only inserts the menu items for the correct UI (given as a parameter to the show_menu template tag).
A nicer way would be if it was possible to have multiple instances of the CMS app, so that each of them had their own database tables as well. But is this possible with django and e.g. django-cms or django-page-cms?
Some further restrictions:
the CMS must support localization
I'd prefer running a single Django instance, to keep the configuration and testing simple
I have not used django-cms so this is just off the top of my head.
There's a section of the docs called Extending the menu that looked promising. It may be unfortunate that so much of their configuration is in settings.py because it looks like you could manipulate their CMS_TEMPLATES to use different base templates (etc.) for different users. One way of getting around this (assuming that there is not a more direct route) is to add something to the UserProfile that identifies a user as consumer/producer. Then in your base.html you do:
{% if user.get_profile.consumer %}
...
{% else %}
...
{% endif %}
This effectively gives you two entirely different look/feel options based on user type. I'll also note that {% extends %} can take either a string constant or a string variable, so you could use a context_processor to set the name of template you are extending.
What you need is show_menu_below_id tag of django-cms. Create the pages consumers and producers with their respective id (advanced fieldset, on bottom of the page form) and then start building the page hierarchy for each one.
Then in the templates uses the tags:
<ul>
{% if user.get_profile.consumer %}
{% show_menu_below_id "consumer" %}
{% else %}
{% show_menu_below_id "provider" %}
{% endif %}
</ul>