there is any way to make a Break Page with django for using with large articles?
thanks
Run your content through a template filter, similar to Linebreaks (as an example)
So, you could make a "parsemore" filter, or something like that.
Documentation for making filters can be found here: http://docs.djangoproject.com/en/1.1/howto/custom-template-tags/
Related
I may not be understanding something obvious, but I'm struggling to add a (top-level) menu item to my Wagtail based menu that hooks to a page rendered by an included app that doesn't know about Wagtail. Ideally, it is just a normal Django TemplateView with standard urlconf, though I may need to add some custom code.
If I use the custom URL in the menu editor, I get a not found from Wagtails core.serve. I've looked at snippets, wagtail hooks, RoutablePageMixin, and the custom URL in the menu editor and none seem like it accomplishes what I'm trying to do.
It may well be that I'm simply misunderstanding the docs, but is there a simple example of someone doing this? The closest I've found so far is https://www.caktusgroup.com/blog/2016/02/15/wagtail-2-steps-adding-pages-outside-cms/. I've also searched https://docs.wagtail.io/en/v2.4/advanced_topics/third_party_tutorials.html to now avail. Any guidance appreciated.
Thx,
--Don
Hope this is useful, but it seems that my problem was not the mixing of Wagtail and non-Wagtail items - it was in my URLConf - Wagtail.core.serve occurred before the Django url I was trying to reach and was trying to respond. Once I reordered the URLConf appropriately, I am getting the view as I wanted.
Sigh...
I have a lot of apps running on my site and I was wanting to make all the views accept a certain kwarg without having to go in and edit them all manually? Is there a way to do this?
I suppose I should add it into the django base view class somewhere, but I am unsure exactly where to add it to in that?
Any ideas?
EDIT:
What I am trying to do is have translations set in my db under a certain model and then have the site default text areas be displayed in a certain language based on the url...
/eng/some/url
/esp/some/url
those two urls would display different languages, but I have to capture the variable and put it into each and every view which is quite cumbersome
Django already has some i18n support in urls, check it out. You need to activate django.middleware.locale.LocaleMiddleware by adding it to your settings.MIDDLEWARE_CLASSES and to tune your urlconf a bit by wrapping your urls with i18n_patterns.
The complete example is given in the docs, I see no sense copying it here.
I'm looking at porting a custom-written PHP CMS into Django. One of the features the CMS currently has is an image upload function. I write an article, tag it with information, then choose a photo for it. If the system has any photos which have been added to articles with tags in common with the new one, it will suggest the photo for that article too. If there are no matches then a new image can be added.
In case this doesn't make sense, let's say I tag an article as Bruce Springsteen, The Beatles and Led Zeppelin. Next time I add an article with the tag The Beatles, it should suggest I use the image added for the first article.
What would be the best Django-applicable way to implement this? I've looked at the Photologue app and have integrated it, and I know it has tagging support (the problem here is that I'm using django-taggit, whereas Photologue supports django-tagging). One approach could be simply building it myself -- when a user uploads an article, I run a hook after they save it to associate the tags with the image. I'm just not sure how to then autosuggest an image in the admin tools based on that info.
Any ideas/approaches greatly appreciated.
This is almost certainly something you're going to have to build yourself. Django has a moderate number of libraries out there (that you've clearly already found). Unlike other solutions, it doesn't have a lot of things that get you 100% to your desired solution (whereas something like Drupal might get you 100% of the way there).
What you will probably need to do (at a high level) is something like this:
Create an AJAX view that takes the current tags as an argument and does a query on the existing posts to see what tags match and returns images from those posts.
Use jQuery/javascript on your view to call your AJAX view on the page as tags are added
Use jQuery to update a <div> on your page and show all the images that your view returned
Here is a similar example that might help get you started.
You might look into django-ajax as a helper library for your requests, but it definitely isn't necessary.
The hook between the your image module and any other django module can be implemented using django's contenttypes framework which also provides some useful instance methods for returning related/hooked objects.
I'm working on a Django site with a basic three column design. Left column navigation, center column content and right column URL specific content blocks.
My question is about the best method of controlling the URL specific content blocks in the right column.
I am thinking of something along the lines of the Flatpages app that will make the content available to the template context if the URL matches a pre-determined pattern (perhaps regex?).
Does anyone know if such an app already exists?
If not, I am looking for some advice about the best way to implement it. Particularly in relation to the matching of patterns to the current URL. Is there any good way to re-use parts of the Django URL dispatcher for this use?
Django CMS is a good suggestion, it depends on how deep you want to go. If this is just the beginning of different sorts of dynamic content you want then you should go that way for sure.
A simple one-off solution would be something like this:
You would just need to write a view and add some variables on the end of the URL that would define what showed up there. Depending on how fancy you need to get, you could just create a simple models, and just map the view to the model key
www.example.com/content/sidecontent/jokes/
so if "jokes" was your block of variable sidecontent (one of many in your sides model instances) the urls.py entry for that would be
(r'^content/sidecontent/(?P<side>)/$,sides.views.showsides),
and then in your sides app you have a view with a
def showsides(request, side):
Sides.objects.get(pk=side)
etc...
For something like this I personally would use Django CMS. It's like flatpages on steroids.
Django CMS has a concept of Pages, Templates, and Plugins. Each page has an associated template. Templates have placeholders where you can insert different plugins. Plugins are like mini-applications that can have dynamic model-based content.
Although Django-CMS is an interesting suggestion, there are quite a few projects that do specifically what you've requested - render blocks of content based on a URL. The main one that I know about is django-flatblocks.
I want to combine pagination with filtering. Since I have a lot of filters I do not want to send them per GET request, since the URLs get really ugly.
Since django pagination uses GET request to pass the page parameters, I do not know how I can combine these two approaches.
Any idea?
Great add-on would be: How can I combine this approach with table sort? :-)
Edit:
Actually it should work like the pagination of stackoverflow - user questions. If a user clicks on a page number one is shown the correct page, without showing the get parameters in the url.
This is the url called.
https://stackoverflow.com/api/userquestions.html?page=2&pagesize=10&userId=237690&sort=Recent
But the url shown in the browser is neat and short.
Seems to be ajax. Anybody an idea how to implement this? :)
If the URL is not shown in the browser`s address bar, I do not care about whether it is beautiful or not.
Edit: The solution:
Make an ajax update with all filter parameters passed to the view. This should help you get started with implementing ajax for your site: link
Thus the GET parameters never show up in the address bar.
have you checked the paginate application for django?
it may help you a lot, use it all the time :D
http://code.google.com/p/django-pagination/
Have you considered django-tables2? It gives you django-admin style tables without you having to write the logic yourself.
maybe you can use the urs, something like:
http://oursite.com/something/filter1/filter2/3/
the doc -> http://docs.djangoproject.com/en/1.1/topics/http/urls/
I figured out two solutions:
Instead of using just hyperlinks use it inside a POST form, i dont have any example now but i remember have used that for REST functions in Ruby on rails
Save the query info in a session.
Hope this help.