I can't find solution for inserting hyperlink into generated pdf from Django easy_pdf. Link is displayed on pdf but as text. How to make this link clickable?
part of template.html:
{{item.campaign_name}}
part of view.py:
from easy_pdf.rendering import render_to_pdf_response
return render_to_pdf_response(request, template_name, context)
campaign_url is in context dictictionary
Thank you for your time
I hope this helps you.
The url will depend on how you stored it in the database. If it is something like https://www.something.com you can use
if you decide to use django urls then something like
You might want to look at the django documentation for more clarity. Django documentation
Related
how can i build dynamic link in Django?
I have a blog and each post need to have a unique link.
content of blog's post must show on linked page.
I searched in google but i'm was a little confused.
Thanks
The answer involves setting a slug on your model instance, then referencing it within your URL, and then consuming it in your view in order to determine what blog post to pass into your context. Honestly, this is unfortunately a loaded question, but one that is gone over in pretty much any starter tutorial on Django.
Check out this video as it covers exactly what you're asking (he's building a blog).
https://www.youtube.com/watch?v=Bmvd1O5pNIY
The Answer is:
steps:
in views:
def article_content_view(request,article_id):
content = Article.objects.get(id=article_id).content
context = {
'content':content
}
return render(request,'article.html',context)
in urls:
path('article/<int:article_id>/', views.article_content_view,name= 'article_content'),
in template html file:
<a href="{% url 'article_content' article.article_id %}"</a>
and my problem solved by add an id for each post in view:
'article_id': article.pk
you can see my github repo for this project.
you can get a unique random URL by using
import uuid
x = uuid.uuid4().hex
get this link in localhost:8000/?url=x
in urls.py
add path in urlpatterns
path(<url>/,viewname)
I've included Product and services in Navbar links,
If I click on Product link it should redirect to services page.
I'm Using Django-Oscar. So I don't want to mess around with the urls.py and view.py.
If you don't want to mess with urls and views rather just use javascript. So just add something similar to this on the navlink/navitem.
onclick="window.location.href = '{% url 'yoururl' %}';"
On the index page of my Django site I generate a list of urls that allow the user to navigate to a detail page. Once the detail page appears the navigation is no longer visible.
What I am trying to achieve is that the navigation list appears on the detail page (and every page that is added to my site).
What I have tried is the following: (the first line in each view is duplicated)
def index(request):
**collection_urls = Collection.objects.order_by('the_year')**
return render(request, 'index.html', {'collection_url': collection_urls})
def originalexample(request, collection_id):
**collection_urls = CarCollection.objects.order_by('the_year')**
car = get_object_or_404(CarCollection, pk=collection_id)
return render(request, 'detail.html', {'originalexample': car, 'collection_url': collection_urls})
Whilst this works, I know it is not right as I running the query twice. My next thought would be to perform the query once and then pass it to the pages as they are rendered. If so how would I do that? or is there a more pythonic method?
In case of need I'm using Django 1.6.2 and the list of urls does not change frequently (They can only be changed via the Django admin screens)
Thanks in advance.
You don't want to pass them from one to another, you simply want them to appear everywhere. That is usually done via context processors or custom tags.
So I am struggling a bit, with something that logically seems so simple but due to my limited understanding of Django I am not sure where to look and how to formulate a solution.
Basically I have a Blog app set up and it shows the complete(all the content including disqus discussion) latest post on the home page. The post has a further link to the posts own page as well. I have set up Disqus and need to get some key information to use for the disqus_url and disqus_identifier. I have set up the model as follows with a method for get_absolute_url as follows:
def get_absolute_url(self):
return reverse('blog.views.renderBlog',args=[str(self.id),str(self.slug)])
My view is set up as follows:
def renderBlog(request,postid=1,slug=None):
template = 'blog_home.html'
if(postid == 1 and slug == None):
post = Post.objects.latest('date_created')
else:
post = Post.objects.get(slug=slug, id=postid)
data = {
'post':post,
}
return render(request, template, data)
As you can see the view is set up to handle both URL's as follows:
url(r'^$', 'renderBlog', name='blogHome'),
url(r'^post/(?P<postid>\d{1,4})/(?P<slug>[\w-]+)/$', 'renderBlog', name='blogPostPage'),
In my template I am setting disqus_identifier = '{{ post.get_absolute_url }}' and I am hardcoding the domain portion in the meantime as disqus_url = 'http://127.0.0.1{{ post.get_absolute_url }}';.. Same goes for the comment count <a href="" data-disqus-identifier.
I dont like doing things in a hackish manner, what would be the best method for me to get the full absolute url. I have looked at request.get_absolute_uri but am not sure on how to actually use it to get what I want.
Thanks
The way I like to do it is configure a context_processor:
from django.contrib.sites.models import Site
def base_context_processor(request):
return {
'BASE_URL': "http://%s" % Site.objects.get_current().domain
}
# or if you don't want to use 'sites' app
return {
'BASE_URL': request.build_absolute_uri("/").rstrip("/")
}
in settings.py:
TEMPLATE_CONTEXT_PROCESSORS = (
...
'path.to.base_context_processor',
...
)
(In newer versions of Django, modify context_processors under TEMPLATES, OPTIONS instead.)
then in templates:
Object Name
Another solution would be to use request.build_absolute_uri(location), but then you would have to create a template tag that takes a request object and a location or an object that has get_absolute_uri method. Then you would be able to use it templates like that: {% get_full_uri request=request obj=post %}. Here is documentation on how to write custom tags.
This question is pretty old but I think its still relevant.
To get_absolute_url with domain in Django template you can do the following:
<li>Home</li>
First check if the request is https or not and then get the request of the host and pass the absolute url.
This way you will get full URL with domain in Django template.
I know the question is old, but I'm not sure the best answer provided is the correct one.
You should just use
as the Django docs point out.
https://docs.djangoproject.com/en/4.0/ref/models/instances/#get-absolute-url
The url provided comes with the domain as well.
Regards
I have a view (views.loaditems) which runs some algorithm and passes items to a template(product.html) where the items are loaded, and with each item, I have a "add_to_shortlist" link. On clicking this link, the item is added in the user's shortlist (for which I have a function). I want that on click, the page is not reloaded and has its items, but simply add that item to the user's shortlist. Also, where should I define this shortlist function?
I'm new to Django, and any help would be much appreciated. Thanks.
Update: Here's my code:
views.py
def loaditems(request):
#some code
ourdeals = SDeals.objects.filter(**{agestring3:0})
sorteddeals = ourdeals.order_by('-total_score')
user = request.user
context = {'deals': sorteddeals, 'sl_products':sl_products, 'user':user,}
template='index.html'
return render_to_response(template, context, context_instance=RequestContext(request))
def usersl(request, id, id2):
userslt = User_Shortlist.objects.filter(id__iexact=id)
products = SDeals.objects.filter(id__iexact=id2)
product = products[0]
if userslt:
userslt[0].sdeals.add(product)
sl = userslt[0].sdeals.all()
return render_to_response('slnew.html', {'sl':sl}, context_instance=RequestContext(request))
in my index.html I have:
<div class="slist"></div>
which in urls.py takes me to views.usersl:
url(r'^usersl/(?P<id>\d+)/(?P<id2>\d+)/$', views.usersl),
I don't want to go to slnew.html, instead be on index.html without reloading it, and on click 'slist', just run the function to add to shortlist.
In order to make changes on the server and in a page without navigating with the browser you need to look at JavaScript solutions. Read up about Ajax. In essence you need to use some JavaScript to send the update to the server, and to change the HTML.
JQuery is one popular library that will help you to do this. A more sophisticated example is AngularJS. On the Django side you'll write some views that handle these small update tasks used in the page. Libraries like Django REST framework or Django Slumber will help you with that.