Django Link to class based views url - python

I created with django a page where I can delete my blog articles using the class based views, I want to add link to each article but I'm struggling to do it.
here is the code to link to the class based view
views.py
user_delete_article = Article.objects.filter(user=request.user)
template.html
{% for article in user_delete_article %}
<p><a class="readmore" href="{% url "article.views.DeleteView" article.id %}">{{ article.titre }}</a></p>
{% endfor %}
urls.py
url(r'^delete/(?P<id>\d+)/$', DeleteView.as_view(), name="DeleteView"),
How can I make this work?

Assuming this is django 1.8
<a href="{% url "DeleteView" id=article.id %}">

Related

How to link an ID generated URL to the <a href> tag

I am making my own portfolio website using Django.. So the idea is: Make posts containing websites that i've developed. All these posts will be displayed in a page called "Projects" and then you can access a single project to read about it. Every project i upload have an ID, so i used this code to acess each project page:
# Single project page
path('projects/<int:post_id>/', views.project, name='project'),
Then in my PROJECTS page i have a simple(for now) HTML code to show all my projects and link them to the single project page. What i want to know is how to LINK this ID generated URL in this code:
<ul>
{% for project in projects %}
<li>{{ project }}</li>
{% empty %}
<li>No project have been added yet.</li>
{% endfor %}
</ul>
I tried using:
<li><a href="{% url 'main_portfolio/projects/<int:post_id>/' %}</a></li
And some other things but nothing i try is working.
Maybe try <a href="{{ project.get_absolute_url }}">, or
<a href="projects/{{ project.id }}/">
You can work with the {% url … %} template tag [Django-doc]:
<li>some text</li>

How do I search for a certain item in multiple models django?

I am new to django and I am trying to build web site for my friend how makes handmade lamps.
My problem is that I have 3 different models that contains different types of lamps, and I want to get access to certain picture in any of this 3 models and display a picture and description on the other page,but it shows only some pictures from first model and for others throws an error.
this is my html and views.py codes.
{% for q in project1.reverse|slice:":2"%}
<div class="image-selected__lamps">
<a href="{% url 'project_detail' q.pk%}">
<img src="{{q.image.url }}">
</a>
</div>
{%endfor%}
{% for e in project2.reverse|slice:":2"%}
<div class="image-selected__lamps">
<a href="{% url 'project_detail' e.pk %}">
<img src="{{e.image.url}}">
</a>
</div>
{%endfor%}
{% for s in project3.reverse|slice:":2"%}
<div class="image-selected__lamps">
<a href="{% url 'project_detail' s.pk %}">
<img src="{{s.image.url}}">
</a>
</div>
{%endfor%}
enter image description here
def project_detail(request, pk):
project = (LampType1.objects.get(pk=pk), LampType2.objects.get(pk=pk), LampType3.objects.get(pk=pk))
context = {
'project': project,
}
return render(request, 'project_detail.html', context)
Edward this might be happening because you are passing the same id/pk to all the models and you got a picture from the first model and not from the other because an object with that id/pk does not exit in the other models.
To check, register your models in the admin and check weather an object with that particular id exit or not.
check if these steps have been done
model
class LampType1(models.Model):
title = models.CharField(max_length=40)
description = models.TextField()
image = models.ImageField(upload_to='media')
#not
#image = ImageField(upload_to='media')
settings.py
MEDIA_URL = 'media/'
MEDIA_ROOT = Path.joinpath(BASE_DIR,'media')
urls.py
from django.conf.urls.static import static
from django.conf import setting
urlpatterns = [.....
] + static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)
from django.contrib import admin
html
{% for p in project %}
<img src="{{ p.image.url }}">
{% endfor %}

Wagtail modeltranslation language switcher doesn't work on /search page

I have added search url to i18n_patterns, but the language switcher doesn't work on that page.
urls.py:
urlpatterns += i18n_patterns(
path("search/", search_views.search, name="search"),
path("", include(wagtail_urls)),
)
language switcher:
{% get_available_languages_wmt as languages %}
<div class="nav-item dropdown float-right">
<p class="nav-link dropdown-toggle m-auto" data-toggle="dropdown" role="button" aria-expanded="false">
{{ request.LANGUAGE_CODE|upper }}</p>
<div class="dropdown-menu w-25">
{% for language in languages %}
{% if language != request.LANGUAGE_CODE %}
<a class="dropdown-item" href="{% change_lang language page %}">{{ language|upper }}</a>
{% endif %}
{% endfor %}
</div>
</div>
Furthermore, when i add search url above root one the search page raises 404 page.
How can i make the language switcher work on the search page?
The language switcher uses the Wagtail page variable. See the change_lang template tag:
<a ... href="{% change_lang language page %}">...</a>
And search is a Django view, not a Wagtail page. The page variable is not defined.
You can make the switcher work on the search view by setting the href yourself:
{% for language in languages %}
<a href="/{{ language.code }}/search/"
{% endfor %}
Alternatively, you can create a Wagtail search page:
class SearchPage(Page):
def get_context(self, request):
context = super().get_context(request)
... # Copy the code from the current search view.
# Update and return the context
context.update({
'search_query': search_query,
'search_results': search_results,
})
return context
https://docs.wagtail.io/en/latest/topics/pages.html#customising-template-context
You also have to:
Rename (search_page.html) and rework (page.variable_name) your search template.
Remove the search url from urls.py
Add the SearchPage via the Wagtail admin interface to the page tree.

How to render data to a {% included a.html %} template in Django

I have a rank.html which is a publicly sharing template for many other templates through {% include rank.html %} method.
This template will display the 48 hours hot news base on the click number.
Here is the view.py:
def rank(self, request):
hot_news_48h = h_mostViewed(48, News, '-pv')
return render(request, "rank.html", {
'hot_news_48h': hot_news_48h,})
h_mostViewed(48, News, '-pv') is a function,that can fetch most viewed(clicked) post within 48 hours.It works.
Here is the rank.html:
<ul>
{% for hot_view in hot_news_48h %}
<li>
<a href="{% url 'news:news_detail' hot_view.pk %}" >
<img src="{{ MEDIA_URL }}{{ hot_view.image }}" >
</a>
<a href="{% url 'news:news_detail' hot_view.pk %}">
<h6>{{ hot_view.title }}</h6>
</a>
</div>
</li>
{% endfor %}
</ul>
Here is the url.py:
path('hot_news', views.rank, name="hot_news")
The problem is,I can only get the html ,but can't receive the data.
But if I give up {% include rank.html %} method and insert the rank.html's code directly inside each template which need this function, I can get the data.
Take new_detail.html template as an example:
Here is the view.py:
def newsDetailView(request, news_pk):
news = get_object_or_404(News, id=news_pk)
all_comments = NewsComments.objects.filter(news=news)
news.comment_nums = all_comments.count()
news.save()
News.objects.filter(id=news_pk).update(pv=F('pv') + 1)
hot_news_48h = h_mostViewed(48, News, '-pv')
relative_news = News.objects.filter(tag__id__in=news.tag.all()).exclude(id=news_pk)[:6]
return render(request, "news_detail.html", {
'news': news,
'all_comments': all_comments,
'hot_news_48h': hot_news_48h,
'relative_news': relative_news
})
Here is the urls.py:
path('-<int:news_pk>', views.newsDetailView, name="news_detail"),
So above,I directly inserted rank.html's code into new_detail.html and it works I can get the data.
My question is what should I do or correct,so that I can get the data in {% include rank.html %} method. Because {% include rank.html %} is simple and flexible.I don't want to repeat the same code in several same template.
Thank you so much for your patience!
How about this:
- Create a folder "templatetags" in your application and add a file "news_tags.py" or name it what you want. Then you can define the tags you need:
from django.template import Library
from your_app.models import your_model
register = Library()
#register.inclusion_tag('your_app/your_template.html')
def hot_news(num, order):
objects = News.objects.order_by(order)[:num]
result['objects'] = objects
return result
In your templates you then do the following:
{% load news_tags %}
{% hot_news 48 '-pv' %}
Then create a template as your already did and reference it in the inclusion tag. Then it should work properly.
If you want it to work for multiple models you can have a look at this: https://docs.djangoproject.com/el/2.1/ref/applications/
The apps framework allows you to fetch models from a string input.
I finally solved the issue by Creating custom context processor.https://www.youtube.com/watch?v=QTgkGBjjVYM

How to add new created page url to modified django-oscar

I'm a beginner in django-oscar and I try to manage a new view on the page.
I've already created two pages with django-oscar dashboard,
https://ibb.co/cM9r0v
and made new buttons in the templates:
Lib/site-packages/oscar/templates/oscar/partials/nav_primary.html
https://gist.github.com/Kalinar/076fc8144869c3b50fc0bc9e52f825e4
I have no idea how to make a good a href="???" to new pages in buttons ... can someone help?
Maybe there is better way to do it, can you explain it to me?
Oscar uses Django's flatpages app, so you can use their template tags to dynamically add in navigation links to the pages you create.
{% load flatpages %}
{% get_flatpages as flatpages %}
{% for page in flatpages %}
<li class="dropdown active">
<a href="{{ page.url }}" class="dropdown-toggle" {% if not expand_dropdown %} data-toggle="dropdown"{% endif %}>
{% trans page.title %}
</a>
</li>
{% endfor %}
You can find more information on the flatpages app in the Django flatpage documentation.
You already know the title and of course the id of the page, and each page is an instance of the model PagePromotion, just query the model for such name/id and use the attribute page_url of the returned instance to send it within a variable in the context of the view that renders the menu.
Then in the template:
href="{{ variable_containing_url }}"

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