In my Django project, I have created a custom admin page for an app via the get_urls() method. I'd like to add a link to the app's main model index view that will take users to this custom page - however, I'm having some trouble creating this link element correctly and I don't seem to be able to piece together the right way to do it - I'm just left with a Reverse for 'export' not found. 'export' is not a valid view function or pattern name. error.
I've set up the admin for the app like so:
# my_project/observations/admin.py
from django.template.response import TemplateResponse
from django.urls import path
class ObservationAdmin(SimpleHistoryAdmin, SoftDeletionModelAdmin):
change_list_template = 'export_link.html'
def get_urls(self):
urls = super().get_urls()
custom_urls = [
path('export/', self.admin_site.admin_view(self.export_view), name='export')
]
return custom_urls + urls
def export_view(self, request):
context = dict(
self.admin_site.each_context(request),
)
return TemplateResponse(request, 'export.html', context)
and the two templates that are referenced:
# my_project/observations/templates/export.html
{% extends "admin/base_site.html" %}
{% block content %}
<div>
Some custom content
</div>
{% endblock %}
# my_project/observations/templates/export_link.html
{% extends 'admin/change_list.html' %}
{% block object-tools-items %}
<li>
Export
</li>
{{ block.super }}
{% endblock %}
Navigating directly to http://localhost:8000/admin/observations/observation/export/ works perfectly, I see the custom content page exactly as I want it... so the issue I'm striking is with the link template - I get the Reverse... error when I navigate to the model index page.
Perhaps the argument I'm passing to url is incorrect, or I need to register that URL elsewhere - but I don't quite know. The other examples of link elements like this that I've been able to find don't reference URLs created via the admin class' get_urls() method - so any guidance on this would be greatly appreciated.
Thanks very much, let me know if there's any other info that I can provide to help sort this out.
I think the problems is in missing namespace in your export_link.html template. Instead of:
Export
try:
Export
Related
I want to include some basic statistics about a model in a stats.html file. The variables don't show in the html. What am I doing wrong?
from django.shortcuts import render, get_object_or_404, redirect
from django.db.models import Avg, Sum, Count
from .models import Production
def statistics(request):
nr_of_plays = Production.objects.count()
nr_of_actors = Production.objects.aggregate(num_actors=Sum('nr_actors'))
nr_of_audience = Production.objects.aggregate(num_audience=Sum('est_audience'))
context = {
'nr_of_plays': nr_of_plays,
'nr_of_actors': nr_of_actors['num_actors'],
'nr_of_audience': nr_of_audience['num_audience'],
'test':'abc'
}
return render(request, 'stats.html', context)
The model:
class Production(models.Model):
title = models.CharField(max_length=200)
nr_actors = models.IntegerField(default=0)
est_audience = models.IntegerField(default=0)
...
urls.py:
path('stats/', views.statistics, name='stats'),
the relevant section of base.html:
<copyright class="text-muted">
<div class="container text-center">
<p>© One World Theatre - {% now "Y" %} {% include 'stats.html' with test=test %} </p>
</div>
</copyright>
And the stats.html template:
{% load static %}
{{ test }} - Stats: {{ nr_of_plays }} plays produced, involving {{ nr_of_actors }} actors, seen by {{ nr_of_audience }} people.
the output:
© One World Theatre - 2020 - Stats: plays produced, involving actors, seen by people.
EDIT:
I didn't mention that I'm using my template stats.html in my base.html template like this {% include 'stats.html' %}. When I add with test=test to the include tag, the test text shows. But when adding with nr_of_plays=nr_of_plays nothing happens :-/.
I ended up forgetting about trying to {% include 'stats.html' %} in my base template and just added those variables where I need them, works great. Not DRY, but what to do... .
EDIT 2:
I was too quick to cry victory. Edited the question with the latest code. Passing the variables in the view that handles the main content block works, but that means I would have to add them in every single view (not DRY). Still not getting what doesn't work with my setup. example.com/stats.html renders exactly what I want, but doesn't show the variables when I include it in my base.html. with test=test doesn't do anything. Clueless (and thankful for the help sofar).
Aggregate returns a dictionary.
You need to access its value via the key
context = {
'nr_of_plays': nr_of_plays,
'nr_of_actors': nr_of_actors['nr_actors_sum'],
'nr_of_audience': nr_of_audience['est_audience_sum']
}
Alternatively you can specify a custom key name instead of the default composite one:
nr_of_actors = Production.objects.aggregate(num_actors=Sum('nr_actors'))
nr_of_audience = Production.objects.aggregate(num_audience=Sum('est_audience'))
Note: .all() is redundant and can be removed
Base on your latest confession and symptoms, you don't seem to be going to your statistics view.
Looks like the url is rendering another view, which also extends base.html confuses you that you are in the right view.
One way to test it is to put a print statement in your statistics view and see if it prints anything in the console:
def statistics(request):
print(111111111111111111111111111111)
...
return render(request, 'stats.html', context)
Second thing is, if your base.html includes stats.html, you shouldn't be rendering the stats.html directly, you should pass the context to a template that extends base.html.
Third thing is, refer to Pynchia's answer to properly get the count of aggregated queryset.
I have a model called Project in an app called projects that I registered with the admin site so the instances can be added/edited/etc. This works as expected. Now I want to add a button for each project in the change list view on the admin site, that links to a custom form that requires a Project instance to do things. I followed a bunch of different tutorials to customize the admin site and managed to add another field to the table of the change list view. However the entries show up outside the table (see image).
I added the custom field by overwriting the admin/change_list.html template and calling a custom template tag custom_result_list within it. This tag adds a table field to the change list and then calls the admin/change_list_results.html template to render it. I have confirmed with a debugger that the item is added to the entries of the change list before the template is rendered (see image).
I cannot explain why the table is not rendered correctly even though the additional field has the same structure as the auto-generated ones. I have to admit I have resorted to Cargo Cult Programming, because I do not understand how this is supposed to work, despite spending too many hours trying to solve this simple problem.
Here's the relevant code.
In file /projects/templatetags/custom_admin_tags.py:
from django import template
from django.contrib.admin.templatetags.admin_list import result_list as admin_result_list
def custom_result_list(chl):
extended_cl = {}
extended_cl.update(admin_result_list(chl))
extended_cl["result_headers"].append({
'class_attrib': r' class="column-__str__"',
'sortable': False,
'text': 'Configure Project'
})
idx = 0
snippet = '<td class="action-button">{}</td>'
for project in chl.result_list:
extended_cl["results"][idx].append(snippet.format(project.id, project.unmod_name))
idx += 1
return extended_cl
register = template.Library()
register.inclusion_tag('admin/change_list_results.html')(custom_result_list)
In file templates/admin/projects/project/change_list.html:
{% extends "admin/change_list.html" %}
{% load i18n admin_urls static admin_list %}
{% load custom_admin_tags %}
{% block result_list %}
{% if action_form and actions_on_top and cl.show_admin_actions %}{% admin_actions %}{% endif %}
{% custom_result_list cl %}
{% if action_form and actions_on_bottom and cl.show_admin_actions %}{% admin_actions %}{% endif %}
{% endblock %}
To fix your issue:
from django.utils.html import format_html
replace your snippet.format(...) with format_html(snippet,...)
Explanation:
in django, all strings you pass from python are automatically HTML escaped. which here means, all your tags will not be considered as HTML. Such limitation is added to avoid any potential exploits by hackers. In your case, use of a template to render html is highly recommended. However, you can also send raw html from python using format_html helper function.
I am trying to include a page editable through the mezzanine admin in all the pages on my site. I read through the Mezzanine doc and the source and cannot figure out how to do this.
From the docs, I thought I could pass my page as an extra context, something like:
mezzanine.pages.views.page(request, slug, template=u'pages/page.html', extra_context={'mypage':<get_page_by_its_slug>})
But the doc says that the extra context is a mezzanine.pages.middleware.PageMiddleware object, which sets the slug from the request.
Do I need to write a context processor to do this? How do a load a specific page by its slug?
Just in case this helps someone out there, I created a context processor to solve this:
# context_processors.py
from mezzanine.pages.models import Page
def featured(request):
# editable page, get by ID or slug or title...
featured_page = Page.objects.get(id=49)
return {'featured_page': featured_page}
added the context processor into my settings.py:
TEMPLATE_CONTEXT_PROCESSORS = (
"django.contrib.auth.context_processors.auth",
## ...etc...,
"myapp.context_processors.featured",
)
and included the featured content in the base.html template:
{% block right_panel %}
<div>
{% editable featured_page.richtextpage.content %}
{{ featured_page.richtextpage.content|richtext_filter|safe }}
{% endeditable %}
</div>
{% endblock %}
If you know of a simpler way to do this, I'd love to hear your solution!
In my website, I want to let the admins reset the password of any user.
With reset I mean exactly what the password_reset view does (under contrib.auth): Send a confirmation link to that user email.
How would be the best way of doing that? Is there an already app/snippet that does that?
Edit:
Let's suppose user john is an admin. What I want is to let john reset any user's password through the admin interface. For example, to reset max password, he will just go to the max user, and click on any link to reset his password.
What I finally did was to add a custom ModelAdmin:
from django.contrib.auth.forms import PasswordResetForm
from django.contrib.auth.admin import UserAdmin
class CustomUserAdmin(UserAdmin):
...
def reset_password(self, request, user_id):
if not self.has_change_permission(request):
raise PermissionDenied
user = get_object_or_404(self.model, pk=user_id)
form = PasswordResetForm(data={'email': user.email})
form.is_valid()
form.save(email_template_name='my_template.html')
return HttpResponseRedirect('..')
def get_urls(self):
urls = super(UserAdmin, self).get_urls()
my_urls = patterns('',
(r'^(\d+)/reset-password/$',
self.admin_site.admin_view(self.reset_password)
),
)
return my_urls + urls
and I also had to override the change_form.html template, like this:
{% extends "admin/change_form.html" %}
{% load i18n %}
{% block object-tools %}
{% if change %}{% if not is_popup %}
<ul class="object-tools">
{# You can also give a name to that pattern and refer to it below using 'url' #}
<li>Reset password</li>
<li>{% trans "History" %}</li>
{% if has_absolute_url %}
<li><a href="../../../r/{{ content_type_id }}/{{ object_id }}/" class="viewsitelink">
{% trans "View on site" %}</a>
</li>
{% endif%}
</ul>
{% endif %}{% endif %}
{% endblock %}
The result looks like this:
If you want a more detailed explanation, I blogged about it.
The passreset app just exposes the django views via urls.py, and adjusts the login template to show a "Forgot my password" link.
The built-in django password reset views and templates are meant for self-reset. I guess the reset form could be prepopulated with a different user's email address (in the query string) but you'd still need to make adjustments such as changing the email template - "You're receiving this e-mail because you requested a password reset for your user account" is probably not what you want:
https://code.djangoproject.com/browser/django/trunk/django/contrib/admin/templates/registration/password_reset_email.html
Therefore you should expose the views at different URLs if you want to include self-reset as well.
Hook the django views into urls.py like so:
urlpatterns += patterns('django.contrib.auth.views',
url(r'^accounts/password/reset/$',
'password_reset',
name='password-reset'),
url(r'^accounts/password/reset/done/$',
'password_reset_done',
name='password-reset-done'),
url(r'^accounts/password/reset/confirm/(?P<uidb36>[-\w]+)/(?P<token>[-\w]+)/$',
'password_reset_confirm',
name='password-reset-confirm'),
url(r'^accounts/password/reset/complete/$',
'views.password_reset_complete',
name='password-reset-complete')
)
and where you want to make adjustments, pass in e.g. your own email template:
url(r'^/accounts/password/reset/$',
'password_reset',
{'email_template_name': 'my_templates/password_reset_email.html'}
name='password-reset'),
The "password_reset" view has more parameters you can tweak:
https://docs.djangoproject.com/en/dev/topics/auth/#module-django.contrib.auth.views
("post_reset_redirect" comes to mind as another one for your purposes)
To show a corresponding link you'd either change the User admin (careful, already registered - unregister then register your own, subclassed plus additional link field) or the change_form template itself.
I'm unaware of an app that provides this out-of-the-box, so I upvoted the question :-).
Yep, there is an app for that. Check here:
https://github.com/bendavis78/django-passreset
I'm sure I've seen this question on Stack Overflow before, but I couldn't find it by my life, so here goes nothing.
I have a normal Django menu which uses the {% url %} tag and static names for the menu items. Now I want to have a different style for the menu item which has been selected. But the menu is being rendered in the base template, so how do I figure out which menu item it is?
You could surely do this with some ugly template code, but a better more globally known way is to use a CSS selector. This lets CSS do all of the work automatically for you.
Here's how it works:
You simply put an id in your body depending on which page you are on.
Then in css you do something like this:
#section-aboutme #nav-aboutme,
#section-contact #nav-contact
/* ... put one of these per body/menu item ... */
{
font-color: red;
}
You put the nav-aboutme, and nav-contact ids on each of your menu items.
The style will automatically be selected by CSS depending on which body id they are inside of.
I normally do it the way Brian suggested, but to accommodate for a template which a designer gave me which used the more common class="selected" method, I wrote a {% nav %} template tag.
Your HTML navigation template will look something like:
{% block nav %}
<ul class="nav">
<li{% if nav.home %} class="selected"{% endif %}>Home</li>
<li{% if nav.about %} class="selected"{% endif %}>About</li>
</ul>
{% endblock %}
To set the navigation in a child template, do:
{% include "base.html" %}
{% load nav %}
{% block nav %}
{% nav "about" %}
{{ block.super }}
{% endblock %}
How about a custom tag which you use to generate your nav item?
The following takes the name of the url for which a nav item should be generated and the text it should display. It generates a li tag with a class of "selected" if the named url's path is the same as the current url (requires 'django.core.context_processors.request' in your TEMPLATE_CONTEXT_PROCESSORS). Within the li, it generates an a tag with the path of the url specified by the url_name. It has the contents specified by contents.
Obviously, this could be tweaked to generate different markup for the nav item, as required.
The rest can be done using CSS.
Advantages:
Easy to use
Little code required
DRY
Could be made to be more flexible
Disadvantages:
Requires 'django.core.context_processors.request'
Requires urls to be named e.g. urlpatterns = patterns('django.views.generic.simple',
...
(r'^$', 'direct_to_template', {'template': 'index.html'}, 'index'),
...
). This could potentially be done differently (e.g. pass in url).
Doesn't cope with pages not exactly equal to the specified and therefore will not apply the selected class to the li when on a page lower in the url heirarchy. For example, if I'm on /products/, it will highlight the nav item directing to /products/. If I'm on /products/myProduct/, it will not highlight the /products/ link. This could be coded around, but it would force people to use sensible urls. For example, change the additionalAttrs assignment to additionalAttrs = ' class=selected' if (context['request'].path.startswith(path) and path != '/') or (context['request'].path == path) else ''.
Code:
from django import template
from django.core.urlresolvers import reverse
register = template.Library()
class NavNode(template.Node):
def __init__(self, url_name, contents):
self.url_name = url_name
self.contents = contents
def render(self, context):
path = reverse(self.url_name)
additionalAttrs = ' class=selected' if path == context['request'].path else ''
return '<li'+additionalAttrs+'>'+self.contents+'</li>'
#register.tag
def nav_link(parser, token):
bits = token.split_contents()
if len(bits) == 3:
contents = bits.pop()
url_name = bits.pop()
else:
raise template.TemplateSyntaxError, "%r tag requires a single argument" % bits[0]
if contents[0] == contents[-1] and contents[0] in ('"', "'"):
contents = contents[1:-1]
return NavNode(url_name, contents)
You can pass request.path to your template
from django.shortcuts import render_to_response
from django.template import RequestContext
return render_to_response('templ.html', {'page':request.path}, context_instance=RequestContext(request))
then use an ugly if template tag to add a CSS class to your menu item
Let's say one has an app named "stackoverflow" and inside of that app folder one has a templates folder with a stackoverflow.html file that extends from the base template.
One way to achieve it is by defining a variable in the views.py of ones stackoverflow app, like
def stackoverflow(request):
return render(request,
'stackoverflow/stackoverflow.html',
{'section': 'stackoverflow'})
Then, in the base template
<li {% if section == "stackoverflow" %} class="selected" {% endif %}>
StackOverflow
</li>
Essentially having the variable section allows to figure the section. Note that one needs a space between section and ==... if one doesn't respect that, then one will get a Django Template Error saying
Could not parse the remainder.