I have a page in my Django app that needs to do one of the following depending on the status of the logged in user in relation to a group (not a Django user group; something custom to my app) represented on the page:
If the user can join the group, display a link to join the group.
If the user is in the group, display a link to leave the group.
If the user can't join the group, don't display either link.
One way of doing this would be to create three templates (one with the join link, one with the leave link, and one with no link) and choose the appropriate one in the view. I feel like it may be overkill to have three different templates that will only differ in one line of code, so I have not gone that route as of yet.
Displaying the correct content for conditions 1 and 2 exclusively using a template is not possible, and if it were I do not think it would be advisable. Users to groups is a many-to-many relationship, and determining group membership requires passing a user to the group or passing a group to the user.
Since Django templates don't allow passing function arguments, I am trying to solve this by passing a context variable to the template using get_context_data.
def get_context_data(self, **kwargs):
context = super(NetworkDetails, self).get_context_data(**kwargs)
user = ???
context['in_group'] = user.in_group(context['group_detail'])
return context
If I do that, how can I get the currently logged in user within that method? If that isn't possible, where else can I get that information outside of the template? Is there an accepted method of doing something like this?
Thanks!
One way of doing this would be to create three templates (one with the
join link, one with the leave link, and one with no link) and choose
the appropriate one in the view
That's funny because if you can choose what template to include, you can just choose what html to display. Instead of:
{% if join_link %}
{% include 'join_link.html' %}
{% endif %}
{% if leave_link %}
{% include 'leave_link.html' %}
{% endif %}
{% if not join_link and not leave_link %}
you can't join
{% endif %}
You could just have:
{% if join_link %}
join
{% endif %}
{% if leave_link %}
leave
{% endif %}
{% if not join_link and not leave_link %}
you can't join
{% endif %}
So, I don't understand why you want to use template inclusion.
If I do that, how can I get the currently logged in user within that method?
self.request.user
self.request.user.is_authenticated() # return True if the user is logged in
You can determine the condition in your view and pass appropriate flag(s) to the template using context.
If there are multiple views/templates that need this info, you could implement custom context processor which can add this info in context and its available in each template.
Or If you have any OneToOne or any such relationship with User in your app, you can implement method in that model.
You can check if a user is logged in by checking permissions
https://docs.djangoproject.com/en/dev/topics/auth/ also you can do the similar for your other needs.
Related
I have Django model with CharFields 'flname1', 'date1', and 'time1'. My goal in my HTML is to have a {{ forloop }} that runs through only the 'date1' and 'time1' fields and displayed all of them. My Problem is that in my views file I can't find a way to create a python variable that only contains two of the three fields from one model. Ie tried a lot but what I'm trying to avoid is...
posts = DocPost.objects.all()
This puts all three fields into a variable and displays them in my for loop which I don't want. I've also tried a lot of work with filters and go things that other people on the internet had success with that didn't work for me like...
posts = DocPost.objects.filter('flname1').only('date1', 'time1')
This didn't work and didn't section off date1 and time1 and pack them away in a variable that I could loop through. Ive tried a lot more than this at this point to no prevail. Thank for any help.
There are two things you can do to only get certain fields in a query and iterate over them. The template for both is pretty much the same
First, you can use only() to generate a queryset where each object only has certain fields populated and all the rest are deferred
# View
context['posts'] = DocPost.objects.only('date1', 'time1')
# Template
{% for post in posts %}
{{ post.date1 }}
{{ post.time1 }}
{% endfor %}
Second, you can use values() to generate a queryset of dictionaries that only contain the fields specified
# View
context['posts'] = DocPost.objects.values('date1', 'time1')
# Template
{% for post in posts %}
{{ post.date1 }}
{{ post.time1 }}
{% endfor %}
Suppose I have a Item model, where Item objects can either be public (accessible to all users) or private (accessible only to authenticated users):
class Item(models.Model):
title = models.CharField(max_length=100)
is_public = models.BoleanField(default=True)
created_at = models.DateTimeField(auto_now_add=True)
#...
secret_key = ...
class Meta:
# I need to keep items in order:
ordering = ('-created_at',)
What I need is to list all items using a generic.ListView - keeping the order - but hide the secret_key of those items with is_public=False for anonymous users.
So in the template, I hide the secret_key if the user is not authenticated, like:
{% if request.user.is_authenticated %}
<p>{{ item.title }} - {{ item.secret_key }}</p>
{% else %}
<p>{{ item.title }} - This item is private. Sign up to see the secret_key!</p>
{% endif %}
and the ListView is like:
class ItemListView(ListView):
model = Item
paginate_by = 10
I'm aware that I can send two separate querysets for non logged-in users to the template, one for public items and the other for private ones; but I'm not sure how can I keep the order ('-created_at') in this approach.
The question is:
Is it safe to send all the secret_keys to the template and hide them for non logged-in users there?
(if it is safe, then) Is there a more efficient way of doing this?
I tried overriding the get_queryset method of my ItemListView and move the if condition from template to there (I think this would increase the performance, right?). I handled the situation where the users is authenticated (simply return all the objects); but for non logged-in users, I thought about somehow joining two separate querysets, one holding the public items and the other holding only the title and created_at of private items; but I didn't find to keep the order in this approach:
class ItemListView(ListView):
model = Item
paginate_by = 10
def get_queryset(self):
if self.request.user.is_authenticated:
return Item.objects.all()
else:
# ???
This was only a minimal-reproducible-example; Actually in the project, I have multiple access_levels; Each user has an access_level, based on their plan (e.g. basic, normal, pro, etc.) and each Item has an access_level; And an I'm dealing with about +100K objects, fetched from different databases (postgresql - some cached on redis) so the performance really matters here. Also the system is up-and-running now; so I prefer less fundamental solutions.
Thanks for your time. Your help is greatly appreciated.
Is it safe to send all the secret_keys to the template and hide them for non logged-in users there?
Your template is rendered server-side, and the client only get the rendered markup, so yes, it is totally safe. Well, unless someone in your team messes with the template code of course ;-)
(if it is safe, then) Is there a more efficient way of doing this?
Just filter the queryset in your view - you don't need two distinct querysets, and filtering the queryset will not change it's ordering.
def get_queryset(self):
qs = super(ItemListView, self).get_queryset()
if not self.request.user.is_authenticated:
qs = qs.filter(is_private=False)
return qs
and in your template:
{# avoids doing the same constant lookup within the loop #}
{% with is_auth=request.user.is_authenticated %}
{# I assume the queryset is named "objects" in the context ?#}
{% for item in objects %}
<p>{{ item.title }}{% if is_auth %} - {{ item.secret_key }}{% endif %}</p>
{% endfor %}
{% endwith %}
EDIT: bdoubleu rightly mentions in his answer that his solution makes testing easier. If you only need fields from your model (no method call), you can also use QuerySet.values() instead:
def get_queryset(self):
qs = super(ItemListView, self).get_queryset()
fields = ["title", "created_at"]
if self.request.user.is_authenticated:
fields.append("secret_key")
else:
qs = qs.filter(is_private=False)
return qs.values(*fields)
This will also make your code a bit more efficient since it doesn't have to build full model instances.
Another option is to annotate the queryset to add an extra attribute for display_secret_key which is going to be more efficient than checking the user access level for each item in the queryset while templating.
from django.db.models import F, Value as V
class ItemListView(ListView):
queryset = Item.objects.all()
paginate_by = 10
def get_queryset(self):
annotations = {
'display_secret_key': V('')
}
if self.request.user.access_level == 'PRO':
annotations['display_secret_key'] = F('secret_key')
return (
super().get_queryset()
.annotate(**annotations)
)
Then in your template:
<p>{{ item.title }} - {{ item.display_secret_key }}</p>
You could use 2 Templates, one for the authenticated user one for the unauthenticated. (just overwrite the get_template_names() for authentication check and add something like _sectempl.html to the found name and add the appropriate copy of the template with the secret data)
But I would say with bruno desthuilliers that if you switched off the debug mode there could be no constellation where unauthenticated users see content within
{% with authenticated=request.user.is_authenticated %}
{% if authenticated %}
do secret stuff
{% endif %}
{% endwith %}
or
{% if request.user.is_authenticated %}
hide secret stuff for all the others
{% endif %}
If you got a complex user-grouping-combination outside the standard django user right management (where you could ask for user-permissions in templates) then I would write the user_status (your "plan" or accesslevel) into the user-session (while authentication) and check for this user_status in the output-function of the attribute of the object.
Sketch:
Use in template:
{% for i in object_list %}
{{ i.name}}, {{ i.print_secret }}
{% endfor %}
In the model you create a "print_secret"-method witch returns the secret according to the previous recorded user_status in the session-data.
I'm trying to filter objects across two templates. One (the parent) should display the five most recently updated, and the other should display all of them.
I have the latter working perfectly with the following code:
views.py:
...
class ChannelProjectList(generic.ListView):
context_object_name = 'projects_by_channel'
template_name = 'channels/projects_by_channel.html'
def get_queryset(self):
self.channel = get_object_or_404(Channel, slug=self.kwargs['slug'])
return Project.objects.filter(channel=self.channel)
...
HTML:
{% for project in projects_by_channel %}
{{project.name}}
{% endfor %}
But when I go to "include" it on the parent page it breaks. After some research I understand why that is happening and why that isnt the proper way to do it. I dug around and found this, which seems to be exactly what I'm trying to do but when I implemented it, not only did it not work it but also broke the page that is working.
This feels like a pretty simple thing, but since this is my first project I'm running into new things every day and this is one of them.
Final Solution:
With the help of this I realised I needed to copy in the same get_queryset into the second template view, which I was then able to call into the template using "view.channel_projects"
You have two possibilities. First you could define two context variables (like its done in your linked solution) or you could slice the qs in the template.
1. Option slice:
This one would display all:
{% for project in projects_by_channel %}
{{project.name}}
{% endfor %}
This on only displays 5 entries:
{% for project in projects_by_channel|slice:":5" %}
{{project.name}}
{% endfor %}
2. Option define two query sets:
(views.py)
def get_queryset(self):
self.channel = get_object_or_404(Channel, slug=self.kwargs['slug'])
self.channel2 = get_object_or_404(Channel, id=12)#whatever
context["list"] = Project.objects.filter(channel=self.channel)
context["list2"] = Project.objects.filter(channel=self.channel2)[0:5] #this slices the query set for the first entries. If you want to order them first(by date or whatever) use "order_by()"
return context
(html)
{% for project in list %}
{{project.name}}
{% endfor %}
{% for project in list2 %}
{{project.name}}
{% endfor %}
If you want to display a single qs but one time the whole thing and in another template just the first 5 you are better suited with using the slice argument in the template. It keeps the view clean and simple and you don't have to query two times.
I hope that helps if not leave a comment.
I have the following code in my template:
{% for req in user.requests_made_set.all %}
{% if not req.is_published %}
{{ req }}
{% endif %}
{% empty %}
No requests
{% endfor %}
If there are some requests but none has the is_published = True then how could I output a message (like "No requests") ?? I'd only like to use Django templates and not do it in my view!
Thanks
Even if this might be possible to achieve in the template, I (and probably many other people) would advise against it. To achieve this, you basically need to find out whether there are any objects in the database matching some criteria. That is certainly not something that belongs into a template.
Templates are intended to be used to define how stuff is displayed. The task you're solving is determining what stuff to display. This definitely belongs in a view and not a template.
If you want to avoid placing it in a view just because you want the information to appear on each page, regardless of the view, consider using a context processor which would add the required information to your template context automatically, or writing a template tag that would solve this for you.
I have build a web site for a client which has a number of applications. Now he has a new URL registered which he wants to point to the same site, but he wants the look and feel changed. That's basically he wants a new home.html and base.html for the new web site. I can easily add the new site to settings and then change the view for the home page, to display a new home2.html.
However how do I do something like this as expressed in psuedo code in base.html
{% if site_id equals 1 %}
{% include "base1.html" %}
{% endif %}
{% if site_id equals 2 %}
{% include "base2.html" %}
{% endif %}
Any ideas. There are 100s of views on the site and nearly 50 models. I cannot recreate models, and mess around. This needs to be a quick fix.
Thanks in advance
You can create a context processor to automatically add site_id to the context: http://docs.djangoproject.com/en/dev/ref/templates/api/#writing-your-own-context-processors
But I would opt for a different solution. You can simply add an extra template directory per site so Django will try the templates specifically for that site first and fall back to the normal templates if they're not available.
To extend the idea of WoLph with the context processor, I would maybe even add the switching of the template to the context processor which would clean up your templates, as otherwise you may have to repeat the if clause quite often:
from django.contrib.sites.models import Site
def base_template(request):
site = Site.objects.get_current()
template = "base%s.html" % str(site.pk)
return {'BASE_TEMPLATE': template}
And in your template: {% include BASE_TEMPLATE %}
Looks nicer to me than the switching in the templates!
Another solution would be writing a Middleware to set ´request.site´ the current site id.