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Django How do I show Author (user) name instea of ID?

I have a custom user model (CustomUsers) on one app called users, and then I have a model called Bookings from which I have created a ModelfForm called BookingsForm.
In the Bookings model (and ultimately the BookingsForm), I have a field called booking_author which has a ForeignKey inherited from the CustomUsers model.
Now I have been able to successfully call the booking_author field into my bookingscreate view and make it uneditable/read only as I had wanted. The issue now is that the field is displaying the id of the author instead of the name of the author. Is anyone able to help me resolve this?
views.py
#login_required
def bookingscreate(request):
bookings_form = BookingsForm(initial={'booking_author': request.user })
context = {
'bookings_form': bookings_form
}
return render(request, 'bookings_create.html', context)

There are multiple ways of getting the users name. To obtain the user name in this case you could use
request.user.get_full_name()
request.user.get_short_name()
Explanation:
request.user is the User object, and thus you have access to all user methods. If the above is not what you're looking for you could create a method in your user class and then call that method in this view.

Django REST Framework hyperlink URL failing to resolve

I am trying to set up hyperlinking in my Django REST Framework API, and for the life of me I can't find out where my error is.
My model:
class Franchise(models.Model):
id = models.AutoField(primary_key=True)
name = models.CharField(max_length=255)
# Other fields
My serializer
class FranchiseListSerializer(serializers.HyperlinkedModelSerializer):
url = serializers.HyperlinkedIdentityField(
view_name='franchise_details',
lookup_field='id',
lookup_url_kwarg='franchiseid'
)
class Meta:
model = Franchise
fields = ('id', 'name', 'url')
My URLs:
url(r'^db/franchise/$', views.franchise_index, name='db_franchise_index'),
url(r'^db/franchise/(?P<franchiseid>[0-9]+)/$', views.franchise_details, name='db_franchise_details')
Note that this is an included url conf, all my api functionality goes within an /api/ url
My views:
#api_view(['GET'])
def franchise_index(request, format=None):
franchise_list = Franchise.objects.all()
serializer = FranchiseListSerializer(franchise_list, many=True, context={'request': request})
return Response(serializer.data)
#api_view(['GET'])
def franchise_details(request, franchiseid, format=None):
franchise = Franchise.objects.get(id=franchiseid)
serializer = FranshiseDetailSerializer(franchise)
return Response(serializer.data)
Note that FranshiseDetailSerializer seen above works just fine.
Summary:
URL /api/db/franchise/ goes to the view franchise_index, which returns data serialized by FranchiseListSerializer.
URL /api/db/franchise/<franchiseid>/ goes to the view franchise_details, which returns data serialized by FranchiseDetailSerializer (Works fine)
As you can see, I have added a url field to FranchiseListSerializer, which I supposed to link to the corresponding franchise details page. Before I added the url field, the serializer only returned id and name, which was the correct and expected behaviour at the time.
When I go to /api/db/franchise/ now, I get the error:
ImproperlyConfigured at /api/db/franchise/
Could not resolve URL for hyperlinked relationship using view name "franchise_details". You may have failed to include the related model in your API, or incorrectly configured the `lookup_field` attribute on this field.
Following this hint, I set up the arguments in the url field of FranchiseListSerializer, and as far as I can tell, they are correct. I have checked and double checked the DRF documentation, here, here and here, but have found no solution.
Following other similar issues on Stackoverflow, I tried changing view_name='franchise_details' to view_name='api:franchise_details' (the name of the Django app the relevant files are in) and view_name='api:franchise_details-detail', but to no avail.
Any and all help is appreciated, cheers.

Thanks to #AKS's promting, I figured it out. view_name is actually the name of the URL, not the view. From the way I read the documentation (and the fact that it is view_name not url_name), it seemed to say that it was supposed to be the name of the view.
I had actually tried using view_name='db_franchise_details' (my urls name) before, but that did not work. After AKS promted my with that comment, I tried again, and also tried using view_name='api:db_franchise_details', which does work!

Making a complicated query in Django all my follows posts

Imagine that:
I have "Posts", "Users", "UserFollowUser" models in a Django app...
I want to get all the POSTS from my POST model but only for people that I FOLLOW.
Every POST has a USER id of the user who is the author.
I mean, there's a register that says:
"Cris" follows "Larry"
I want to get the posts from Larry and some other people I follow.
I tried:
follows = UserFollowUser.objects.get(follower = request.user) #I AM request.user
posts = Post.objects.filter(user = follows.followed).order_by('-id')
But I can only get posts from ONE person because of the "GET" function in objects object.

What you need to do is get an array of user ids for your followers. Then use the __in syntax to make a query where you are asking for all posts whose user id is in the set of user ids of your followers. The code would look something like this:
follower_user_ids = UserFollowUser.objects.filter(follower__user_id = request.user.id)\
.values_list('follower__user_id', flat=True)\
.distinct()
posts = Post.objects.filter(user_id__in=follower_user_ids)
*this assumes UserFollowUser.follower is a foreign key to the User table.

This is code if you want to show posts of all users logged in user follows also your own posted posts in the feed:
follows_users = user.profile.follows.all()
follows_posts = Post.objects.filter(author_id__in=follows_users)
user_posts = Post.objects.filter(author=user)
post_list = (follows_posts|user_posts).distinct().order_by('-date_posted')
You can always take reference for social media django clone from https://github.com/uttampatel007/nccbuddy

How to pass values to defined template from views.py in Django-python

I have a model like this:
class EventTypeCategory(models.Model):
name = models.CharField(max_length=50, verbose_name="Name")
user = models.ForeignKey(User, verbose_name="User")
Message_slug = models.SlugField(blank=True, verbose_name="Message")
def __unicode__(self):
return self.name
In urls.py:
url(r'^categ/$',
'eventcateg_detail', name='eventcateg_detail'),
In views.py:
def eventcateg_detail(request,event_categ_id=None, event_categ_slug=None):
I want to add/edit/delete(CRUD) above defined value i.e name and Message_slug by template level. I am not getting any hint how to relate url.py with views.py and what should be definition of eventcateg_detail function.How this function will pass values to template (template name will be categ.html)
I am newbie in Django :)
want your help

You need to allow the URL to accept parameters to allow you specify which event category you want to view:
/categ/outdoor-events/
/categ/catered-events/
...
Do do this, you use a named URL pattern in your url scheme:
url(r'^categ/(?P<slug>[-\w]+)/$','eventcateg_detail', name='eventcateg_detail'),
and in your view:
from django.shortcuts import get_object_or_404, render
def eventcateg_detail(request,slug):
return render(request, "categ.html", {
'obj' : get_object_or_404(EventCateg, Message_slug =slug) # You should change Message_slug to just slug
})
and in your template:
<h1>{{ obj.name }}</h1>
So when a user enters a URL like we have outlined above, it gets matched to our URL pattern and the slug part of the url (catered-events) gets passed as a parameter to our view.

It's better that you follow the Django tutorial first, this is all covered in there. See for example part 3 of the tutorial for more information on how to relate urls.py with views.py and part 4 discusses passing variables to the template.

I believe that a view function is only passed an httprequest when it is called by the Django framework, the other two parameters of the function will only be useful if you call the function yourself but will not be useful through the web.
As pointed out in the comments I was mistaken in my belief, extra parameters can be passed as dynamic urls (i.e. urls designated like this url(r'^polls/(?P<poll_id>\d+)/$', 'polls.views.detail'),. See this link and the answer by #pastylegs
The Django Admin will allow you to edit all model fields if this is what you are after. Instructions on setting it up can be found in the Django documentation.
However I think what you are asking is how to enable CRUD editing through the web to users who are not admin level users. In that case you have many options. One of those options is to use a pre-built framework for Django like piston. Another way would be to use generic views
The other option is to build views yourself enabling operations on your model. In that case all of Django is available to you. You can pass parameters to your custom functions within the httprequest, for example as POST data.

Inline-like solution for Django Admin where Admin contains ForeignKey to other model

I have several Customers who book Appointments. Each Appointment has exactly one customer, though a customer can be booked for multiple appointments occurring at different times.
class Customer(model.Model):
def __unicode__(self):
return u'%s' % (self.name,)
name = models.CharField(max_length=30)
# and about ten other fields I'd like to see from the admin view.
class Appointment(models.Model):
datetime = models.DateTimeField()
customer = models.ForeignKey("Customer")
class Meta:
ordering = ('datetime',)
Now when an admin goes to browse through the schedule by looking at the Appointments (ordered by time) in the admin, sometimes they want to see information about the customer who has a certain appointment. Right now, they'd have to remember the customer's name, navigate from the Appointment to the Customer admin page, find the remembered Customer, and only then could browse their information.
Ideally something like an admin inline would be great. However, I can only seem to make a CustomerInline on the Appointment admin page if Customer had a ForeignKey("Appointment"). (Django specifically gives me an error saying Customer has no ForeignKey to Appointment). Does anyone know of a similar functionality, but when Appointment has a ForeignKey('Customer')?
Note: I simplified the models; the actual Customer field currently has about ~10 fields besides the name (some free text), so it would be impractical to put all the information in the __unicode__.

There is no easy way to do this with django. The inlines are designed to follow relationships backwards.
Potentially the best substitute would be to provide a link to the user object. In the list view this is pretty trivial:
Add a method to your appointment model like:
def customer_admin_link(self):
return 'Customer' % reverse('admin:app_label_customer_change %s') % self.id
customer_admin_link.allow_tags = True
customer_admin_link.short_description = 'Customer'
Then in your ModelAdmin add:
list_display = (..., 'customer_admin_link', ...)
Another solution to get exactly what you're looking for at the cost of being a bit more complex would be to define a custom admin template. If you do that you can basically do anything. Here is a guide I've used before to explain:
http://www.unessa.net/en/hoyci/2006/12/custom-admin-templates/
Basically copy the change form from the django source and add code to display the customer information.

Completing #John's answer from above - define what you would like to see on the your changelist:
return '%s' % (
reverse('admin:applabel_customer_change', (self.customer.id,)),
self.customer.name # add more stuff here
)
And to add this to the change form, see: Add custom html between two model fields in Django admin's change_form

In the ModelAdmin class for your Appointments, you should declare the following method:
class MySuperModelAdmin(admin.ModelAdmin):
def get_form(self, request, obj=None, **kwargs):
if obj:
# create your own model admin instance here, because you will have the Customer's
# id so you know which instance to fetch
# something like the following
inline_instance = MyModelAdminInline(self.model, self.admin_site)
self.inline_instances = [inline_instance]
return super(MySuperModelAdmin, self).get_form(request, obj, **kwargs)
For more information, browser the source for that function to give you an idea of what you will have access to.
https://code.djangoproject.com/browser/django/trunk/django/contrib/admin/options.py#L423

There is a library you can use it.
https://github.com/daniyalzade/django_reverse_admin
But if you want to use link to object in showing table you can like this code:
def customer_link(self, obj):
if obj.customer:
reverse_link = 'admin:%s_%s_change' % (
obj.customer._meta.app_label, obj.customer._meta.model_name)
link = reverse(reverse_link, args=[obj.customer.id])
return format_html('More detail' % link)
return format_html('<span >-</span>')
customer_link.allow_tags = True
customer_link.short_description = 'Customer Info'
And in list_display:
list_display = (...,customer_link,...)