In admin add view I would like to able to select dynamically field A_permission from selected A-object while creating new B-object
A(models.Model):
name = models.CharField(primary_key=True, max_length=50)
permission = models.CharField(max_length=50)
B(models.Model):
A_name = models.ForeignKey(A)
A_permissions = models.ManyToManyField(A)
So I have objectA1 and objectA2 for example. While creating object B I should be able to first select one of the A objects, and then just get its permission for selection in field A_permissions
Can someone please point me how to do this? I feel like I tried everything
Further to the comments I made to your question, below is how I would allow selecting multiple permissions from A to one record in B.
In models.py:
class A(models.Model):
name = models.CharField(primary_key=True, max_length=50)
permission = models.CharField(max_length=50)
def __str__(self):
return f"{self.name}: {self.permission}"
class B(models.Model):
A_name = models.ManyToManyField(A)
In admin.py
from .models import A, B
admin.site.register(A)
admin.site.register(B)
Go to admin
You should be able to select multiple Permissions from A in one record for B using shift+click or ctrl+click.
There are many good django tutorials out there. I find the mozilla django tutorial to be quite good.
Related
I have these two models :
class Project(models.Model):
name = models.CharField(max_length=50)
users = models.ManyToManyField(User)
class User(models.Model):
name = models.CharField(max_length=25)
I want to get a queryset containing all the Projects with a specific user in the 'users' ManyToManyField
I tried this : Project.objects.filter(users__contains=user), but it's not working
Does someone know how can I do it ?
if filtering with id:
Project.objects.filter(users=search_id)
if filtering with user name:
Project.objects.filter(users__name__icontains=search_input)
I am creating my own users, Restaurant and Customer. I have extended the AbstractUser class and then created a OneToOneField field for each user. I am wondering if I need to add the AUTH_USER_MODEL in my settings.py. And also wondering what that does exactly...
What I was planning on doing was adding to my settings.py:
AUTH_USER_MODEL = 'myapp.Customer','myapp.Restaurant'
Do I have the right idea here?
My models.py:
class User(AbstractUser):
is_restaurant = models.BooleanField(default=False)
is_customer = models.BooleanField(default=False)
class Restaurant(models.Model):
user = models.OneToOneField(User, primary_key=True, on_delete=models.CASCADE)
restaurant_name = models.CharField(max_length=50)
def __str__(self):
return self.restaurant_name
class Customer(models.Model):
user = models.OneToOneField(User, primary_key=True, on_delete=models.CASCADE)
address = models.CharField(max_length=200)
def __str__(self):
return self.user.get_full_name()
No. AUTH_USER_MODEL isn't expecting a tuple, so this won't work.
In any case, Restaurant and Customer are not your user model; your subclassed User is. That's what you should be putting in that setting.
I would suggest create single user table instead of three different tables and add type as restaurant, customer, admin etc. And add only one table into settings file. this won't lead any further issues authentication etc. Having single user table is always robust. In your case having three tables seems not good to maintain.
========== UPDATE ===========
Create model for user named as CustomUser (or name which you feel better) and extends to User Model of Django using AbstractBaseUser,PermissionsMixin. like
class CustomUser(AbstractBaseUser): have all fields which user table has already. and add your desired table to bifurcate type of restaurant and
customer have type field with choices option.
For further help you can check section https://docs.djangoproject.com/en/1.11/topics/auth/customizing/#substituting-a-custom-user-model
I've got a two part question regarding Django Admin.
Firstly, I've got a Django model Classified that has a foreign key field from another table Address. On setting data, I've got no issues with any of the fields and all fields get saved correctly.
However, if I want to edit the foreign field in the entry in Classified, it doesn't display the old/existing data in the fields. Instead it shows empty fields in the popup that opens.
How do I get the fields to display the existing data on clicking the + so that I can edit the correct information?
Secondly, I'm sure I've seen search fields in Django Admin. Am I mistaken? Is there a way for me to implement search in the admin panel? I have over 2 million records which need to be updated deleted from time to time. It's very cumbersome to manually go through all the pages in the admin and delete or edit those.
Adding Model Code:
Classified
class Classified(models.Model):
id = models.AutoField(primary_key=True)
name = models.CharField(max_length=256)
contact_person = models.CharField(max_length=300, blank=True)
email = models.CharField(max_length=100, blank=True)
address = models.ForeignKey(Address)
subcategory = models.ForeignKey(Subcategory)
Address
class Address(models.Model):
id = models.AutoField(primary_key=True)
build_add = models.CharField(max_length=255)
street_add = models.CharField(max_length=255)
area = models.CharField(max_length=255)
city = models.ForeignKey(Cities)
The + means just that - add a new instance of the related object and relate the object you're editing to that. Because you're adding a new object it will be blank to start. If you want to be able to edit existing related objects from another object's admin you need to use inlines.
In your app's admin.py have something like:
from django.contrib import admin
from yourapp.models import Address, Classified
class AddressInline(admin.TabularInline):
model = Address
class ClassifiedAdmin(admin.ModelAdmin):
inlines = [AddressInline,]
admin.site.register(Classified, ClassifiedAdmin)
Adding search from there is really easy.
...
class ClassifiedAdmin(admin.ModelAdmin):
inlines = [AddressInline,]
search_fields = [
'field_you_want_to_search',
'another_field',
'address__field_on_relation',
]
...
Note the double underscore in that last one. That means you can search based on values in related objects' fields.
EDIT: This answer is right in that your foreignkey relationship is the wrong way round to do it this way - with the models shown in your question Classified would be the inline and Address the primary model.
I've got a weird problem in django admin list_display. Whenever I add a foreign key to a list_display the whole change list view goes blank showing only the total no of entries.
models.py:
class Organization(models.Model):
org_id = models.AutoField(primary_key=True)
org_name = models.CharField(max_length=288)
def __unicode__(self):
return self.org_name
class Meta:
db_table = u'organization'
class Server(models.Model):
server_id = models.AutoField(primary_key=True)
server_name = models.CharField(max_length=135,verbose_name="Server Name")
org = models.ForeignKey(Organization,verbose_name="Organization")
def __unicode__(self):
return self.server_name
class Meta:
db_table = u'server'
admin.py:
class ServerAdmin(admin.ModelAdmin):
list_display = ('server_name','org')
admin.site.register(Server,ServerAdmin)
Now I'd expect this code to show me the organization name in the ChangeList View, But instead I get this:
If I remove the org in the list_display of ServerAdmin class, I get this:
I didn't modify the template or override any ModelAdmin methods. I'm using Mysql(5.1.58) as my database that comes with ubuntu 11.10 repository.
I'll be really glad if I could a get a sloution for this problem guys. Thanks in advance.
I second Stefano on the fact that null=True, blank=True is to be added. But, I think you only need to add it to the org_name field of the Organization model. That should make your way through. It has to be done because you have run inspectdb to create models from your legacy DB. And probably the organization table in the DB has an empty string stored. So, adding the above would allow the Admin to have a blank field/column displayed.
Moreover, you can also try using callbacks in situations where you don't want to make changes to your model definition like the above.
Try adding null=True, blank=True to all your model fields.
Usually django admin will silenty fail (thus show no records in the list) if the row does not validate the model constraints.
See: https://stackoverflow.com/a/163968/1104941
Does the following work for you?
admin.py:
class ServerAdmin(admin.ModelAdmin):
list_display = ('server_name','org__org_name')
admin.site.register(Server,ServerAdmin)
I had a similar problem and solved it like this (using your example):
class ServerAdmin(admin.ModelAdmin):
list_display = ('server_name', 'get_org')
def get_org(self, obj):
return obj.org.org_name
get_org.short_description = 'Org'
admin.site.register(Server,ServerAdmin)
I am working on creating a simple contest submission system using django. This is my first real django project. Basically each user can view a list of problems, submit a file, and view a results page.
Each problem can be associated with multiple contests, and different contests can use the same problem. Because of this, both problem and contest have a manyToManyField with each other. This is what is causing my problem.
Here is the initial models.py implementation I am going with:
startfile
from django.db import models
class User(models.Model):
username = models.CharField(max_length=50)
firstname = models.CharField(max_length=50)
lastname = models.CharField(max_length=50)
class Problem(models.Model):
name = models.CharField(max_length=50)
filename = models.CharField(max_length=300)
contests = models.ManyToManyField(Contest)
class Contest(models.Model):
name = models.CharField(max_length=50)
problems = models.ManyToManyField(Problem)
date = models.DateField()
class Submission(models.Model):
user = models.ForeignKey(User)
problem = models.ForeignKey(Problem)
filename = models.CharField(max_length=300)
endfile
Is there a simple way to fix this? Or should I rethink my entire layout? I tried breaking each class into its own django app but I don't think thats how I should do it. The error I get is that Contest can not be found (because it exists lower in the file).
All advice is appreciated!
You don't need a ManyToManyField in both Contest and Problem. Many-to-many fields are already bidirectional. Just put it on one - doesn't matter which.
Djano will automatically create the reverse relation for you, so you only need to create it one end, eg.
class Problem(models.Model):
name = models.CharField(max_length=50)
filename = models.CharField(max_length=300)
contests = models.ManyToManyField(Contest, related_name='problems')
related_name gives you the possibility to assign a name to the reverse relation. Without defining the relation on the Contest model, you can then access eg. a_contest.problems.all()!