Consider the following model:
class FPModel(models.Model):
# The user who created
author = models.ForeignKey(auth.models.User, null=False)
# The user who last edited
editor = models.ForeignKey(auth.models.User, null=True)
# Create Time
created_at = models.DateTimeField(auto_now_add=True)
# Modify Time
edited_at = models.DateTimeField(auto_now=True)
class Meta:
abstract = True
I will be auto-populating the author and editor fields from django admin.
When I sync the database I am getting the following error:
(pinax-env)gautam#Aspirebuntu:$
python manage.py syncdb
Error: One or more models did not validate:
FP.fpmodel: Accessor for field 'author' clashes with related field 'User.fpmodel_set'. Add a related_name argument to the definition for 'author'.
FP.fpmodel: Accessor for field 'editor' clashes with related field 'User.fpmodel_set'. Add a related_name argument to the definition for 'editor'.
I am using django 1.2.5 and pinax 0.7.2.
What should I do to solve this?
I found the answer from the docs, specifically here and here.
I have to use
author = models.ForeignKey(auth.models.User , null = False ,related_name="%(class)s_related_author" ) # The user who created
editor = models.ForeignKey(auth.models.User , null = True,related_name="%(class)s_related_editor" ) # The user who last edited
Related
I have a simple model:
class Item(models.Model):
user = ForeignKey(User, related_name="user_items")
users = ManyToManyField(User, related_name="users_items")
I want it so that when a user creates an Item via ViewSet, that user is automatically assigned to the user and users fields.
I typically do this for ForeignKey's via the ViewSet's perform_create:
class ItemViewSet(viewsets.ModelViewSet):
...
def perform_create(self, serializer):
if self.request.user.is_authenticated:
serializer.save(user=self.request.user)
else:
serializer.save()
When I try to do it for the ManyToManyField too, I get this error:
{'users': [ErrorDetail(string='This list may not be empty.', code='empty')]}
I've tried the following in perform_create():
# 1
serializer.save(user=self.request.user, users=self.request.user)
# 2
serializer.save(user=self.request.user, users=[self.request.user])
# 2
serializer.save(user=self.request.user, users=[self.request.user.id])
How can I update a ManyToManyField via a ViewSet's perform_create?
Edit:
I guess the following works:
obj = serializer.save(user=self.request.user)
obj.users.add(self.request.user)
Is there no way to have the M2M field when the object is initially created though?
when you want set a list to m2m field one of the things you can do is this:
item_obj.users.set(user_list)
probably you need first get your item_obj.
for this you can get your object id from item_id = serializer.data.get('id') , and then item_obj = Item.objects.get(id = item_id)
I created a Custom User midway through my project, Called OUser, in an app called accounts. I have some existing groups defined in the default AuthUserGroups. How do I add OUsers to a group? When I try, I get an error, Invalid column name 'ouser_id'. What do I need to do to allow OUsers to be added to my predefined groups? Do I have to create a custom group model to put my OUsers? Is there anyway that I can over tell Auth_User_groups that it should be looking to OUsers?
accounts.admin
class OUserAdmin(BaseUserAdmin):
# The forms to add and change user instances
form = UserAdminChangeForm
add_form = UserAdminCreationForm
# The fields to be used in displaying the User model.
# These override the definitions on the base UserAdmin
# that reference specific fields on auth.User.
fieldsets = (
('Personal info', {'fields': ('first_name', 'last_name', 'email',)}),
('Permissions', {'fields': ('is_superuser', 'is_active', 'is_staff',)}),
)
# add_fieldsets is not a standard ModelAdmin attribute. UserAdmin
# overrides get_fieldsets to use this attribute when creating a user.
add_fieldsets = (
(None, {
'classes': ('wide',),
'fields': ('username',)}
),
)
search_fields = ('username',)
ordering = ('username',)
filter_horizontal = ()
list_filter = ('is_superuser',)
admin.site.register(OUser, OUserAdmin)
'''
There are also other dependencies on things, such as Auth_Group_permission, but I assume that it will be a similar fix to switching Auth_Group to using OUser.
How would I make groups contain references to Ouser. I have tried just swapping the Foreign key of the user_id in AuthUserGroups to contain a reference to OUser, but that didn't work. Maybe I missed something?
Thanks for the help!
AuthUserGroups is a table which holds the ManyToMany relationships between User and Group. It has foreign key constraints to these models. You can't have a field which has a foreign key relationship to both OUsers and User
You would need to create a different ManyToMany relationship form OUsers to Group that goes through a separate table.
Here is an example on to add it to the OUser model
from django.contrib.auth.models import Group
class OUsers(models.Model):
...
groups = models.ManyToManyField(Group)
Docs on ManyToMany fields
I'm using django 1.11 and I tried to to create django dynamic models by referring this link https://code.djangoproject.com/wiki/DynamicModels , by executing each and every step it runs without any issue, but How can I see this created table in django admin panel?
action.py
from django.db import models
from django.contrib import admin
def create_model(name, fields=None, app_label='', module='', options=None, admin_opts=None):
"""
Create specified model
"""
class Meta:
# Using type('Meta', ...) gives a dictproxy error during model creation
pass
if app_label:
# app_label must be set using the Meta inner class
setattr(Meta, 'app_label', app_label)
# Update Meta with any options that were provided
if options is not None:
for key, value in options.iteritems():
setattr(Meta, key, value)
# Set up a dictionary to simulate declarations within a class
attrs = {'__module__': module, 'Meta': Meta}
# Add in any fields that were provided
if fields:
attrs.update(fields)
# Create the class, which automatically triggers ModelBase processing
model = type(name, (models.Model,), attrs)
# Create an Admin class if admin options were provided
if admin_opts is not None:
print admin_opts
class Admin(admin.ModelAdmin):
pass
for key, value in admin_opts:
setattr(Admin, key, value)
admin.site.register(model, Admin)
return model
In Console:
from action import create_model
from django.db import models
fields = {
'first_name': models.CharField(max_length=255),
'last_name': models.CharField(max_length=255),
'__str__': lambda self: '%s %s' (self.first_name, self.last_name),
}
options = {
'ordering': ['last_name', 'first_name'],
'verbose_name': 'valued customer',
}
admin_opts = {}
model = create_model('Person', fields,
options=options,
admin_opts=admin_opts,
app_label='form',
module='project.app.model',
)
I can see no. of fields by
len(model._meta.fields)
But I have no idea of, how to register the created model in admin, and what parameter will come inside admin_opts = {} , how can i do makemigrations and migrate,how can I access this model in views.py, from where i will import this model .Can you guys please help me for this , it will be very useful for me and Thanks in advance.
with connection.schema_editor() as editor:
editor.create_model(Model)
This is from github source code , try it instead of sql_model_create and I try to success in my project,and it's true..
I have worked hard for a long time because I don't find django-dynamic-model in "django 1.10".
I think you forgot to execute this function.
def install(model):
from django.core.management import sql, color
from django.db import connection
# Standard syncdb expects models to be in reliable locations,
# so dynamic models need to bypass django.core.management.syncdb.
# On the plus side, this allows individual models to be installed
# without installing the entire project structure.
# On the other hand, this means that things like relationships and
# indexes will have to be handled manually.
# This installs only the basic table definition.
# disable terminal colors in the sql statements
style = color.no_style()
cursor = connection.cursor()
statements, pending = sql.sql_model_create(model, style)
for sql in statements:
cursor.execute(sql)
I am using wagtail ModelAdmin for some of my non page models and want to add some custom validation.
This is some of the code.
class EditPlanningView(EditView):
def publish_url(self):
return self.url_helper.get_action_url('publish', self.pk_quoted)
def unpublish_url(self):
return self.url_helper.get_action_url('unpublish', self.pk_quoted)
def post(self, request, *args, **kwargs):
form = self.get_form()
if form.is_valid():
instance = form.save(commit=False)
if bool(request.POST.get('action-publish')):
try:
instance.publish(commit=True)
except PublishWithoutMeetingError as e:
form.add_error(
'planning_meeting',
e
)
return self.form_invalid(form)
When validation fails the invalid form is returned, but the error I added is not bound to the field. In stead a 'general error message' appears at the top.
Can someone help me out?
Cheers,
Robert
I think the error is in the following lines.
form.add_error(
'planning_meeting',
e
)
Actually can't say anything without knowing about PublishWithoutMeetingError, the type of e. Better to replace e with a string. And make sure the post method is not throwing any exceptions. Other than that, what you have done is correct. Read the following to also to check if you have missed any point.
Long Answer
There are two ways that you can achieve showing an error messages in forms.
Overriding the Form
Overriding the EditView
In both of these cases, you are going to use a method called add_error. That method takes 2 argument, field and error. From these two, error is the most important argument. The field simply state the field of the form that this error applies to. This can be None.
The error argument can be multiple types.
The error argument can be an instance of str. Then wagtail will assign the given error to the given field.
The error argument can be an instance of list of str. Then wagtail will assign the given list of errors to the given field.
The error argument can be an instance of dict with str keys and str or list of str values. In this case field should be None. The keys will be used as the fields for the errors given by values.
The error argument can be an instance of ValidationError exception. You can create a ValidationError using a str, list, or dict, which represent the above three cases.
Overriding the Form
In the form clean method need to be overridden in order to find errors.
from wagtail.admin.forms.models import WagtailAdminModelForm
class ExtraForm(WagtailAdminModelForm):
def clean(self):
cleaned_data = super().clean() # Get the already cleaned data. Same as self.cleaned_data in this case. But this way is better.
title = cleaned_data.get('title') # Get the cleaned title
if title is None: # Title is never None here, but still..
return cleaned_data
title = title.strip() # Do some formatting if needed
if title.startswith('A'): # Validation
self.add_error('title', 'Title cannot start with A') # Validation error
else:
cleaned_data['title'] = title # Use the formatted title
return cleaned_data
class MyModel(models.Model):
id = models.AutoField(primary_key=True)
title = models.CharField(max_length=500, default='', blank=False)
# Or any other fields you have
base_form_class = ExtraForm # Tell wagtail to use ExtraForm instead of the default one
Overriding the EditView
This way is same as the way that you have mentioned in the question. You need to override post method. You need to check if the form associated with the EditView is valid or invalid and return the appropriate form.
To check validity, is_valid method of the form is used by default. That method will clean the form and check if there are errors added to the form.
If form is valid, you need to return self.valid_form and self.invalid_form otherwise.
Unlike overriding the Form, you can access the request here.
class MyEditView(EditView):
def post(self, request, *args, **kwargs):
form = self.get_form() # Get the form associated with this edit view
if form.is_valid(): # Check if the form pass the default checks
my_field = request.POST.get('my_field') # You can access the request
title = form.cleaned_data.get('title') # You can access the form data
if title != my_field: # Validation
form.add_error('title', 'Title must match my_field') # Validation error
return self.form_invalid(form) # Return invalid form if there are validation errors
return self.form_valid(form) # Return the valid form if there are no validation errors
else:
return self.form_invalid(form) # Return invalid form if default check failed
class MyModelAdmin(ModelAdmin):
model = MyModel
menu_label = 'My Model'
list_display = ('id', 'title')
search_fields = (
'title',
)
edit_view_class = MyEditView # Tell wagtail to use MyEditView instead of the default one.
here is the models page
In this picture, only the title shows up on here, I used:
def __unicode__(self):
return self.title;
here is the each individual objects
How do I show all these fields?
How do I show all the fields in each Model page?
If you want to include all fields without typing all fieldnames, you can use
list_display = BookAdmin._meta.get_all_field_names()
The drawback is, the fields are in sorted order.
Edit:
This method has been deprecated in Django 1.10 See Migrating from old API for reference. Following should work instead for Django >= 1.9 for most cases -
list_display = [field.name for field in Book._meta.get_fields()]
By default, the admin layout only shows what is returned from the object's unicode function. To display something else you need to create a custom admin form in app_dir/admin.py.
See here: https://docs.djangoproject.com/en/dev/ref/contrib/admin/#django.contrib.admin.ModelAdmin.list_display
You need to add an admin form, and setting the list_display field.
In your specific example (admin.py):
class BookAdmin(admin.ModelAdmin):
list_display = ('title', 'author', 'price')
admin.site.register(Book, BookAdmin)
If you want to include all but the ManyToManyField field names, and have them in the same order as in the models.py file, you can use:
list_display = [field.name for field in Book._meta.fields if field.name != "id"]
As you can see, I also excluded the id.
If you find yourself doing this a lot, you could create a subclass of ModelAdmin:
class CustomModelAdmin(admin.ModelAdmin):
def __init__(self, model, admin_site):
self.list_display = [field.name for field in model._meta.fields if field.name != "id"]
super(CustomModelAdmin, self).__init__(model, admin_site)
and then just inherit from that:
class BookAdmin(CustomModelAdmin):
pass
or you can do it as a mixin:
class CustomModelAdminMixin(object):
def __init__(self, model, admin_site):
self.list_display = [field.name for field in model._meta.fields if field.name != "id"]
super(CustomModelAdminMixin, self).__init__(model, admin_site)
class TradeAdmin(CustomModelAdminMixin, admin.ModelAdmin):
pass
The mixin is useful if you want to inherit from something other than admin.ModelAdmin.
I found OBu's answer here to be very useful for me. He mentions:
The drawback is, the fields are in sorted order.
A small adjustment to his method solves this problem as well:
list_display = [f.name for f in Book._meta.fields]
Worked for me.
The problem with most of these answers is that they will break if your model contains ManyToManyField or ForeignKey fields.
For the truly lazy, you can do this in your admin.py:
from django.contrib import admin
from my_app.models import Model1, Model2, Model3
#admin.register(Model1, Model2, Model3)
class UniversalAdmin(admin.ModelAdmin):
def get_list_display(self, request):
return [field.name for field in self.model._meta.concrete_fields]
Many of the answers are broken by Django 1.10. For version 1.10 or above, this should be
list_display = [f.name for f in Book._meta.get_fields()]
Docs
Here is my approach, will work with any model class:
MySpecialAdmin = lambda model: type('SubClass'+model.__name__, (admin.ModelAdmin,), {
'list_display': [x.name for x in model._meta.fields],
'list_select_related': [x.name for x in model._meta.fields if isinstance(x, (ManyToOneRel, ForeignKey, OneToOneField,))]
})
This will do two things:
Add all fields to model admin
Makes sure that there is only a single database call for each related object (instead of one per instance)
Then to register you model:
admin.site.register(MyModel, MySpecialAdmin(MyModel))
Note: if you are using a different default model admin, replace 'admin.ModelAdmin' with your admin base class
Show all fields:
list_display = [field.attname for field in BookModel._meta.fields]
Every solution found here raises an error like this
The value of 'list_display[n]' must not be a ManyToManyField.
If the model contains a Many to Many field.
A possible solution that worked for me is:
list_display = [field.name for field in MyModel._meta.get_fields() if not x.many_to_many]
I like this answer and thought I'd post the complete admin.py code (in this case, I wanted all the User model fields to appear in admin)
from django.contrib import admin
from django.contrib.auth.models import User
from django.db.models import ManyToOneRel, ForeignKey, OneToOneField
MySpecialAdmin = lambda model: type('SubClass'+model.__name__, (admin.ModelAdmin,), {
'list_display': [x.name for x in model._meta.fields],
'list_select_related': [x.name for x in model._meta.fields if isinstance(x, (ManyToOneRel, ForeignKey, OneToOneField,))]
})
admin.site.unregister(User)
admin.site.register(User, MySpecialAdmin(User))
I'm using Django 3.1.4 and here is my solution.
I have a model Qualification
model.py
from django.db import models
TRUE_FALSE_CHOICES = (
(1, 'Yes'),
(0, 'No')
)
class Qualification(models.Model):
qual_key = models.CharField(unique=True, max_length=20)
qual_desc = models.CharField(max_length=255)
is_active = models.IntegerField(choices=TRUE_FALSE_CHOICES)
created_at = models.DateTimeField()
created_by = models.CharField(max_length=255)
updated_at = models.DateTimeField()
updated_by = models.CharField(max_length=255)
class Meta:
managed = False
db_table = 'qualification'
admin.py
from django.contrib import admin
from models import Qualification
#admin.register(Qualification)
class QualificationAdmin(admin.ModelAdmin):
list_display = [field.name for field in Qualification._meta.fields if field.name not in ('id', 'qual_key', 'qual_desc')]
list_display.insert(0, '__str__')
here i am showing all fields in list_display excluding 'id', 'qual_key', 'qual_desc' and inserting '__str__' at the beginning.
This answer is helpful when you have large number of modal fields, though i suggest write all fields one by one for better functionality.
list_display = [field.name for field in Book._meta.get_fields()]
This should work even with python 3.9
happy coding
I needed to show all fields for multiple models, I did using the following style:
#admin.register(
AnalyticsData,
TechnologyData,
TradingData,
)
class CommonAdmin(admin.ModelAdmin):
search_fields = ()
def get_list_display(self, request):
return [
field.name
for field in self.model._meta.concrete_fields
if field.name != "id" and not field.many_to_many
]
But you can also do this by creating a mixin, if your models have different fields.