Context
I have the models AppVersion, App & DeployApp. In the AppVersion model users can upload APK files to the filesystem. I am using a pre_save signal to prevent uploading APK files with the same version_code for a specific App like this:
#receiver(pre_save, sender=AppVersion)
def prevent_duplicate_version_code(sender, instance, **kwargs):
qs = AppVersion.objects.filter(app_uuid=instance.app_uuid, version_code=instance.version_code)
if qs.exists():
raise FileExistsError("Version code has to be unique for a specific app")
This signal does what I want, except it also raises the error when I am trying to create an object in the bridge-table DeployApp.
Models
# models.py
class App(models.Model):
app_uuid = models.UUIDField(primary_key=True, default=uuid.uuid4, editable=False, db_index=True)
app_name = models.CharField(max_length=100)
class AppVersion(models.Model):
app_version_uuid = models.UUIDField(primary_key=True, default=uuid.uuid4, editable=False, db_index=True)
app_uuid = models.ForeignKey(App, on_delete=models.CASCADE, related_name='app_versions')
app_version_name = models.CharField(max_length=100)
version_code = models.IntegerField(blank=True, null=True, editable=False)
source = models.FileField(upload_to=get_app_path, storage=AppVersionSystemStorage())
class DeployApp(models.Model):
deploy_app_uuid = models.UUIDField(primary_key=True, default=uuid.uuid4, editable=False, db_index=True)
app_version = models.ForeignKey(AppVersion, on_delete=models.CASCADE)
device_group = models.ForeignKey(DeviceGroup, on_delete=models.CASCADE)
release_date = UnixDateTimeField()
My guess is that when creating an object of DeployApp the related AppVersion is also saved and thus the pre_save signal is called and raises the Exception.
I also tried to override the save() method for the AppVersion model but the results are the same.
How do I make sure that the Exception only happens upon creating a new AppVersion instance and does not happen when adding or editing a DeployApp instance?
Solved it thanks to Bear Brown his suggestion. I removed the signal and added UniqueConstraint to the AppVersion model like this:
class Meta:
db_table = 'app_version'
constraints = [
models.UniqueConstraint(fields=['app_uuid', 'version_code'], name='unique appversion')
]
Related
I have three models like those:
class Ambulance(models.Model):
ambulance_id = models.UUIDField(primary_key=True, default=uuid.uuid4, editable=False)
ambulance_name = models.CharField(max_length=255)
forms = models.ManyToManyField(Form)
class Form(models.Model):
form_id = models.UUIDField(primary_key=True, default=uuid4, editable=False)
form_title = models.CharField(max_length=255)
class Page(models.Model):
page_id = models.UUIDField(primary_key=True, default=uuid4, editable=False)
page_title = models.CharField(max_length=100, default='Untitled Page')
form = models.ForeignKey(
Form,
related_name='pages',
on_delete=models.CASCADE,
null=True
)
And the serializers as follow:
class AmbulanceSerializer(serializers.ModelSerializer):
forms = FormSerializer(read_only=True, many=True)
class Meta:
model = Ambulance
fields = ['ambulance_id', 'ambulance_name', 'forms']
class FormSerializer(ModelSerializer):
pages = PageSerializer(many=True, read_only=True)
class Meta:
model = Form
fields = ['form_id', 'form_title', 'pages']
class PageSerializer(ModelSerializer):
class Meta:
model = Page
fields = '__all__'
They were working and give me the nested object that I wanted but after adding many-to-many relationships, it gives me an internal server error 500 and it didn't tell me what the problem was, after debugging I found the problem occurs when calling AmbulanceSerializer(ambulance) or FormSerializer(form)
Could someone give me some hint? and what the problem actually!
Your code is wrong:
Can you please check this line?
fields = fields = ['ambulance_id', 'ambulance_name', 'forms']
which is wrong syntax.
You can try with this:
fields = ['ambulance_id', 'ambulance_name', 'forms']
Following your comment, where you state that you are using these serializers for "all types of requests", the answer is simple: remove read_only=True whenever you are using a nested serializer.
The read_only parameter will stop writes i.e. POST, PUT, etc... requests.
I'm making a task tracker webapp (the full source code is also available) and I have a database structure where each task has a title, a description, and some number of instances, that can each be marked incomplete/incomplete:
class Task(models.Model):
title = OneLineTextField()
description = models.TextField(blank=True)
class TaskInstance(models.Model):
task = models.ForeignKey(Task, on_delete=models.CASCADE)
is_complete = models.BooleanField()
The task and the instances can be shared separately, although access to the instance should imply read access to the task. This is intended for classroom situations, where the teacher creates a task and assigns it to their students.
class TaskPermission(models.Model):
task = models.ForeignKey(Task, on_delete=models.CASCADE, related_name='permissions')
user = models.ForeignKey(User, on_delete=models.CASCADE, related_name='task_permissions_granted')
shared_by = models.ForeignKey(User, on_delete=models.PROTECT, null=True, related_name='task_permissions_granting')
can_edit = models.BooleanField(default=False)
class Meta:
unique_together = 'task', 'user', 'shared_by',
class TaskInstancePermission(models.Model):
task_instance = models.ForeignKey(TaskInstance, on_delete=models.CASCADE, related_name='permissions')
user = models.ForeignKey(User, on_delete=models.CASCADE, related_name='task_instance_permissions_granted')
shared_by = models.ForeignKey(User, on_delete=models.PROTECT, null=True, related_name='task_instance_permissions_granting')
can_edit = models.BooleanField(default=False)
class Meta:
unique_together = 'task_instance', 'user', 'shared_by',
My question is how to create a form for TaskInstances with fields for its is_complete, as well as its Task's title and description. Would something like this work? Or would I need to implement my own save and clean methods?
class TaskForm(ModelForm):
class Meta:
model = TaskInstance
fields = ('is_complete', 'task__title', 'task__description')
I think inlineformset_factory is what I'm looking for!
Actually, it does not seem to be useful: it is for multiple forms of the same type, not different types...
I want to change the model used for the default LogEntry class so it creates a varchar rather than a clob in the database for the attribute "object_id"
The original model is defined in
django/contrib/admin/models.py
class LogEntry(models.Model):
action_time = models.DateTimeField(_('action time'), auto_now=True)
user = models.ForeignKey(settings.AUTH_USER_MODEL)
content_type = models.ForeignKey(ContentType, blank=True, null=True)
object_id = models.TextField(_('object id'), blank=True, null=True)
object_repr = models.CharField(_('object repr'), max_length=200)
action_flag = models.PositiveSmallIntegerField(_('action flag'))
change_message = models.TextField(_('change message'), blank=True)
I want to change the definition of object_id to
object_id = models.Charfield(_('object id'), max_length=1000, blank=True, null=True)
I'm aware that this could cause issues if someone defined a primary key on an entity which is larger than a varchar(1000), but I wouldn't want an entity with a PK defined like that so I'm happy with the limitation.
This will greatly improve the efficiency of the queries when accessing the history log.
I don't really want to hack the actual model definition, but I can't find out how to elegantly override the model definition.
Any ideas?
Django’s model fields provide an undocumented contribute_to_class method.
The other feature of Django we can use is the class_prepared signal.
from django.db.models import CharField
from django.db.models.signals import class_prepared
def add_field(sender, **kwargs):
"""
class_prepared signal handler that checks for the model named
MyModel as the sender, and adds a CharField
to it.
"""
if sender.__name__ == "MyModel":
field = CharField("New field", max_length=100)
field.contribute_to_class(sender, "new_field")
class_prepared.connect(add_field)
To override field you can simply delete original field from model:
from django.db.models import CharField
from django.db.models.signals import class_prepared
def override_field(sender, **kwargs):
if sender.__name__ == "LogEntry":
field = CharField('object id', max_length=1000, blank=True, null=True)
sender._meta.local_fields = [f for f in sender._meta.fields if f.name != "object_id"]
field.contribute_to_class(sender, "object_id")
class_prepared.connect(override_field)
I have just tested this solution by placing this code in __init__.py of my app. You will also need to write a custom migration:
from django.db import migrations, models
class Migration(migrations.Migration):
def __init__(self, name, app_label):
# overriding application operated upon
super(Migration, self).__init__(name, 'admin')
dependencies = [
('my_app', '0001_initial'),
]
operations = [
migrations.AlterField(
model_name='logentry',
name='object_id',
field=models.CharField('object id', max_length=1000, blank=True, null=True),
),
]
Looks like it works but use it on your own risk.
You can read more here.
I have created a app using the following Model
models.py
class Vendor(models.Model):
name = models.CharField(max_length=100, unique=True, blank=False)
def __str__(self):
return self.name
class Model(models.Model):
name = models.CharField(max_length=100, unique=True, blank=False)
def __str__(self):
return self.name
class Request(models.Model):
job_reference = models.CharField(max_length=100, unique=True, blank=False)
def __str__(self):
return self.job_reference
class Device(models.Model):
Vendor = models.ForeignKey('Vendor')
Model = models.ForeignKey('Model')
device_id = models.CharField(max_length=255, unique=True, blank=True)
is_encrypted = models.BooleanField()
is_medical = models.BooleanField()
request_job_reference = models.ForeignKey('Request')
submitted = models.DateTimeField(default=timezone.now)
When i go to the admin page I can add new devices which displayed each of the fields and the "Vendor" and "Model" allows me to either select an existing entry or has a plus icon to add a new entry (which is great)
Django_Admin_form
When i create a form for my app
forms.py
from django import forms
from . models import Device
class AddDevice(forms.ModelForm):
class Meta:
model = Device
fields = ('Vendor', 'Model', 'device_id', 'is_encrypted', 'is_medical', 'submitted')
The form on my webpage display ok however there is no option to insert a new entry to "Vendor" or "Model".
Webpage Form
I have looked on other posts on here as users have had the same issue and it's been suggested to use "ModelChoiceField" but unfortunately it still doesn't make any sense to me. Either i'm completely missing something or I have setup my models in a way which is making things harder for myself.
Can anyone explain how I can go about doing this?
I have an abstract model that all my other models inherit from, it looks like this.
class SupremeModel(models.Model):
creator = models.ForeignKey(User, related_name="%(class)s_creator")
created = models.DateTimeField(null=True, blank=True)
deleted = models.BooleanField(default=False)
modified = models.DateTimeField(null=True,blank=True)
class Meta:
abstract = True
I then have a bunch of other models that inherit from this model, with something along these lines...
class ExampleModel(SupremeModel):
name = models.TextField(null=False, blank=False)
description = models.TextField(null=False, blank=False)
class AnotherModel(SupremeModel):
title = models.TextField(null=False, blank=False)
location = models.TextField(null=False, blank=False)
I want to create a Django model form for nearly all of my custom models that look similar to ExampleModel, but I always want the fields in SupremeModel to be excluded in the form...
How can I create a ModelForm that can be used to inherit the exclude parameters that will hide creator,created,deleted, and modified but show all of the other fields (in this case name and description or title and location).
you may try this
class ExcludedModelForm(ModelForm):
class Meta:
exclude = ['creator', 'created', 'deleted', 'modified']
class ExampleModelForm(ExcludedModelForm):
class Meta(ExcludedModelForm.Meta):
model = ExampleModel