In a django model I would like to modify the parameters of a queryset before running it.
I'm creating an object MyModel with :
m = MyModel(name="name", value="value")
m.save()
In my save method I hash name and value before inserting them. This is working fine.
After this I would like to do :
MyModel.objects.get(name="name", value="value")
to know I name and value are already in my database.
I tryed with manager by I don't find a way to get the queryset params
oau
the answer is manager :
class ModelManager(models.Manager):
def get(self, *args, **kwargs):
// modify args here
return super().get(*args, **kwargs)
Related
let's suppose i have two models like below:
class Client(models.Model):
login = models.CharField(max_length=100)
password = models.CharField(max_length=100)
class Users(models.Model):
user_login = models.CharField(max_length=100)
user_pass = models.CharField(max_length=100)
One model [Users] is filled with data, second [Client] is empty. First i can populate Client based on Users. Now all Django data changes will by held only on Client model.
Question:
How to dynamically connect this two models fields? For example if some data is changed on model Client then according data will by changed also on Users model.
Solution suggestion:
I can update or populate Client data and simultaneously update or populate Users data. Is there another solution to do this?
if login is unique then no need to maintain foreign key relations, You can override the save() methods in both models else You have to maintain the foreign key relation and override the save() methods.
With help of this post i manage to link two models with below code:
class Client(models.Model):
login = models.CharField(max_length=100)
password = models.CharField(max_length=100)
# save previous instance
def __init__(self, *args, **kwargs):
super(Invoice, self).__init__(*args, **kwargs)
self.original_login = self.login
# override save method to update another model fields
def save(self, *args, **kwargs):
# import in place to avoid circular imports
from app.models import Users
if self.login != self.original_login:
Users.objects.filter(id=self.id).update(user_login=self.login)
# save changed data on Client model
super(Client, self).save(*args, **kwargs)
The same method goes to the Users model.
Note that this two model are in different apps.
I have an API endpoint that return a list of objects of one of my models.
I have a very trivial serializer that currently returns all the model fields.
What I need now is to add to this serializer another field which is calculated by applying a logic on one of the model field and another data structure that is also calculated dynamicly - but should be calculated only once for each call to the API.
What is the way to supply this structure to the serializer for every API call but not for every object the serializer returns?
My ModelViewSet (without the extra field) is quite trivial:
class MyModelViewset(ModelViewSet):
queryset = SomeModel.objects.all()
serializer_class = SomeModelSerializer
Some suggestions on how you can achieve this:
To calculate something only once and use it in the serializer you should add it to the serializer context.
You can create #property in your SomeModel class and it can be exposed via serializer fields as read only. Django has also #cached_property so that value is only calculated once and persist as long as the instance does
You can use SerializerMethodField field if you want to calculate something (you have access the serializer context in there).
Just a quick example on how to use this.
class MyModelViewset(ModelViewSet):
queryset = SomeModel.objects.all()
serializer_class = SomeModelSerializer
def get_serializer_context(self):
context = super().get_serializer_context()
context['my_value'] = [] # calculate something here, you have access to self.request
return context
class SomeModelSerializer(serializers.ModelSerializer):
my_data = serializers.SerializerMethodField()
class Meta:
model = SomeModel
def get_my_data(self, obj):
# you have access to self.context['my_value']
# you have access to obj.some_property
return my_data
I have a Django app where users submit orders for payment. Clearly, security is important. I want to minimise the amount of code that I have to write, to avoid introducing any security holes, and ease maintenance.
The model is simple:
class Order(models.Model):
user = models.ForeignKey(User)
created = models.DateTimeField()
paid = models.DateTimeField(null=True, blank=True)
items = models.ManyToManyField(Item)
I'm using a CreateView to create instances of Order:
class OrderView(CreateView):
model = Order
form_class = OrderForm
I want to enforce values for certain fields in those instances. For example, I want the instance user field set to the current logged-in user. I don't want any possibility that the user can change the value of this field, so I don't want it to appear in the form at all. Therefore I use a custom ModelForm to remove these fields from the form:
class OrderForm(forms.ModelForm):
class Meta:
model = Order
# For security, we control exactly which fields are placed
# in the form, rather than excluding some:
fields = ('items',)
Now I want the newly created Order instances to have the user field set to the current logged-in user. I can't find any documentation about what is the best way to do this.
(A) I can override the form's save() method to modify the object before saving, but it feels like this code doesn't belong in the form, which doesn't know anything about the user field. I also don't have access to the request here, which I'd need to determine the current user. But it might look like this:
class OrderForm(forms.ModelForm):
def save(self, commit=True):
instance = super(OrderForm, self).save(commit=False)
instance.user = get_request_magic().user
if commit:
instance.save()
return instance
(B) I can override the view's form_valid method to save the object with commit=False, like a class-based version of this question. But I can't call the superclass method directly, because it saves the object with no way to disable commit, so I have to manually skip a generation of form_valid which is nasty. Apart from that complaint, this does look like the best way I've found so far:
class OrderView(CreateView):
def form_valid(self, form):
self.object = form.save(commit=False)
self.object.user = self.request.user
self.object.save()
return super(ModelFormMixin, self).form_valid(form)
(C) I could write a replacement for CreateView that adds a hook to allow objects to be changed before saving them. But that feels like more boilerplate and duplication.
(D) I can't provide an initial value, because there's no form field to put it in, so it will be ignored.
Any other ideas? If (B) the best option, is there any way around the hacky way of manually specifying which superclass' form_valid method I want to call?
Django user Charettes answered the question for me:
You can achieve this by overriding form_valid:
class OrderCreateViewMixin(CreateView):
def form_valid(self, form):
form.instance.user = request.user
return super(OrderCreateViewMixin, self).form_valid(form)
Which pointed me towards the right part of the documentation:
class AuthorCreate(CreateView):
form_class = AuthorForm
model = Author
def form_valid(self, form):
form.instance.created_by = self.request.user
return super(AuthorCreate, self).form_valid(form)
This is definitely the simplest and cleanest answer I've found so far. It doesn't require modifying the form in any way, although it does directly access its instance member which is a bit ugly. However, at least it's officially documented, so it's unlikely to break.
There are probably multiple approaches to this. I would do this:
Create a constructor in your form which takes the request:
def __init__(self, *args, **kwargs):
request = kwargs.pop('request', None)
super(OrderForm, self).__init__(*args, **kwargs)
self.request = request
When creating your form for POST processing, instantiate it as follows:
form = OrderForm(data=request.POST, request=request)
Now, in your save() method, you have access to the user on the request by referencing self.request.user and can set it accordingly on your model.
The way I've gone about handling this situation with CBVs, is to pass in an unsaved instance of the model to the form. This is how I've done it:
class OrderView(CreateView):
def get_form_kwargs(self):
self.object = Order(user=self.request.user)
return super(OrderView, self).get_form_kwargs()
Both CreateView and UpdateView will add instance to the form kwargs, setting it to the value of self.object.
The only other way, besides what you've already mentioned, is to construct your view class from the same elements that CreateView does, and then change the get and post methods to populate self.object there. I've done that when I have needed a lot of create views in my project:
class OrderView(SingleObjectTemplateResponseMixin, ModelFormMixin, ProcessFormView):
template_name_suffix = '_form'
def get(self, request, *args, **kwargs):
self.object = Order(user=request.user)
return super(OrderView, self).get(request, *args, **kwargs)
def post(self, request, *args, **kwargs):
self.object = Order(user=request.user)
return super(OrderView, self).post(request, *args, **kwargs)
Here is a more generalized version to be reused: https://gist.github.com/4439975
I'm experimenting with django-nonrel on appengine and trying to use a djangotoolbox.fields.ListField to implement a many-to-many relation. As I read in the documentation a ListField is something that you can use to make a workaround for djamgo-nonrel not supporting many-to-many relations.
This is an excerpt from my model:
class MyClass(models.Model):
field = ListField(models.ForeignKey(AnotherClass))
So if I am getting this right I am creating a list of foreign keys to another class to show a relationship with multiple instances of another class
With this approach everything works fine ... No Exceptions. I can create `MyClass' objects in code and views. But when I try to use the admin interface I get the following error
No form field implemented for <class 'djangotoolbox.fields.ListField'>
So I though I would try something that I haven't done before. Create my own field. Well actually my own form for editing MyClass instances in the admin interface. Here is what I did:
class MyClassForm(ModelForm):
field = fields.MultipleChoiceField(choices=AnotherClass.objects.all(), widget=FilteredSelectMultiple("verbose_name", is_stacked=False))
class Meta:
model = MyClass
then I pass MyClassForm as the form to use to the admin interface
class MyClassAdmin(admin.ModelAdmin):
form = MyClassForm
admin.site.register(MyClass, MyClassAdmin)
I though that this would work but It doesn't. When I go to the admin interface I get the same error as before. Can anyone tell what I am doing wrong here ... or if you have any other suggestions or success stories of using the ListField, SetField, etc. from djangotoolbox.fields in the admin interface it would be very much appreciated.
OK, here is what I did to get this all working ...
I'll start from the beginning
This is what what my model looked like
class MyClass(models.Model):
field = ListField(models.ForeignKey(AnotherClass))
I wanted to be able to use the admin interface to create/edit instances of this model using a multiple select widget for the list field. Therefore, I created some custom classes as follows
class ModelListField(ListField):
def formfield(self, **kwargs):
return FormListField(**kwargs)
class ListFieldWidget(SelectMultiple):
pass
class FormListField(MultipleChoiceField):
"""
This is a custom form field that can display a ModelListField as a Multiple Select GUI element.
"""
widget = ListFieldWidget
def clean(self, value):
#TODO: clean your data in whatever way is correct in your case and return cleaned data instead of just the value
return value
These classes allow the listfield to be used in the admin. Then I created a form to use in the admin site
class MyClassForm(ModelForm):
def __init__(self, *args, **kwargs):
super(MyClasstForm,self).__init__(*args, **kwargs)
self.fields['field'].widget.choices = [(i.pk, i) for i in AnotherClass.objects.all()]
if self.instance.pk:
self.fields['field'].initial = self.instance.field
class Meta:
model = MyClass
After having done this I created a admin model and registered it with the admin site
class MyClassAdmin(admin.ModelAdmin):
form = MyClassForm
def __init__(self, model, admin_site):
super(MyClassAdmin,self).__init__(model, admin_site)
admin.site.register(MyClass, MyClassAdmin)
This is now working in my code. Keep in mind that this approach might not at all be well suited for google_appengine as I am not very adept at how it works and it might create inefficient queries an such.
As far as I understand, you're trying to have a M2M relationship in django-nonrel, which is not an out-of-the-box functionality. For starters, if you want a quick hack, you can go with this simple class and use a CharField to enter foreign keys manually:
class ListFormField(forms.Field):
""" A form field for being able to display a djangotoolbox.fields.ListField. """
widget = ListWidget
def clean(self, value):
return [v.strip() for v in value.split(',') if len(v.strip()) > 0]
But if you want to have a multiple selection from a list of models normally you'd have to use ModelMultipleChoiceField, which is also not functional in django-nonrel. Here's what I've done to emulate a M2M relationship using a MultipleSelectField:
Let's say you have a M2M relationship between 2 classes, SomeClass and AnotherClass respectively. You want to select the relationship on the form for SomeClass. Also I assume you want to hold the references as a ListField in SomeClass. (Naturally you want to create M2M relationships as they're explained here, to prevent exploding indexes if you're working on App Engine).
So you have your models like:
class SomeClass(models.Model):
another_class_ids = ListField(models.PositiveIntegerField(), null=True, blank=True)
#fields go here
class AnotherClass(models.Model):
#fields go here
And in your form:
class SomeClassForm(forms.ModelForm):
#Empty field, will be populated after form is initialized
#Otherwise selection list is not refreshed after new entities are created.
another_class = forms.MultipleChoiceField(required=False)
def __init__(self, *args, **kwargs):
super(SomeClassForm,self).__init__(*args, **kwargs)
self.fields['another_class'].choices = [(item.pk,item) for item in AnotherClass.objects.all()]
if self.instance.pk: #If class is saved, highlight the instances that are related
self.fields['another_class'].initial = self.instance.another_class_ids
def save(self, *args, **kwargs):
self.instance.another_class_ids = self.cleaned_data['another_class']
return super(SomeClassForm, self).save()
class Meta:
model = SomeClass
Hopefully this should get you going for the start, I implemented this functionality for normal forms, adjust it for admin panel shouldn't be that hard.
This could be unrelated but for the admin interface, be sure you have djangotoolbox listed after django.contrib.admin in the settings.. INSTALLED_APPS
You could avoid a custom form class for such usage by inquiring for the model object
class ModelListField(ListField):
def __init__(self, embedded_model=None, *args, **kwargs):
super(ModelListField, self).__init__(*args, **kwargs)
self._model = embedded_model.embedded_model
def formfield(self, **kwargs):
return FormListField(model=self._model, **kwargs)
class ListFieldWidget(SelectMultiple):
pass
class FormListField(MultipleChoiceField):
widget = ListFieldWidget
def __init__(self, model=None, *args, **kwargs):
self._model = model
super(FormListField, self).__init__(*args, **kwargs)
self.widget.choices = [(unicode(i.pk), i) for i in self._model.objects.all()]
def to_python(self, value):
return [self._model.objects.get(pk=key) for key in value]
def clean(self, value):
return value
setattr allows you to dynamically name attributes in Python classes. I'm trying to do something similar with an App Engine Model:
class MyModel(db.Model):
def __init__(self, *args, **kwargs):
super(MyModel, self).__init__(*args, **kwargs)
# Doesn't fully work
setatr(self, 'prop1', db.ListProperty(db.Key))
setatr(self, 'prop2', db.StringListProperty())
# Works fully
# prop1 = db.ListProperty(db.Key))
# prop2 = db.StringListProperty())
This code compiles, but when I call model.prop1.append(key) later on, I get this error:
AttributeError: 'ListProperty' object has no attribute 'append'
I suspect this is because prop1 is declared in models instead of self.prop1, but I don't fully understand the syntax's significance.
Has anyone accomplished this, or does anyone have any insight into syntactic differences?
I think you're looking for the db.Expando class (instead of db.Model).
This SO question:
Adding a user supplied property (at runtime) to an instance of Expando class in google app engine?
gives an example of what you want to do.
Not sure but would this work maybe:
class MyModel(db.Model):
#classmethod
def new(cls, *args, **kwargs):
setattr(cls, 'props1', db.ListProperty(db.Key))
setattr(cls, 'props2', db.StringListProperty())
mymodel = cls(*args, **kwargs)
delattr(cls, 'props1')
delattr(cls, 'props2')
return mymodel