Is there a sane way to simulate virtual inheritance in Django models? - python

I want to log actions made by users. In most OO languages, I would implement this via a LoggedAction class, having several child classes like LoginActionand LogoutAction. I could then iterate over a list of LoggedActions and get the specific child behaviour through virtual inheritance. This does not work using Django models however.
Example models.py:
class LoggedAction(models.Model):
user = models.ForeignKey(User)
timestamp = models.DateTimeField(auto_now_add=True)
def __unicode__(self):
return "%s: %s %s" % (unicode(self.timestamp), unicode(self.user), unicode(self.action()))
def action(self):
return ""
class LoginAction(LoggedAction):
def action(self):
return "logged in"
class LogoutAction(LoggedAction):
def action(self):
return "logged out"
Then I'd like to do [unicode(l) for l in LoggedAction.objects.all()] and get a list of messages like u'2012-02-18 18:47:09.105840: knatten logged in'.
As expected, this does not work, since what I get from all() is a list of LoggedAction objects having either a loginaction member or a logoutaction member. (The output is a list of messages like u'2012-02-18 18:47:09.105840: knatten, with no mention of the action.)
Is there a sane way to get the behaviour I'm after, or am I trying to apply the wrong paradigm here? (I guess I am, and that I should just have the specific action as a member in LoggedAction)

Yes, this is probably the wrong paradigm. It's easy to be misled by the object-relational mapper (ORM) - database tables don't really map all that well to objects, and this difference is known as the object-relational impedance mismatch.
What you actually need is to make action a field. This field can take a choices parameter which represents the possible values of that field - ie logged in or logged out:
class LoggedAction(models.Model):
ACTIONS = (
('I', 'logged in'),
('O', 'logged out')
)
user = models.ForeignKey(User)
timestamp = models.DateTimeField(auto_now_add=True)
action = models.CharField(max_length=1, choices=ACTIONS)
def __unicode__(self):
return u"%s: %s %s" % (self.timestamp, self.user, self.get_action_display())
Note that I've used arbitrary single-character strings to represent the actions, and the get_action_display() magic method to get the full description.

Have a look at InheritanceManager from django-model-utils. It allows you to get the concrete subclasses.

Related

How to have 2 Types with the same model in Graphene?

I created a model Checkout on my project, with a CheckoutType to handle the requests, but now i need a Profile, that is basically just getting many of the fields on Checkout. The problem is that Checkout and Profile will be retrieved by users with very different permissions, and the while the first one will have the right ones, the second one must not have them. so i went with creating 2 types:
Checkout:
class CheckoutType(ModelType):
class Meta:
model = Checkout
interfaces = [graphene.relay.Node]
connection_class = CountableConnection
permissions = ['app.view_checkout']
filter_fields = {
'zone': ['exact'],
'vehicle__mark': ['exact'],
'status': ['exact']
}
Profile:
class ProfileFilter(django_filters.FilterSet):
class Meta:
model = Checkout
fields = ['zone','status']
#property
def qs(self):
# The query context can be found in self.request.
return super(ProfileFilter, self).qs.filter(salesman=self.request.user)
class ProfileType(ModelType):
class Meta:
model = Checkout
interfaces = [graphene.relay.Node]
connection_class = CountableConnection
filterset_class = ProfileFilter
The thing here is that, the first one shouldn't filter, and just be a regular schema, while the second one should filter by the user that made the request, that and the permissions is the reason i use 2, but as soon as i implemented, all the tests i did for the Checkout Type started to fail, since it seems it tries to use the ProfileType. I searched a little, and it seems that relay only allows a type per model in Django, so this approach doesn't seems possible, but i'm not sure how to overwrite the CheckoutType on another schema, or how to make a second Type with different permissions and different filters. Does someone knows if this is possible?
Just in case someone is on the same boat, i think i found a way to make it work, but with a different approach, i just modified the CheckoutType a little:
class CheckoutType(ModelType):
# Meta
#classmethod
def get_queryset(cls, queryset, info):
if info.context.user.has_perm('app.view_checkout'):
return queryset
return queryset.filter(salesman=info.context.user)
Basically here i remove the permission from the Meta, since i don't want to check that there, and then i overwrite the get_queryset() to check if the user has the perms, if that's the case, then just return the normal query, but if not just filter(And any additional thing you want to do for people without the permission). I'm not sure if there's a better way, but definitely did the job.

Django save behaving randomly

I have a Story model with a M2M relationship to some Resource objects. Some of the Resource objects are missing a name so I want to copy the title of the Story to the assigned Resource objects.
Here is my code:
from collector import models
from django.core.paginator import Paginator
paginator = Paginator(models.Story.objects.all(), 1000)
def fix_issues():
for page in range(1, paginator.num_pages + 1):
for story in paginator.page(page).object_list:
name_story = story.title
for r in story.resources.select_subclasses():
if r.name != name_story:
r.name = name_story
r.save()
if len(r.name) == 0:
print("Something went wrong: " + name_story)
print("done processing page %s out of %s" % (page, paginator.num_pages))
fix_issues()
I need to use a paginator because I'm dealing with a million objects. The weird part is that after calling fix_issues() about half of my resources that had no name, now have the correct name, while the other half still has no name. I can call fix_issues() again and again and every time more objects receive a name. This seems really weird to me, why would an object not be updated the first time but only the second time?
Additional information:
The "Something went wrong: " message is never printed.
I'm using select_subclasses from django-model-utils to iterate over all resources (any type).
The story.title is never empty.
No error message is printed, when I run these commands.
I did not override the save method of the Resource model (only the save method of the Story model).
I tried to use #transaction.atomic but the result was the same.
My Model:
class Resource(models.Model):
name = models.CharField(max_length=200)
# Important for retrieving the correct subtype.
objects = InheritanceManager()
def __str__(self):
return str(self.name)
class CustomResource(Resource):
homepage = models.CharField(max_length=3000, default="", blank=True, null=True)
class Story(models.Model):
url = models.URLField(max_length=3000)
resources = models.ManyToManyField(Resource)
popularity = models.FloatField()
def _update_popularity(self):
self.popularity = 3
def save(self, *args, **kwargs):
super(Story, self).save(*args, **kwargs)
self._update_popularity()
super(Story, self).save(*args, **kwargs)
Documentation for the select_subclasses:
http://django-model-utils.readthedocs.io/en/latest/managers.html#inheritancemanager
Further investigation:
I thought that maybe select_subclasses did not return all the objects. Right now every story has exactly one resource. So it was easy enough to check that select_subclasses always returns one item. This is the function I used:
def find_issues():
for page in range(1, paginator.num_pages + 1):
for story in paginator.page(page).object_list:
assert(len(story.resources.select_subclasses()) == 1)
print("done processing page %s out of %s" % (page, paginator.num_pages))
But again, this executes without any problems. So I don't thing the select_subclasses is to blame. I also checked if paginator.num_pages is right and it is. If i divide by 1000 (items per page) I get exactly the number of stories I have in my database.
I think I know what is happening:
The Paginator loads a Queryset and gives me the first n items. I process these and update some of the values. But for the next iteration the order of the items in the queryset changes (because I updated some of them and did not define an order). So I'm skipping over items that are now on the first page. I can avoid it by specifying an order (pk for example).
If you think I'm wrong, please let me know. Otherwise I will accept this as the correct answer. Thank you.

Checking uniqueness contraint during form validation in App Engine

I am using Flask and WTforms in App Engine, trying to implement uniqueness contraint on one of the field. The question is big, please be patient and I have been stuck here from many hours, need some help from you people. Started learning App Engine, Flask and WTForms a month ago. Thanks in advance.
Application has model 'Team' as shown below:
class Team(db.Model):
name = db.StringProperty(required=True)
-- some other fields here --
Requirement: Name of the team has to be unique.
I have followed the links
http://www.codigomanso.com/en/2010/09/solved-anadir-claves-unicas-en-google-app-engine-en-3-lineas/
http://squeeville.com/2009/01/30/add-a-unique-constraint-to-google-app-engine/
http://csimms.botonomy.com/2012/07/there-are-only-two-ways-to-enforce-unique-constraints-in-google-app-engine.html
Have come up with the following code:
models.py: Created a separate table 'Unique' as given in the link:
class Unique(db.Model):
""" Handles uniqueness constriant on a field """
#classmethod
def unique_check(cls, form_name, field_data):
def tx(form_name, field_data):
key_name = "%s%s" % (form_name, field_data)
uk = Unique.get_by_key_name(key_name)
app.logger.debug("UK:" + str(uk))
if uk:
return False
uk = Unique(key_name=key_name)
uk.put()
return True
ret_val = db.run_in_transaction(tx, form_name, field_data)
app.logger.debug("ret_val:" + str(ret_val))
return ret_val
forms.py: I have overridden the __init__() and validate_on_submit() function in which uniqueness is checked and if it is not unique, error is attached to that field and validation error will be raised in the same way as wtforms's validators.
class TeamForm(wtf.Form):
def __init__(self, *args, **kwargs):
super(TeamForm, self).__init__(*args, **kwargs)
if kwargs.get('edit', None):
self.old_name = self.name.data.lower()
def validate_on_submit(self, edit=False):
if not super(TeamForm, self).validate_on_submit():
return False
if edit:
if self.old_name and self.old_name != self.name.data.lower():
Unique.delete_entity(self.__class__.__name__, self.old_name)
if not Unique.unique_check(self.__class__.__name__, self.name.data.lower()):
self.name.errors.append("Value '%s' is not unique" % self.name.data)
return False
else:
if not Unique.unique_check(self.__class__.__name__, self.name.data.lower()):
self.name.errors.append("Value '%s' is not unique" % self.name.data)
return False
return True
**---- Form fields declaration ----**
The above code works when new team is inserted.I mean it checks uniqueness properly. The problem occurs, when user edits the team information. Following two scenarios are problematic:
When the user tries to submit the form, application will throw "Not unique" error, it is obvious because "Unique" table has "key_name" for this team.
If user changes "team name", application has to delete the previous team name from the "Unique" table and has to check uniqueness for the "changed team name". I am not able to handle these two scenarios.
My edit_team function looks like this:
#app.route('/team/edit/<key>', methods=['GET','POST'])
#login_required
def edit_team(key):
k = db.Key(key)
team = db.get(k)
form = TeamForm(obj = team, edit=True) # to save old name, doesn't work.
if form.validate_on_submit(edit=True): # edit=True is given only in edit function
team.name = form.name.data
-- others fields are updated in the similar way --
team.put()
return redirect(url_for('teams_list'))
return render_template('edit_team.html', form=form)
Problem can be easily solved if I am able to find out 'old name' of the team, so that I can delete it from the "Unique" table. As you can see I am saving old name of the team in TeamForm __init__() function, but __init__() is called during GET(old name is saved) and also in POST(modified name will get saved!!). So, I cannot find out old name at all and it remains in the "Unique" table, nobody can use this "old team name" anymore.
I tried to explain as much as possible, please let me know if you want more info.
Edit: didn't answer your question properly the first time.
Separate instances of the Form object will be instantiated for the GET and POST requests, so you can't save the old_name to self.
You'll need to pass the old_name to the browser in the form, and have the browser submit the old_name back in the POST request.
The easyish way to do this is to create a hidden form field that the user doesn't see, but will get submitted by the POST request. I'm not too familiar with WTForms but I assume you can initialize the old_name field value in your the GET request handler.

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,...)

Django admin list_display weirdly slow with foreign keys

Django 1.2.5
Python: 2.5.5
My admin list of a sports model has just gone really slow (5 minutes for 400 records). It was returning in a second or so until we got 400 games, 50 odd teams and 2 sports.
I have fixed it in an awful way so I'd like to see if anyone has seen this before. My app looks like this:
models:
Sport( models.Model )
name
Venue( models.Model )
name
Team( models.Model )
name
Fixture( models.Model )
date
sport = models.ForeignKey(Sport)
venue = models.ForeignKey(Venue)
TeamFixture( Fixture )
team1 = models.ForeignKey(Team, related_name="Team 1")
team2 = models.ForeignKey(Team, related_name="Team 2")
admin:
TeamFixture_ModelAdmin (ModelAdmin)
list_display = ('date','sport','venue','team1','team2',)
If I remove any foreign keys from list_display then it's quick. As soon as I add any foreign key then slow.
I fixed it by using non foreign keys but calculating them in the model init so this works:
models:
TeamFixture( Fixture )
team1 = models.ForeignKey(Team, related_name="Team 1")
team2 = models.ForeignKey(Team, related_name="Team 2")
sport_name = ""
venue_name = ""
team1_name = ""
team2_name = ""
def __init__(self, *args, **kwargs):
super(TeamFixture, self).__init__(*args, **kwargs)
self.sport_name = self.sport.name
self.venue_name = self.venue.name
self.team1_name = self.team1.name
self.team2_name = self.team2.name
admin:
TeamFixture_ModelAdmin (ModelAdmin)
list_display = ('date','sport_name','venue_name','team1_name','team2_name',)
Administration for all other models are fine with several thousand records at the moment and all views in the actual site is functioning fine.
It's driving me crazy. list_select_related is set to True, however adding a foreign key to User in the list_display generates one query per row in the admin, which makes the listing slow. Select_related is True, so the Django admin shouldn't call this query on each row.
What is going on ?
The first thing I would look for, are the database calls. If you shouldn't have done that already, install django-debug-toolbar. That awesome tool lets you inspect all sql queries done for the current request. I assume there are lots of them. If you look at them, you will know where to look for the problem.
One problem I myself have run into: When the __unicode__ method of a model uses a foreign key, that leads to one database hit per instance. I know of two ways to overcome this problem:
use select_related, which usually is your best bet.
make your __unicode__ return a static string and override the save method to update this string accordingly.
This is a very old problem with django admin and foreign keys. What happens here is that whenever you try to load an object it tries to get all the objects of that foreign key. So lets say you are trying to load a fixture with a some teams (say the number of teams is about 100), its going to keep on including all the 100 teams in one go. You can try to optimize them by using something called as raw_fields. What this would do is instead of having to calling everything at once, it will limit the number of calls and make sure that the call is only made when an event is triggered (i.e. when you are selecting a team).
If that seems a bit like a UI mess you can try using this class:
"""
For Raw_id_field to optimize django performance for many to many fields
"""
class RawIdWidget(ManyToManyRawIdWidget):
def label_for_value(self, value):
values = value.split(',')
str_values = []
key = self.rel.get_related_field().name
for v in values:
try:
obj = self.rel.to._default_manager.using(self.db).get(**{key: v})
x = smart_unicode(obj)
change_url = reverse(
"admin:%s_%s_change" % (obj._meta.app_label, obj._meta.object_name.lower()),
args=(obj.pk,)
)
str_values += ['<strong>%s</strong>' % (change_url, escape(x))]
except self.rel.to.DoesNotExist:
str_values += [u'No input or index in the db']
return u', '.join(str_values)
class ImproveRawId(admin.ModelAdmin):
raw_id_fields = ('created_by', 'updated_by')
def formfield_for_dbfield(self, db_field, **kwargs):
if db_field.name in self.raw_id_fields:
kwargs.pop("request", None)
type = db_field.rel.__class__.__name__
kwargs['widget'] = RawIdWidget(db_field.rel, site)
return db_field.formfield(**kwargs)
return super(ImproveRawId, self).formfield_for_dbfield(db_field, **kwargs)
Just make sure that you inherit the class properly. I am guessing something like TeamFixture_ModelAdmin (ImproveRawIdFieldsForm). This will most likely give you a pretty cool performance boost in your django admin.
I fixed my problem by setting list_select_related to the list of related model fields instead of just True

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