Let's say there is ndb.Model that looks like this:
class Foo(ndb.Model):
bar = ndb.StringProperty()
My question is, if my only input is the Foo.query() how can I get the model as an object that this query belongs to?
def query_to_model(query):
# some magic
return model
The Foo.query().kind return the model's name as a string, but I didn't manage to find a way to get it as an object.
The following works using eval, but only when the model is defined in the same file:
def query_to_model(query):
return eval(query.kind)
I want something more general than that.
After you have imported code with this model definition, the list ndb.Model._kind_map should contain it. Here is the magic:
def query_to_model(query):
return ndb.Model._kind_map[query.name]
I use this code to find the model class if you have the kind name:
model_module = KIND_MODULES(kind_name)
mod = __import__(model_module, globals(), locals(), [kind_name], -1)
model_class = getattr(mod, kind_name)
The KIND Modules dict holds the modules to import the models from:
KIND_MODULES = { 'Users' : 'models', 'Comments' : 'models', 'Cities' : 'examples.models' }
Related
I have a Django-tastypie resource that represents a banner and has a field called impression that I increment whenever the banner appears on the site.
class BannerResource(ModelResource):
owner = fields.ForeignKey('advertisment.api.AdvertiserResource', 'owner', full=True)
class Meta:
queryset = Banner.objects.all()
resource_name = 'banner'
authorization = Authorization()
I would like to get the banner that has the minimum impression, in the official documentation there is nothing like
filtering = {'impressions': ('min',)}
I'm using BackboneJS in the front end and I could get all the banners with Backbone collection and do the filtering with JavaScript but I'm looking for a quicker way to do it.
Any ideas?
Thanks
If you'd like to retrieve banners with number of impressions greater than X you need to things. For one you need to define possible filtering operations on your resource like so (given your model has impressions field):
class BannerResource(ModelResource):
owner = fields.ForeignKey('advertisment.api.AdvertiserResource', 'owner', full=True)
class Meta:
queryset = Banner.objects.all()
resource_name = 'banner'
authorization = Authorization()
filtering = { 'impressions' : ALL }
for available options take a look at Tastypie's documentation on filtering.
Then if you made the following request:
GET http://<your_host>/v1/banners?impressions__gte=X
you should get what you need.
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.
In my CakePHP application, I have a model like this:
class Duck extends AppModel {
var $name = 'Duck';
function get_table_name() {
$tbl_name = //compute default table name for this model
}
}
I would like to write the function get_table_name() that outputs the default table name for the model. For the example above, it should output ducks.
EDIT:
Several people have pointed out the use of $this->table.
I did small testing and found out the following:
In the question as I have put above, $this->table indeed contains the table name.
However, actually, my code looked more like this:
class Duck extends Bird {
var $name = 'Duck';
function get_table_name(){
$tbl_name = //comput default table name for this model
}
}
class Bird extends AppModel {
}
In this case $this->table is empty string.
I went with this approach because I wanted to share some code between two of my models. Looks like this is not a good way to share code between models which need some common functionality.
You're looking for the Inflector class.
Inflector::tableize($this->name)
(tableize calls two Inflector methods to generate the table name: underscore() and pluralize())
Edit:
According to the source code, $this->table should contain the name of the table that CakePHP will use for the model, but in my experience this isn't always set. I'm not sure why.
To get the name of the table that the model is currently using, you can use: $this->table. If you don't manually change the model's table conventions, this may be the most useful in the case of CakePHP ever changing its conventions to use table names using something other than Inflector.
CakePHP's Inflector
function get_table_name() {
$tbl_name = Inflector::pluralize($this->name);
}
OR the tableize method
function get_table_name() {
$tbl_name = Inflector::tableize($this->name);
}
Edit
This also addresses the apparent "ghost" issue with $this->table in the Model.
Digging around in the __construct for Model I discovered two things:
Cake uses Inflector::tableize() to get the table name. This alone is enough to warrant using tableize over pluralize. You'll get consistent results.
$this->table is not set by the Model::__construct() unless $this->useTable === false AND $this->table === false.
It appears that if you know you haven't set $this->useTable to false you should be able to use this over $this->table. Admittedly though I only briefly scanned the source and I haven't really dug deep enough to say why $this->table isn't working sometimes.
To get the full table name for a model you have to take the table prefix into account.
$table = empty($this->table) ? Inflector::tableize($this->name) : $this->table;
$fullTableName = $this->tablePrefix . $table;
I used to use inflector to get the table name from model's name
$tableName = Inflector::pluralize(Inflector::underscore($model));
but this is not really universal, using useTable looks better, by default it will contain table's name by convention, and if you have a table that does not match the conventions, then you should manually specify it by useTable. So, in both cases the result will be correct
$this->User->useTable
I just have a hunch about this. But if feels like I'm doing it the wrong way. What I want to do is to have a db.StringProperty() as a unique identifier. I have a simple db.Model, with property name and file. If I add another entry with the same "name" as one already in the db.Model I want to update this.
As of know I look it up with:
template = Templates.all().filter('name = ', name)
Check if it's one entry already:
if template.count() > 0:
Then add it or update it. But from what I've read .count() is every expensive in CPU usage.
Is there away to set the "name" property to be unique and the datastore will automatic update it or another better way to do this?
..fredrik
You can't make a property unique in the App Engine datastore. What you can do instead is to specify a key name for your model, which is guaranteed to be unique - see the docs for details.
I was having the same problem and came up with the following answer as the simplest one :
class Car(db.Model):
name = db.StringProperty(required=True)
def __init__(self,*args, **kwargs):
super(Car, self).__init__(*args, **kwargs)
loadingAnExistingCar = ("key" in kwargs.keys() or "key_name" in kwargs.keys())
if not loadingAnExistingCar:
self.__makeSureTheCarsNameIsUnique(kwargs['name'])
def __makeSureTheCarsNameIsUnique(self, name):
existingCarWithTheSameName = Car.GetByName(name)
if existingCarWithTheSameName:
raise UniqueConstraintValidationException("Car should be unique by name")
#staticmethod
def GetByName(name):
return Car.all().filter("name", name).get()
It's important to not that I first check if we are loading an existing entity first.
For the complete solution : http://nicholaslemay.blogspot.com/2010/07/app-engine-unique-constraint.html
You can just try to get your entity and edit it, and if not found create a new one:
template = Templates.gql('WHERE name = :1', name)
if template is None:
template = Templates()
# do your thing to set the entity's properties
template.put()
That way it will insert a new entry when it wasn't found, and if it was found it will update the existing entry with the changes you made (see documentation here).
An alternative solution is to create a model to store the unique values, and store it transationally using a combination of Model.property_name.value as key. Only if that value is created you save your actual model. This solution is described (with code) here:
http://squeeville.com/2009/01/30/add-a-unique-constraint-to-google-app-engine/
I agree with Nick. But, if you do ever want to check for model/entity existence based on a property, the get() method is handy:
template = Templates.all().filter('name = ', name).get()
if template is None:
# doesn't exist
else:
# exists
I wrote some code to do this. The idea for it is to be pretty easy to use. So you can do this:
if register_property_value('User', 'username', 'sexy_bbw_vixen'):
return 'Successfully registered sexy_bbw_vixen as your username!'
else:
return 'The username sexy_bbw_vixen is already in use.'
This is the code. There are a lot of comments, but its actually only a few lines:
# This entity type is a registry. It doesn't hold any data, but
# each entity is keyed to an Entity_type-Property_name-Property-value
# this allows for a transaction to 'register' a property value. It returns
# 'False' if the property value is already in use, and thus cannot be used
# again. Or 'True' if the property value was not in use and was successfully
# 'registered'
class M_Property_Value_Register(db.Expando):
pass
# This is the transaction. It returns 'False' if the value is already
# in use, or 'True' if the property value was successfully registered.
def _register_property_value_txn(in_key_name):
entity = M_Property_Value_Register.get_by_key_name(in_key_name)
if entity is not None:
return False
entity = M_Property_Value_Register(key_name=in_key_name)
entity.put()
return True
# This is the function that is called by your code, it constructs a key value
# from your Model-Property-Property-value trio and then runs a transaction
# that attempts to register the new property value. It returns 'True' if the
# value was successfully registered. Or 'False' if the value was already in use.
def register_property_value(model_name, property_name, property_value):
key_name = model_name + '_' + property_name + '_' + property_value
return db.run_in_transaction(_register_property_value_txn, key_name )
i'm experimenting with django and the builtin admin interface.
I basically want to have a field that is a drop down in the admin UI. The drop down choices should be all the directories available in a specified directory.
If i define a field like this:
test_folder_list = models.FilePathField(path=/some/file/path)
it shows me all the files in the directory, but not the directories.
Does anyone know how i can display the folders?
also i tried doing
test_folder_list = models.charField(max_length=100, choices=SOME_LIST)
where SOME_LIST is a list i populate using some custom code to read the folders in a directory. This works but it doesn't refresh. i.e. the choice list is limited to a snapshot of whatever was there when running the app for the first time.
thanks in advance.
update:
after some thinking and research i discovered what i want may be to either
1. create my own widget that is based on forms.ChoiceField
or
2. pass my list of folders to the choice list when it is rendered to the client
for 1. i tried a custom widget.
my model looks like
class Test1(models.Model):
test_folder_ddl = models.CharField(max_length=100)
then this is my custom widget:
class FolderListDropDown(forms.Select):
def __init__(self, attrs=None, target_path):
target_folder = '/some/file/path'
dir_contents = os.listdir(target_folder)
directories = []
for item in dir_contents:
if os.path.isdir(''.join((target_folder,item,))):
directories.append((item, item),)
folder_list = tuple(directories)
super(FolderListDropDown, self).__init__(attrs=attrs, choices=folder_list)
then i did this in my modelForm
class test1Form(ModelForm):
test_folder_ddl = forms.CharField(widget=FolderListDropDown())
and it didn't seem to work.What i mean by that is django didn't want to use my widget and instead rendered the default textinput you get when you use a CharField.
for 2. I tried this in my ModelForm
class test1Form(ModelForm):
test_folder_ddl = forms.CharField(widget=FolderListDropDown())
test_folder_ddl.choices = {some list}
I also tried
class test1Form(ModelForm):
test_folder_ddl = forms.ChoiceField(choices={some list})
and it would still render the default char field widget.
Anyone know what i'm doing wrong?
Yay solved. after beating my head all day and going through all sorts of examples by people i got this to work.
basically i had the right idea with #2. The steps are
- Create a ModelForm of our model
- override the default form field user for a models.CharField. i.e. we want to explcitly say use a choiceField.
- Then we have to override how the form is instantiated so that we call the thing we want to use to generate our dynamic list of choices
- then in our ModelAdmin make sure we explicitly tell the admin to use our ModelForm
class Test1(models.Model):
test_folder_ddl = models.CharField(max_length=100)
class Test1Form(ModelForm):
test_folder_ddl = forms.choiceField()
def __init__(self, *args, **kwargs):
super(Test1Form, self).__init__(*args, **kwargs)
self.fields['test_folder_ddl'].choices = utility.get_folder_list()
class Test1Admin(admin.ModelAdmin):
form = Test1Form
I use a generator:
see git://gist.github.com/1118279.git
import pysvn
class SVNChoices(DynamicChoice):
"""
Generate a choice from somes files in a svn repo
""""
SVNPATH = 'http://xxxxx.com/svn/project/trunk/choices/'
def generate(self):
def get_login( realm, username, may_save ):
return True, 'XXX', 'xxxxx', True
client = pysvn.Client()
client.callback_get_login = get_login
return [os.path.basename(sql[0].repos_path) for sql in client.list(self.SVNPATH)[1:]]