New to django. I'm doing my best to implement CRUD using Django, mongodb, and mongoengine. I'm able to query the database and render my page with the correct information from the database. I'm also able to change some document fields using javascript and do an Ajax POST back to the original Django View class with the correct csrf token.
The data payload I'm sending back and forth is a list of each Document Model (VirtualPageModel) serialized to json (each element contains ObjectId string along with the other specific fields from the Model.)
This is where it starts getting murky. In order to update the original document in my View Class post function I do an additional query using the object id and loop through the dictionary items, setting the respective fields each time. I then call save and any new data is pushed to the Mongo collection correctly.
I'm not sure if what I'm doing to update existing documents is correct or in the spirit of django's abstracted database operations. The deeper I get the more I feel like I'm not using some fundamental facility earlier on (provided by either django or mongoengine) and because of this I'm having to make things up further downstream.
The way my code is now I would not be able to create a new document (although that's easy enough to fix). However what I'm really curious about is how I would know when to delete a document which existed in the initial query, but was removed by the user/javascript code? Am I overthinking things and the contents of my POST should contain a list of ObjectIds to delete (sounds like a security risk although this would be an internal tool.)
I was assuming that my View Class might maintain either the original document objects (or simply ObjectIds) it queried and I could do my comparisions off of that set, but I can't seem to get that information to persist (as a class variable in VolumeSplitterView) from its inception to when I received the POST at the end.
I would appreciate if anyone could take a look at my code. It really seems like the "ease of use" facilities of Django start to break when paired with Mongo and/or a sufficiently complex Model schema which needs to be directly available to javascript as opposed to simple Forms.
I was going to use this dev work to become django battle-hardened in order to tackle a future app which will be much more complicated and important. I can hack on this thing all day and make it functional, but what I'm really interested in is anyone's experience in using Django + MongoDB + MongoEngine to implement CRUD on a Database Schema which is not vary Form-centric (think more nested metadata).
Thanks.
model.py: uses mongoengine Field types.
class MongoEncoder(JSONEncoder):
def default(self, o):
if isinstance(o, VirtualPageModel):
data_dict = (o.to_mongo()).to_dict()
if isinstance(data_dict.get('_id'), ObjectId):
data_dict.update({'_id': str(data_dict.get('_id'))})
return data_dict
else:
return JSONEncoder.default(self, o)
class SubTypeModel(EmbeddedDocument):
filename = StringField(max_length=200, required=True)
page_num = IntField(required=True)
class VirtualPageModel(Document):
volume = StringField(max_length=200, required=True)
start_physical_page_num = IntField()
physical_pages = ListField(EmbeddedDocumentField(SubTypeModel),
default=list)
error_msg = ListField(StringField(),
default=list)
def save(self, *args, **kwargs):
print('In save: {}'.format(kwargs))
for k, v in kwargs.items():
if k == 'physical_pages':
self.physical_pages = []
for a_page in v:
tmp_pp = SubTypeModel()
for p_k, p_v in a_page.items():
setattr(tmp_pp, p_k, p_v)
self.physical_pages.append(tmp_pp)
else:
setattr(self, k, v)
return super(VirtualPageModel, self).save(*args, **kwargs)
views.py: My attempt at a view
class VolumeSplitterView(View):
#initial = {'key': 'value'}
template_name = 'click_model/index.html'
vol = None
start = 0
end = 20
def get(self, request, *args, **kwargs):
self.vol = self.kwargs.get('vol', None)
records = self.get_records()
records = records[self.start:self.end]
vp_json_list = []
img_filepaths = []
for vp in records:
vp_json = json.dumps(vp, cls=MongoEncoder)
vp_json_list.append(vp_json)
for pp in vp.physical_pages:
filepath = get_file_path(vp, pp.filename)
img_filepaths.append(filepath)
data_dict = {
'img_filepaths': img_filepaths,
'vp_json_list': vp_json_list
}
return render_to_response(self.template_name,
{'data_dict': data_dict},
RequestContext(request))
def get_records(self):
return VirtualPageModel.objects(volume=self.vol)
def post(self, request, *args, **kwargs):
if request.is_ajax:
vp_dict_list = json.loads(request.POST.get('data', []))
for vp_dict in vp_dict_list:
o_id = vp_dict.pop('_id')
original_doc = VirtualPageModel.objects.get(id=o_id)
try:
original_doc.save(**vp_dict)
except Exception:
print(traceback.format_exc())
I am having a difficult time getting autocomplete to work with haystack and Solr in a search form. Following the instructions here Auto-complete i was able to create my index in the following way.
class PersonIndex(indexes.RealTimeSearchIndex, indexes.Indexable):
text = CharField(document=True, use_template=True)
first_name = CharField(model_attr='first_name')
last_name = CharField(model_attr='last_name')
first_name_auto = indexes.EdgeNgramField(model_attr='first_name')
def index_queryset(self):
"""Used when the entire index for model is updated."""
return self.get_model().objects.all().order_by('first_name')
def get_model(self):
return Person
And with the way my URL route is set up, i dont have a view request that get directed to, the search method works.
url(r'^search/person/', search_view_factory(
view_class=SearchView,
template='index.html',
form_class=ModelSearchForm
), name='haystack_search'),
The instructions say that we can perform the query in this fashion
from haystack.query import SearchQuerySet
sqs = SearchQuerySet().filter(content_auto=request.GET.get('q', ''))
but where do we put this SearchQuerySet, i am not sure what to override, how to modify my url to route correctly. My search currently works out of the box this way but i want to try auto complete with EdgeNgramField ?
You'll need to define your own custom search form and tell it how to generate the SearchQuerySet it returns to the view, and then tell your search_view_factory to use that form instead of the ModelSearchForm.
Specify the way you want to generate the SearchQuerySet used by your view by overriding the ModelSearchForm search method:
from haystack.forms import ModelSearchForm
class AutocompleteModelSearchForm(ModelSearchForm):
def search(self):
if not self.is_valid():
return self.no_query_found()
if not self.cleaned_data.get('q')
return self.no_query_found()
sqs = self.searchqueryset.filter(first_name_auto=self.cleaned_data['q'])
if self.load_all
sqs = sqs.load_all()
return sqs
This will now perform a filter on the form's SearchQuerySet on the first_name_auto field rather than the auto_query that it would usually do on the text field (see haystack/forms.py to see what the original search function looks like).
You specify that you want to use this form in the argument list to your search_view_factory
from path.to.your.forms import AutocompleteModelSearchForm
url(r'^search/person/', search_view_factory(
view_class=SearchView,
template='index.html',
form_class=AutocompleteModelSearchForm
), name='haystack_search'),
I am writing an application in Django, which uses [year]/[month]/[title-text] in the url to identitfy news items. To manage the items I have defined a number of urls, each starting with the above prefix.
urlpatterns = patterns('msite.views',
(r'^(?P<year>[\d]{4})/(?P<month>[\d]{1,2})/(?P<slug>[\w]+)/edit/$', 'edit'),
(r'^(?P<year>[\d]{4})/(?P<month>[\d]{1,2})/(?P<slug>[\w]+)/$', 'show'),
(r'^(?P<year>[\d]{4})/(?P<month>[\d]{1,2})/(?P<slug>[\w]+)/save$', 'save'),
)
I was wondering, if there is a mechanism in Django, which allows me to preprocess a given request to the views edit, show and save. It could parse the parameters e.g. year=2010, month=11, slug='this-is-a-title' and extract a model object out of them.
The benefit would be, that I could define my views as
def show(news_item):
'''does some stuff with the news item, doesn't have to care
about how to extract the item from request data'''
...
instead of
def show(year, month, slug):
'''extract the model instance manually inside this method'''
...
What is the Django way of solving this?
Or in a more generic way, is there some mechanism to implement request filters / preprocessors such as in JavaEE and Ruby on Rails?
You need date based generic views and create/update/delete generic views maybe?
One way of doing this is to write a custom decorator. I tested this in one of my projects and it worked.
First, a custom decorator. This one will have to accept other arguments beside the function, so we declare another decorator to make it so.
decorator_with_arguments = lambda decorator: lambda * args, **kwargs: lambda func: decorator(func, *args, **kwargs)
Now the actual decorator:
#decorator_with_arguments
def parse_args_and_create_instance(function, klass, attr_names):
def _function(request, *args, **kwargs):
model_attributes_and_values = dict()
for name in attr_names:
value = kwargs.get(name, None)
if value: model_attributes_and_values[name] = value
model_instance = klass.objects.get(**model_attributes_and_values)
return function(model_instance)
return _function
This decorator expects two additional arguments besides the function it is decorating. These are respectively the model class for which the instance is to be prepared and injected and the names of the attributes to be used to prepare the instance. In this case the decorator uses the attributes to get the instance from the database.
And now, a "generic" view making use of a show function.
def show(model_instance):
return HttpResponse(model_instance.some_attribute)
show_order = parse_args_and_create_instance(Order, ['order_id'])(show)
And another:
show_customer = parse_args_and_create_instance(Customer, ['id'])(show)
In order for this to work the URL configuration parameters must contain the same key words as the attributes. Of course you can customize this by tweaking the decorator.
# urls.py
...
url(r'^order/(?P<order_id>\d+)/$', 'show_order', {}, name = 'show_order'),
url(r'^customer/(?P<id>\d+)/$', 'show_customer', {}, name = 'show_customer'),
...
Update
As #rebus correctly pointed out you also need to investigate Django's generic views.
Django is python after all, so you can easily do this:
def get_item(*args, **kwargs):
year = kwargs['year']
month = kwargs['month']
slug = kwargs['slug']
# return item based on year, month, slug...
def show(request, *args, **kwargs):
item = get_item(request, *args, **kwargs)
# rest of your logic using item
# return HttpResponse...
I'm currently working with django generic views and I have a problem I can't figure out.
When using delete_object I get a TypeError exception:
delete_object() takes at least 3 non-keyword arguments (2 given)
Here is the code (I have ommited docstrings and imports):
views.py
def delete_issue(request, issue_id):
return delete_object(request,
model = Issue,
object_id = issue_id,
template_name = 'issues/delete.html',
template_object_name = 'issue')
urls.py
urlpatterns = patterns('issues.views',
(r'(?P<issue_id>\d+)/delete/$', 'delete_issue'),
)
The other generic views (object_list, create_object, etc.) work fine with those parameters. Another problem I have is when using the create_object() function, it says something about a CSRF mechanism, what is that?
You need to provide post_delete_redirect, this means url, where user should be redirected after object is deleted. You can find this in view signature:
def delete_object(request, model, post_delete_redirect, object_id=None,
slug=None, slug_field='slug', template_name=None,
template_loader=loader, extra_context=None, login_required=False,
context_processors=None, template_object_name='object'):
There is a lot of documentation on how to serialize a Model QuerySet but how do you just serialize to JSON the fields of a Model Instance?
You can easily use a list to wrap the required object and that's all what django serializers need to correctly serialize it, eg.:
from django.core import serializers
# assuming obj is a model instance
serialized_obj = serializers.serialize('json', [ obj, ])
If you're dealing with a list of model instances the best you can do is using serializers.serialize(), it gonna fit your need perfectly.
However, you are to face an issue with trying to serialize a single object, not a list of objects. That way, in order to get rid of different hacks, just use Django's model_to_dict (if I'm not mistaken, serializers.serialize() relies on it, too):
from django.forms.models import model_to_dict
# assuming obj is your model instance
dict_obj = model_to_dict( obj )
You now just need one straight json.dumps call to serialize it to json:
import json
serialized = json.dumps(dict_obj)
That's it! :)
To avoid the array wrapper, remove it before you return the response:
import json
from django.core import serializers
def getObject(request, id):
obj = MyModel.objects.get(pk=id)
data = serializers.serialize('json', [obj,])
struct = json.loads(data)
data = json.dumps(struct[0])
return HttpResponse(data, mimetype='application/json')
I found this interesting post on the subject too:
http://timsaylor.com/convert-django-model-instances-to-dictionaries
It uses django.forms.models.model_to_dict, which looks like the perfect tool for the job.
There is a good answer for this and I'm surprised it hasn't been mentioned. With a few lines you can handle dates, models, and everything else.
Make a custom encoder that can handle models:
from django.forms import model_to_dict
from django.core.serializers.json import DjangoJSONEncoder
from django.db.models import Model
class ExtendedEncoder(DjangoJSONEncoder):
def default(self, o):
if isinstance(o, Model):
return model_to_dict(o)
return super().default(o)
Now use it when you use json.dumps
json.dumps(data, cls=ExtendedEncoder)
Now models, dates and everything can be serialized and it doesn't have to be in an array or serialized and unserialized. Anything you have that is custom can just be added to the default method.
You can even use Django's native JsonResponse this way:
from django.http import JsonResponse
JsonResponse(data, encoder=ExtendedEncoder)
It sounds like what you're asking about involves serializing the data structure of a Django model instance for interoperability. The other posters are correct: if you wanted the serialized form to be used with a python application that can query the database via Django's api, then you would wan to serialize a queryset with one object. If, on the other hand, what you need is a way to re-inflate the model instance somewhere else without touching the database or without using Django, then you have a little bit of work to do.
Here's what I do:
First, I use demjson for the conversion. It happened to be what I found first, but it might not be the best. My implementation depends on one of its features, but there should be similar ways with other converters.
Second, implement a json_equivalent method on all models that you might need serialized. This is a magic method for demjson, but it's probably something you're going to want to think about no matter what implementation you choose. The idea is that you return an object that is directly convertible to json (i.e. an array or dictionary). If you really want to do this automatically:
def json_equivalent(self):
dictionary = {}
for field in self._meta.get_all_field_names()
dictionary[field] = self.__getattribute__(field)
return dictionary
This will not be helpful to you unless you have a completely flat data structure (no ForeignKeys, only numbers and strings in the database, etc.). Otherwise, you should seriously think about the right way to implement this method.
Third, call demjson.JSON.encode(instance) and you have what you want.
If you want to return the single model object as a json response to a client, you can do this simple solution:
from django.forms.models import model_to_dict
from django.http import JsonResponse
movie = Movie.objects.get(pk=1)
return JsonResponse(model_to_dict(movie))
If you're asking how to serialize a single object from a model and you know you're only going to get one object in the queryset (for instance, using objects.get), then use something like:
import django.core.serializers
import django.http
import models
def jsonExample(request,poll_id):
s = django.core.serializers.serialize('json',[models.Poll.objects.get(id=poll_id)])
# s is a string with [] around it, so strip them off
o=s.strip("[]")
return django.http.HttpResponse(o, mimetype="application/json")
which would get you something of the form:
{"pk": 1, "model": "polls.poll", "fields": {"pub_date": "2013-06-27T02:29:38.284Z", "question": "What's up?"}}
.values() is what I needed to convert a model instance to JSON.
.values() documentation: https://docs.djangoproject.com/en/3.0/ref/models/querysets/#values
Example usage with a model called Project.
Note: I'm using Django Rest Framework
from django.http import JsonResponse
#csrf_exempt
#api_view(["GET"])
def get_project(request):
id = request.query_params['id']
data = Project.objects.filter(id=id).values()
if len(data) == 0:
return JsonResponse(status=404, data={'message': 'Project with id {} not found.'.format(id)})
return JsonResponse(data[0])
Result from a valid id:
{
"id": 47,
"title": "Project Name",
"description": "",
"created_at": "2020-01-21T18:13:49.693Z",
}
I solved this problem by adding a serialization method to my model:
def toJSON(self):
import simplejson
return simplejson.dumps(dict([(attr, getattr(self, attr)) for attr in [f.name for f in self._meta.fields]]))
Here's the verbose equivalent for those averse to one-liners:
def toJSON(self):
fields = []
for field in self._meta.fields:
fields.append(field.name)
d = {}
for attr in fields:
d[attr] = getattr(self, attr)
import simplejson
return simplejson.dumps(d)
_meta.fields is an ordered list of model fields which can be accessed from instances and from the model itself.
Here's my solution for this, which allows you to easily customize the JSON as well as organize related records
Firstly implement a method on the model. I call is json but you can call it whatever you like, e.g.:
class Car(Model):
...
def json(self):
return {
'manufacturer': self.manufacturer.name,
'model': self.model,
'colors': [color.json for color in self.colors.all()],
}
Then in the view I do:
data = [car.json for car in Car.objects.all()]
return HttpResponse(json.dumps(data), content_type='application/json; charset=UTF-8', status=status)
Use list, it will solve problem
Step1:
result=YOUR_MODELE_NAME.objects.values('PROP1','PROP2').all();
Step2:
result=list(result) #after getting data from model convert result to list
Step3:
return HttpResponse(json.dumps(result), content_type = "application/json")
Use Django Serializer with python format,
from django.core import serializers
qs = SomeModel.objects.all()
serialized_obj = serializers.serialize('python', qs)
What's difference between json and python format?
The json format will return the result as str whereas python will return the result in either list or OrderedDict
To serialize and deserialze, use the following:
from django.core import serializers
serial = serializers.serialize("json", [obj])
...
# .next() pulls the first object out of the generator
# .object retrieves django object the object from the DeserializedObject
obj = next(serializers.deserialize("json", serial)).object
All of these answers were a little hacky compared to what I would expect from a framework, the simplest method, I think by far, if you are using the rest framework:
rep = YourSerializerClass().to_representation(your_instance)
json.dumps(rep)
This uses the Serializer directly, respecting the fields you've defined on it, as well as any associations, etc.
It doesn't seem you can serialize an instance, you'd have to serialize a QuerySet of one object.
from django.core import serializers
from models import *
def getUser(request):
return HttpResponse(json(Users.objects.filter(id=88)))
I run out of the svn release of django, so this may not be in earlier versions.
ville = UneVille.objects.get(nom='lihlihlihlih')
....
blablablab
.......
return HttpResponse(simplejson.dumps(ville.__dict__))
I return the dict of my instance
so it return something like {'field1':value,"field2":value,....}
how about this way:
def ins2dic(obj):
SubDic = obj.__dict__
del SubDic['id']
del SubDic['_state']
return SubDic
or exclude anything you don't want.
This is a project that it can serialize(JSON base now) all data in your model and put them to a specific directory automatically and then it can deserialize it whenever you want... I've personally serialized thousand records with this script and then load all of them back to another database without any losing data.
Anyone that would be interested in opensource projects can contribute this project and add more feature to it.
serializer_deserializer_model
Let this is a serializers for CROPS, Do like below. It works for me, Definitely It will work for you also.
First import serializers
from django.core import serializers
Then you can write like this
class CropVarietySerializer(serializers.Serializer):
crop_variety_info = serializers.serialize('json', [ obj, ])
OR you can write like this
class CropVarietySerializer(serializers.Serializer):
crop_variety_info = serializers.JSONField()
Then Call this serializer inside your views.py
For more details, Please visit https://docs.djangoproject.com/en/4.1/topics/serialization/
serializers.JSONField(*args, **kwargs) and serializers.JSONField() are same. you can also visit https://www.django-rest-framework.org/api-guide/fields/ for JSONField() details.