I'm trying to setup Django Haystack to search based on some pretty urls. Here is my urlpatterns.
urlpatterns += patterns('',
url(r'^search/$', SearchView(),
name='search_all',
),
url(r'^search/(?P<category>\w+)/$', CategorySearchView(
form_class=SearchForm,
),
name='search_category',
),
)
My custom SearchView class looks like this:
class CategorySearchView(SearchView):
def __name__(self):
return "CategorySearchView"
def __call__(self, request, category):
self.category = category
return super(CategorySearchView, self).__call__(request)
def build_form(self, form_kwargs=None):
data = None
kwargs = {
'load_all': self.load_all,
}
if form_kwargs:
kwargs.update(form_kwargs)
if len(self.request.GET):
data = self.request.GET
kwargs['searchqueryset'] = SearchQuerySet().models(self.category)
return self.form_class(data, **kwargs)
I keep getting this error running the Django dev web server if I try and visit /search/Vendor/q=Microsoft
UserWarning: The model u'Vendor' is not registered for search.
warnings.warn('The model %r is not registered for search.' % model)
And this on my page
The model being added to the query must derive from Model.
If I visit /search/q=Microsoft, it works fine. Is there another way to accomplish this?
Thanks for any pointers
-Jay
There are a couple of things going on here. In your __call__ method you're assigning a category based on a string in the URL. In this error:
UserWarning: The model u'Vendor' is not registered for search
Note the unicode string. If you got an error like The model <class 'mymodel.Model'> is not registered for search then you'd know that you haven't properly created an index for that model. However this is a string, not a model! The models method on the SearchQuerySet class requires a class instance, not a string.
The first thing you could do is use that string to look up a model by content type. This is probably not a good idea! Even if you don't have models indexed which you'd like to keep away from prying eyes, you could at least generate some unnecessary errors.
Better to use a lookup in your view to route the query to the correct model index, using conditionals or perhaps a dictionary. In your __call__ method:
self.category = category.lower()
And if you have several models:
my_querysets = {
'model1': SearchQuerySet().models(Model1),
'model2': SearchQuerySet().models(Model2),
'model3': SearchQuerySet().models(Model3),
}
# Default queryset then searches everything
kwargs['searchqueryset'] = my_querysets.get(self.category, SearchQuerySet())
Is it possible to get object with comments related to it? Right now django comment framework creates query for every object which has related comments and another queries for comments owners. Can I somehow avoid this? I use django 1.4 so prefetch_related is allowed.
You could create a function that caches the count:
from django.contrib.contenttypes.models import ContentType
from django.contrib import comments
def get_comment_count_key(model):
content_type = ContentType.objects.get_for_model(model)
return 'comment_count_%s_%s' % (content_type.pk, model.pk)
def get_comment_count(model):
key = get_comment_count_key(model)
value = cache.get(key)
if value is None:
value = comments.get_model().objects.filter(
content_type = ContentType.objects.get_for_model(model),
object_pk = model.pk,
site__pk = settings.SITE_ID
).count()
cache.set(key, value)
return value
You could extend the Comment model and add get_comment_count there. Or put get_comment_count as a template filter. It doesn't matter.
Of course, you would also need cache invalidation when a new comment is posted:
from django.db.models import signals
from django.contrib import comments
def refresh_comment_count(sender, instance, **kwargs):
cache.delete(get_comment_count_key(instance.content_object))
get_comment_count(instance.content_object)
post_save.connect(refresh_comment_count, sender=comments.get_model())
post_delete.connect(refresh_comment_count, sender=comments.get_model())
You could improve this last snippet, by using cache.incr() on comment_was_posted, and cache.decr() on post_delete but that's left as an exercise for you :)
I have following structure of models in django :
class BodySubPart(models.Model):
body_subpart=models.CharField(max_length=20)
def __str__(self):
return self.body_subpart
class BodyPart(models.Model):
body_part=models.CharField(max_length=20)
body_subpart=models.ManyToManyField(BodySubPart)
def __str__(self):
return self.body_part
Ex:
example,
if BodyPart=head then BodySubPart= face,mouth,eyes,nose.
if BodyPart=arm then BodySubPart= shoulder,fingers,elbow.
like this many body parts are stored.
...
now I want to create a runtime form have two choicefields (BodySubPart and BodyPart) such that when i select the BodyPart it should change the list in BodySubPart.
Ex.
The first choicefield has body parts={head,arm,chest...}
The second choice field should change when i select a particular part
If i select "head" then second choice field should show,
body sub parts={face,mouth,eyes,nose...}
Please help me here.....
What have you tried?? I think you will find people are more willing to help you if you have actually tried something yourself and not just want others to do it for you. It should go something like this:
1) BodyPart.objects.all() # all body parts
2) head = BodyPart.objects.get(body_part='head')
head_subparts = head.body_subpart.all() # all head subparts
django does a great job of explaining how to query these relationships.
https://docs.djangoproject.com/en/dev/topics/db/models/#many-to-many-relationships
In addition there are a number of really great tutorials online regarding djangos' manytomany relationships.
This requires a bit of AJAX, so first step is to create a view to handle that:
from django.core import serializers
from django.http import HttpResponse, HttpResponseBadRequest
from django.shortcuts import get_list_or_404
def ajax_get_bodysubparts(request):
bodypart_id = request.GET.get('bodypart_id')
if bodypart_id:
bodysubparts = get_list_or_404(BodySubPart, bodypart__id=bodypart_id)
data = serializers.serialize('json', bodysubparts)
return HttpResponse(data, mimetype='application/json')
else:
return HttpResponseBadRequest()
Tie that to some URL in urls.py. Then, some JavaScript for your form (assumes jQuery):
$(document).ready(function(){
$('#id_bodypart').change(function(){
var selected = $(this).val();
if (selected) {
$.getJSON('/url/to/ajax/view/', {
'bodypart_id': selected
}, function (data, jqXHR) {
options = [];
for (var i=0; i<data.length; i++) {
options.append('<option value="'+data[i].id+'">'+data[i].body_subpart+'</option>');
}
$('#id_bodysubpart).html(options.join(''));
});
}
});
});
You will probably need a combination of custom form fields and widgets to get what you want.
Check out the django-ajax-filtered-fields project to see if that is close what you are looking for. It will at least provide some guidance if you decide to create your own.
You will need some javascript to make a new request to populate your fields dynamically, so that will also not be available with standard django forms.
I am writing an application in Django, which uses [year]/[month]/[title-text] in the url to identitfy news items. To manage the items I have defined a number of urls, each starting with the above prefix.
urlpatterns = patterns('msite.views',
(r'^(?P<year>[\d]{4})/(?P<month>[\d]{1,2})/(?P<slug>[\w]+)/edit/$', '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...
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.