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  • django getting current user id

    - by dana
    hello, i have a mini app where users can login, view their profile, and follow each other. 'Follow' is a relation like a regular 'friend' relationship in virtual communities, but it is not necessarily reciprocal, meaning that one can follow a user, without the need that the user to be following back that person who follows him. my problem is: if i am a logged in user, and i navigate to a profile X, and push the button follow, how can i take the current profile id ?(current profile meaning the profile that I, the logged in user, am viewing right now.) the view: def follow(request): if request.method == 'POST': form = FollowForm(request.POST) if form.is_valid(): new_obj = form.save(commit=False) new_obj.initiated_by = request.user u = User.objects. what here? new_obj.follow = u new_obj.save() return HttpResponseRedirect('.') else: form = FollowForm() return render_to_response('followme/follow.html', { 'form': form, }, context_instance=RequestContext(request)) thanks in advance!

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  • Django Admin: Many-to-Many listbox doesn't show up with a through parameter

    - by NP
    Hi All, I have the following models: class Message(models.Model): date = models.DateTimeField() user = models.ForeignKey(User) thread = models.ForeignKey('self', blank=True, null=True) ... class Forum(models.Model): name = models.CharField(max_length=24) messages = models.ManyToManyField(Message, through="Message_forum", blank=True, null=True) ... class Message_forum(models.Model): message = models.ForeignKey(Message) forum = models.ForeignKey(Forum) status = models.IntegerField() position = models.IntegerField(blank=True, null=True) tags = models.ManyToManyField(Tag, blank=True, null=True) In the admin site, when I go to add/change a forum, I don't see the messages listbox as you'd expect. However, it shows up if I remove the 'through' parameter in the ManyToManyField declaration. What's up with that? I've registered all three models (plus Tag) to the admin site in admin.py. TIA

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  • Self Authenticating Links in Django

    - by awolf
    In my web app I would like to be able to email self-authenticating links to users. These links will contain a unique token (uuid). When they click the link the token being present in the query string will be enough to authenticate them and they won't have to enter their username and password. What's the best way to do this?

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  • Django - foreignkey problem

    - by realshadow
    Hey, Imagine you have this model: class Category(models.Model): node_id = models.IntegerField(primary_key = True) type_id = models.IntegerField(max_length = 20) parent_id = models.IntegerField(max_length = 20) sort_order = models.IntegerField(max_length = 20) name = models.CharField(max_length = 45) lft = models.IntegerField(max_length = 20) rgt = models.IntegerField(max_length = 20) depth = models.IntegerField(max_length = 20) added_on = models.DateTimeField(auto_now = True) updated_on = models.DateTimeField(auto_now = True) status = models.IntegerField(max_length = 20) node = models.ForeignKey(Category_info, verbose_name = 'Category_info', to_field = 'node_id' The important part is the foreignkey. When I try: Category.objects.filter(type_id = type_g.type_id, parent_id = offset, status = 1) I get an error that get returned more than category, which is fine, because it is supposed to return more than one. But I want to filter the results trough another field, which would be type id (from the second Model) Here it is: class Category_info(models.Model): objtree_label_id = models.AutoField(primary_key = True) node_id = models.IntegerField(unique = True) language_id = models.IntegerField() label = models.CharField(max_length = 255) type_id = models.IntegerField() The type_id can be any number from 1 - 5. I am desparately trying to get only one result where the type_id would be number 1. Here is what I want in sql: SELECT n.*, l.* FROM objtree_nodes n JOIN objtree_labels l ON (n.node_id = l.node_id) WHERE n.type_id = 15 AND n.parent_id = 50 AND l.type_id = 1 Any help is GREATLY appreciated. Regards

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  • ahow can I resolve Django Error: str' object has no attribute 'autoescape'?

    - by Angelbit
    Hi have tried to create a inclusion tag on Django but don't work and return str' object has no attribute 'autoescape' this is the code of custom tag: from django import template from quotes.models import Quotes register = template.Library() def show_quote(): quote = Quotes.objects.values('quote', 'author').get(id=0) return {'quote': quote['quote']} register.inclusion_tag('quotes.html')(show_quote) EDIT: Quote class from django.db import models class Quotes(models.Model): quote = models.CharField(max_length=255) author = models.CharField(max_length=100) class Meta: db_table = 'quotes' quotes.html <blockquote id="quotes">{{ quote }}</blockquote>

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  • How do I write this URL in Django?

    - by alex
    (r'^/(?P<the_param>[a-zA-z0-9_-]+)/$','myproject.myapp.views.myview'), How can I change this so that "the_param" accepts a URL(encoded) as a parameter? So, I want to pass a URL to it. mydomain.com/http%3A//google.com

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  • Django: Update order attribute for objects in a queryset

    - by lazerscience
    I'm having a attribute on my model to allow the user to order the objects. I have to update the element's order depending on a list, that contains the object's ids in the new order; right now I'm iterating over the whole queryset and set one objects after the other. What would be the easiest/fastest way to do the same with the whole queryset? def update_ordering(model, order): """ order is in the form [id,id,id,id] for example: [8,4,5,1,3] """ id_to_order = dict((order[i], i) for i in range(len(order))) for x in model.objects.all(): x.order = id_to_order[x.id] x.save()

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  • How to host 50 domains/sites with common Django code base

    - by Off Rhoden
    I have 50 different websites that use the same layout and code base, but mostly non-overlapping data (regional support sites, not link farm). Is there a way to have a single installation of the code and run all 50 at the same time? When I have a bug to fix (or deploy new feature), I want to deploy ONE time + 1 restart and be done with it. Also: Code needs to know what domain the request is coming to so the appropriate data is displayed.

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  • input_formats in django admin has no effect

    - by pablo
    I'm trying to use input_foramts in the admin but it has no effect. What am I doing wrong? # model class Feedback(models.Model): created_at = models.DateTimeField(auto_now_add=True) # admin form class FeedbackAdminForm(forms.ModelForm): created_at = forms.DateTimeField(input_formats=('%d/%m/%Y',)) class Meta: model = Feedback # admin class FeedbackAdmin(admin.ModelAdmin): form = FeedbackAdminForm admin.site.register(Feedback, FeedbackAdmin) Thanks

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  • what is this 'map' mean..in django

    - by zjm1126
    this is the code: def create(request, form_class=MapForm, template_name="maps/create.html"): map_form = form_class(request.POST or None) if map_form.is_valid(): map = map_form.save(commit=False) and the map_form is : class MapForm(forms.ModelForm): slug = forms.SlugField(max_length=20, help_text = _("a short version of the name consisting only of letters, numbers, underscores and hyphens."), #error_message = _("This value must contain only letters, numbers, underscores and hyphens.")) ) def clean_slug(self): if Map.objects.filter(slug__iexact=self.cleaned_data["slug"]).count() > 0: raise forms.ValidationError(_("A Map already exists with that slug.")) return self.cleaned_data["slug"].lower() def clean_name(self): if Map.objects.filter(name__iexact=self.cleaned_data["name"]).count() > 0: raise forms.ValidationError(_("A Map already exists with that name.")) return self.cleaned_data["name"] class Meta: model = Map fields = ('name', 'slug', 'description')

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  • Django: order by count of a ForeignKey field?

    - by AP257
    This is almost certainly a duplicate question, in which case apologies, but I've been searching for around half an hour on SO and can't find the answer here. I'm probably using the wrong search terms, sorry. I have a User model and a Submission model. Each Submission has a ForeignKey field called user_submitted for the User who uploaded it. class Submission(models.Model): uploaded_by = models.ForeignKey('User') class User(models.Model): name = models.CharField(max_length=250 ) My question is pretty simple: how can I get a list of the three users with the most Submissions? I trued creating a num_submissions method on the User model: def num_submissions(self): num_submissions = Submission.objects.filter(uploaded_by=self).count() return num_submissions and then doing: top_users = User.objects.filter(problem_user=False).order_by('num_submissions')[:3] but this fails, as do all the other things I've tried. Can I actually do it using a smart database query? Or should I just do something more hacky in the views file?

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  • Google App Engine django model form does not pick up BlobProperty

    - by Wes
    I have the following model: class Image(db.Model): auction = db.ReferenceProperty(Auction) image = db.BlobProperty() thumb = db.BlobProperty() caption = db.StringProperty() item_to_tag = db.StringProperty() And the following form: class ImageForm(djangoforms.ModelForm): class Meta: model = Image When I call ImageForm(), only the non-Blob fields are created, like this: <tr><th><label for="id_auction">Auction:</label></th><td><select name="auction" id="id_auction"> <option value="" selected="selected">---------</option> <option value="ahRoYXJ0bWFuYXVjdGlvbmVlcmluZ3INCxIHQXVjdGlvbhgKDA">2010-06-19 11:00:00</option> </select></td></tr> <tr><th><label for="id_caption">Caption:</label></th><td><input type="text" name="caption" id="id_caption" /></td></tr> <tr><th><label for="id_item_to_tag">Item to tag:</label></th><td><input type="text" name="item_to_tag" id="id_item_to_tag" /></td></tr> I want the Blob fields to be included in the form as well (as file inputs). What am I doing wrong?

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  • Writing a custom auth system (like the default django auth system), use it to generate tables in DB

    - by dotty
    Hay all, I've been reading up on middleware and how to use it with a context object. I want to write a simple middleware class which i can use on my own applications, it will essentially be a cut down version of the django one. The problem i seem to have is that if i have INSTALLED_APPS = ('django.contrib.my_auth') in the settings file, all is well. I've also added MIDDLEWARE_CLASSES = ('django.contrib.my_auth.middleware.MyAuthMiddleware') in it and everything is fine. My question is, how would i make my middleware automatically generate tables from a models.py module, much like how the django auth does when i run manage.py syncdb? thanks

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  • Django finding which field matched in a multiple OR query

    - by Greg Hinch
    I've got a couple models which are set up something like this: class Bar(models.Model): baz = models.CharField() class Foo(models.Model): bar1 = models.ForeignKey(Bar) bar2 = models.ForeignKey(Bar) bar3 = models.ForeignKey(Bar) And elsewhere in the code, I end up with an instance of Bar, and need to find the Foo it is attached to in some capacity. Right now I came up with doing a multiple OR query using Q, something like this: foo_inst = Foo.objects.get(Q(bar1=bar_inst) | Q(bar2=bar_inst) | Q(bar3=bar_inst)) What I need to figure out is, which of the 3 cases actually hit, at least the name of the member (bar1, bar2, or bar3). Is there a good way to do this? Is there a better way to structure the query to glean that information?

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  • django model relation definition

    - by Laurent Luce
    Hello, Let say I have 3 models: A, B and C with the following relations. A can have many B and many C. B can have many C Is the following correct: class A(models.Model): ... class B(models.Model): ... a = ForeignKey(A) class C(models.Model): ... a = ForeignKey(A) b = ForeignKey(B)

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  • django sort by manytomany relationship

    - by Marconi
    I have the following model: class Service(models.Model): ratings = models.ManyToManyField(User) Now if I wanna get all the service with ratings sorted in descending order I did something: services_list = Service.objects.filter(ratings__gt=0).distinct() services_list = list(services_list) services_list.sort(key=lambda service: service.ratings.all().count(), reverse=True) As you can see its a three step process and I don't feel right about this. Anybody who knows a better way to do this?

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  • Per instance dynamic fields django model

    - by Roberto Rosario
    I have a model with a JSON field or a link to a CouchDB document. I can currently access the dynamic informaction in a way such as: genericdocument.objects.get(pk=1) == genericdocument.json_field['sample subfield'] instead I would like genericdocument.sample_subfield to maintain compatibility with all the apps the project currently shares.

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  • Can this django query be improved?

    - by Hobhouse
    Given a model structure like this: class Book(models.Model): user = models.ForeignKey(User) class Readingdate(models.Model): book = models.ForeignKey(Book) date = models.DateField() One book may have several readingdates. How do I list books having at least one readingdate within a specific year? I can do this: from_date = datetime.date(2010,1,1) to_date = datetime.date(2010,12,31) book_ids = Readingdate.objects\ .filter(date__range=(from_date,to_date))\ .values_list('book_id', flat=True) books_read_2010 = Book.objects.filter(id__in=book_ids) Is it possible to do this with one queryset, or is this the best way?

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