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  • Using a backwards relation (i.e FOO_set) for ModelChoiceField in Django

    - by Bwmat
    I have a model called Movie, which has a ManyToManyField called director to a model called Person, and I'm trying to create a form with ModelChoiceField like so: class MovieSearchForm(forms.Form): producer = forms.ModelChoiceField(label='Produced by', queryset=movies.models.Person.producer_set, required=False) but this seems to be failing to compile (I'm getting a ViewDoesNotExist exception for the view that uses the form, but it goes away if I just replace the queryset with all the person objects), I'm guessing because '.producer_set' is being evaluated too 'early'. How can I get this work? here are the relevant parts of the movie/person classes: class Person(models.Model): name = models.CharField(max_length=100) class Movie(models.Model): ... producer = models.ForeignKey(Person, related_name="producers") director = models.ForeignKey(Person, related_name="directors") What I'm trying to do is get ever Person who is used in the producer field of some Movie.

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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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  • Django form and i18n

    - by madewulf
    I have forms that I want to display in different languages : I used the label parameter to set a parameter, and used ugettext() on the labels : agreed_tos = forms.BooleanField(label=ugettext('I agree to the terms of service and to the privacy policy.')) But when I am rendering the form in my template, using {{form.as_p}} The labels are not translated. Does somebody have a solution for this problem ?

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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 | capture sub domain as a string

    - by MMRUser
    How do I capture a part of sub-domain name and get that name as a string in my views through a request. ex: user.domain.com developer.domain.com I want to capture the user part of this domain name through a request (lets say when the first time user hits the page). Thanks.

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  • django access to parent

    - by SledgehammerPL
    model: class Product(models.Model): name = models.CharField(max_length = 128) (...) def __unicode__(self): return self.name class Receipt(models.Model): name = models.CharField(max_length=128) (...) components = models.ManyToManyField(Product, through='ReceiptComponent') def __unicode__(self): return self.name class ReceiptComponent(models.Model): product = models.ForeignKey(Product) receipt = models.ForeignKey(Receipt) quantity = models.FloatField(max_length=9) unit = models.ForeignKey(Unit) def __unicode__(self): return unicode(self.quantity!=0 and self.quantity or '') + ' ' + unicode(self.unit) + ' ' + self.product.genitive And now I'd like to get list of the most often useable products: ReceiptComponent.objects.values('product').annotate(Count('product')).order_by('-product__count' the example result: [{'product': 3, 'product__count': 5}, {'product': 6, 'product__count': 4}, {'product': 5, 'product__count': 3}, {'product': 7, 'product__count': 2}, {'product': 1, 'product__count': 2}, {'product': 11, 'product__count': 1}, {'product': 8, 'product__count': 1}, {'product': 4, 'product__count': 1}, {'product': 9, 'product__count': 1}] It's almost what I need. But I'd prefer having Product object not product value, because I'd like to use this in views.py for generating list.

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  • How to generate a line break in Django template

    - by Iamamac
    I want to give default value to a textarea. The code is something like this: <textarea>{{userSetting.list | join:"NEWLINE"}}</textarea> where userSetting.list is a string list, each item of whom is expected to show in one line. textarea takes the content between the tags as the default value, preserving its line breaks and not interpreting any HTML tags (which means <br>,\n won't work). I have found a solution: {{userSetting.list | join:" " | wordwrap:0}} (there is no whitespace in the list). But obviously it is NOT a good one. Any help would be appreciated.

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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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  • How to teach Django to non-web programmers? [closed]

    - by Greg
    I've been tasked with providing a workshop for my co-workers to teach them Django. They're all good programmers but they've never done any web programming. I was thinking to just go through the Django tutorial with them, but are there things in there that wouldn't make sense to non-web programmers? Do they need any kind of webdev background first? Any thoughts on a good way to provide the basics so that Django will make sense?

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  • Downloading a csv file in django

    - by spyder
    I am trying to download a CSV file using HttpResponse to make sure that the browser treats it as an attachment. I follow the instructions provided here but my browser does not prompt a "Save As" dialog. I cannot figure out what is wrong with my function. All help is appreciated. dev savefile(request): try: myfile = request.GET['filename'] filepath = settings.MEDIA_ROOT + 'results/' destpath = os.path.join(filepath, myfile) response = HttpResponse(FileWrapper(file(destpath)), mimetype='text/csv' ) response['Content-Disposition'] = 'attachment; filename="%s"' %(myfile) return response except Exception, err: errmsg = "%s"%(err) return HttpResponse(errmsg) Happy Pat's day!

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  • Django Aggregation Across Reverse Relationship

    - by Tom
    Given these two models: class Profile(models.Model): user = models.ForeignKey(User, unique=True, verbose_name=_('user')) about = models.TextField(_('about'), blank=True) zip = models.CharField(max_length=10, verbose_name='zip code', blank=True) website = models.URLField(_('website'), blank=True, verify_exists=False) class ProfileView(models.Model): profile = models.ForeignKey(Profile) viewer = models.ForeignKey(User, blank=True, null=True) created = models.DateTimeField(auto_now_add=True) I want to get all profiles sorted by total views. I can get a list of profile ids sorted by total views with: ProfileView.objects.values('profile').annotate(Count('profile')).order_by('-profile__count') But that's just a dictionary of profile ids, which means I then have to loop over it and put together a list of profile objects. Which is a number of additional queries and still doesn't result in a QuerySet. At that point, I might as well drop to raw SQL. Before I do, is there a way to do this from the Profile model? ProfileViews are related via a ForeignKey field, but it's not as though the Profile model knows that, so I'm not sure how to tie the two together. As an aside, I realize I could just store views as a property on the Profile model and that may turn out to be what I do here, but I'm still interested in learning how to better use the Aggregation functions.

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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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  • Why am I seeing an error about _OBJC_CLASS_$_CPGraphHostingView with Core Plot?

    - by user616281
    I downloaded the Core Plot example application, but when I compile it I saw a few errors. I then added the Core Plot SDK, but in this SDK there is no class named CPGraphHostingView. Therefore, I added the class manually from this link. However, I now see the following error: ERROR - "_OBJC_CLASS_$_CPGraphHostingView", referenced from: How can I work around this to get the sample application to compile?

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