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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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  • 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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  • Django: get count of ForeignKey item in template?

    - by AP257
    Straightforward question - apologies if it is a duplicate, but I can't find the answer if so. I have a User model and a Submission model, like this: class Submission(models.Model): uploaded_by = models.ForeignKey('User') class User(models.Model): name = models.CharField(max_length=250 ) How can I show the number of Submissions made by each user in the template? I've tried {{ user.submission.count }}, like this: for user in users: {{ user.name }} ({{ user.submission.count }} submissions) but no luck...

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  • Django forms I cannot save picture file

    - by dana
    i have the model: class OpenCv(models.Model): created_by = models.ForeignKey(User, blank=True) first_name = models.CharField(('first name'), max_length=30, blank=True) last_name = models.CharField(('last name'), max_length=30, blank=True) url = models.URLField(verify_exists=True) picture = models.ImageField(help_text=('Upload an image (max %s kilobytes)' %settings.MAX_PHOTO_UPLOAD_SIZE),upload_to='jakido/avatar',blank=True, null= True) bio = models.CharField(('bio'), max_length=180, blank=True) date_birth = models.DateField(blank=True,null=True) domain = models.CharField(('domain'), max_length=30, blank=True, choices = domain_choices) specialisation = models.CharField(('specialization'), max_length=30, blank=True) degree = models.CharField(('degree'), max_length=30, choices = degree_choices) year_last_degree = models.CharField(('year last degree'), max_length=30, blank=True,choices = year_last_degree_choices) lyceum = models.CharField(('lyceum'), max_length=30, blank=True) faculty = models.ForeignKey(Faculty, blank=True,null=True) references = models.CharField(('references'), max_length=30, blank=True) workplace = models.ForeignKey(Workplace, blank=True,null=True) the form: class OpencvForm(ModelForm): class Meta: model = OpenCv fields = ['first_name','last_name','url','picture','bio','domain','specialisation','degree','year_last_degree','lyceum','references'] and the view: def save_opencv(request): if request.method == 'POST': form = OpencvForm(request.POST, request.FILES) # if 'picture' in request.FILES: file = request.FILES['picture'] filename = file['filename'] fd = open('%s/%s' % (MEDIA_ROOT, filename), 'wb') fd.write(file['content']) fd.close() if form.is_valid(): new_obj = form.save(commit=False) new_obj.picture = form.cleaned_data['picture'] new_obj.created_by = request.user new_obj.save() return HttpResponseRedirect('.') else: form = OpencvForm() return render_to_response('opencv/opencv_form.html', { 'form': form, }, context_instance=RequestContext(request)) but i don't seem to save the picture in my database... something is wrong, and i can't figure out what :(

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  • How to make django test framework read from live database?

    - by lfborjas
    I realize there's a similar question here, but this one has a different approach: I have a django app that does queries over data indexed with djapian ; I'd like to write unit tests for this app's search component, and, obviously, I'd need the django settings module and all connections with the database active, so the test runner that django provides seems ideal. however, the django testing framework creates a dummy database and I'd hate to dump all my data to a fixture and then index it (the tests would take forever!); My data isn't at risk because the tests would only read from the database, so, how could this be achieved? -I'm new at this whole unit testing thing, so the solution of writing a new test runner I read in that similar question doesn't enlighten me a bit, at least not without some details

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  • Django query filter a set of data

    - by dana
    if a have a query like following = Relations.objects.filter(initiated_by = request.user) in which i'm having all the users followed by the currently logged in user, and i want to display those user's blog posts. Using a query like: blog = New.objects.filter(created_by = following) it only shows me the blog posts of the user with the id = 1 (though the currently logged in user doesn't actually follow him) in template i have : {% for object in blog %} <a href='/accounts/profile_view/{{object.created_by}}/'> {{object.created_by}}</a> <br /> {{object.post}}<br /> {% endfor %} Where am i wrong?

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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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  • 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 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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