Search Results

Search found 26207 results on 1049 pages for 'django users'.

Page 86/1049 | < Previous Page | 82 83 84 85 86 87 88 89 90 91 92 93  | Next Page >

  • Passing list of values to django view via jQuery ajax call

    - by finspin
    I'm trying to pass a list of numeric values (ids) from one web page to another with jQuery ajax call. I can't figure out how to pass and read all the values in the list. I can successfully post and read 1 value but not multiple values. Here is what I have so far: jQuery: var postUrl = "http://localhost:8000/ingredients/"; $('li').click(function(){ values = [1, 2]; $.ajax({ url: postUrl, type: 'POST', data: {'terid': values}, traditional: true, dataType: 'html', success: function(result){ $('#ingredients').append(result); } }); }); /ingredients/ view: def ingredients(request): if request.is_ajax(): ourid = request.POST.get('terid', False) ingredients = Ingredience.objects.filter(food__id__in=ourid) t = get_template('ingredients.html') html = t.render(Context({'ingredients': ingredients,})) return HttpResponse(html) else: html = '<p>This is not ajax</p>' return HttpResponse(html) With Firebug I can see that POST contains both ids but probably in the wrong format (terid=1&terid=2). So my ingredients view picks up only terid=2. What am I doing wrong? EDIT: To clarify, I need the ourid variable pass values [1, 2] to the filter in the ingredients view.

    Read the article

  • 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

    Read the article

  • Limit number of views per day in Django

    - by ariddell
    Is there an easy way to limit the number of times a view can be accessed by a given IP address per day/week? A simplified version of the technique used by some booksellers to limit the number of pages of a book you can preview? There's only one view that this limit need apply to--i.e. it's not a general limit--and it would be nice if I could just have a variable overlimit in the template context. The solution need not be terribly robust, but limiting by IP address seemed like a better idea than using a cookie. I've looked into the session middleware but it doesn't make any references to tracking IP addresses as far as I can tell. Has anyone encountered this problem?

    Read the article

  • 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

    Read the article

  • 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

    Read the article

  • 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()

    Read the article

  • Wordpress & Django -- One domain, two servers. Possible?

    - by DomoDomo
    My question is about hosting Django and Wordpress under one domain, but two physical machines (actually, they are VMs but same diff). Let's say I have a Django webapp at example.com. I'd like to start a Wordpress blog about my webapp, so any blog page rank mojo flows back to my webapp, I'd like the blog address t be example.com/blog. My understanding is blog.example.com would not transfer said page rank mojo. Because I'm worried about Wordpress security flaws compromising my Django webapp, I want to host Django and Wordpress on two physically separate machines. Given all that, is it possible using re-write rules or a reverse proxy server to do this? I know the easy way is to make my Wordpress blog a subdomain, but I really don't want to do that. Has anyone done this in the past, is it stable? If I need a third server to be a dedicated reverse proxy, that's totally fine. Thanks!

    Read the article

  • 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?

    Read the article

  • Django url rewrites and passing a parameter from Javascript

    - by William T Wild
    As a bit of a followup question to my previous , I need to pass a parameter to a view. This parameter is not known until the JS executes. In my URLConf: url(r'^person/device/program/oneday/(?P<meter_id>\d+)/(?P<day_of_the_week>\w+)/$', therm_control.Get_One_Day_Of_Current_Thermostat_Schedule.as_view(), name="one-day-url"), I can pass it this URL and it works great! ( thanks to you guys). http://127.0.0.1:8000/personview/person/device/program/oneday/149778/Monday/ In My template I have this: var one_day_url = "{% url personview:one-day-url meter_id=meter_id day_of_the_week='Monday' %}"; In my javascript: $.ajax({ type: 'GET', url: one_day_url , dataType: "json", timeout: 30000, beforeSend: beforeSendCallback, success: successCallback, error: errorCallback, complete: completeCallback }); When this triggers it works fine except I dont necessarily want Monday all the time. If I change the javascript to this: var one_day_url = "{% url personview:one-day-url meter_id=meter_id %}"; and then $.ajax({ type: 'GET', url: one_day_url + '/Monday/', dataType: "json", timeout: 30000, beforeSend: beforeSendCallback, success: successCallback, error: errorCallback, complete: completeCallback }); I get the Caught NoReverseMatch while rendering error. I assume because the URLconf still wants to rewrite to include the ?P\w+) . I seems like if I change the URL conf that breaks the abailty to find the view , and if I do what I do above it gives me the NoREverseMatch error. Any guidance would be appreciated.

    Read the article

  • 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 :(

    Read the article

  • 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

    Read the article

  • 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?

    Read the article

  • 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.

    Read the article

  • 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?

    Read the article

  • 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.

    Read the article

  • 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 ?

    Read the article

  • 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)

    Read the article

  • 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?

    Read the article

  • 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.

    Read the article

  • 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.

    Read the article

< Previous Page | 82 83 84 85 86 87 88 89 90 91 92 93  | Next Page >