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Search found 3554 results on 143 pages for 'django many to many'.

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• how to save django object using dictionary?

  - by shahjapan

  is there a way that I can save the model by using dictionary for e.g. this is working fine, p1 = Poll.objects.get(pk=1) p1.name = 'poll2' p1.descirption = 'poll2 description' p1.save() but what if I have dictionary like { 'name': 'poll2', 'description: 'poll2 description' } is there a simple way to save the such dictionary direct to Poll

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• Django: FloatField or DecimalFied for Currency ?

  - by Hellnar

  I am curious which one would be better fitting as a currency field ? I will do simple operations such as taking difference, the percentage between old and new prices. I plan to keep two digits after the zero (ie 10.50) and majority of the time if these digits are zero, I will be hiding these numbers and display it as "10"

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• Django multiple many to many fields?

  - by Tom

  Hello everybody I am building a news app for my website. I want to use a sort of tag system. Each news article can have different and multiple tags. All tags are saved in a tag model, and i want to connect the tags to the newsarticle. Now is this possible with: tags = models.ForeignKey( TagsModel ) for one tag, but how i can do this with multiple of them? Thank you!

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• Django show manytomanyfield on form when definition is on other model

  - by John

  Hi I have the definition for my manytomany relationship on one model but want to display the field on a form for the other model. How do I do this? for example: # classes class modelA(models.Model): name = models.CharField(max_length=300) manytomany = models.ManyToManyField(modelA) class modelB(models.Model): name = models.CharField(max_length=300) # forms class modelBForm(forms.ModelForm): class Meta: model = modelB I want to use the form modelBForm but show a select box with a list from modelA (just how it would work if the model was set to modelA in the form class). How can I do this? Thanks

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• How do I make nested regroups in Django?

  - by Marcio Cruz

  I've got the following situation in this system: Each category of products has many subcategories, and each subcategory has many products under it. I'm trying to make a product searh, which returns a list, and in my template, I show an overview of the results, like this: Cellphones Dumbphones (2 results) Smartphones (3 results) Monitors CRT (1 result) LCD (3 results) I'm my template I have only the list of products. I've tryed many combinations of nested regroups, without success. Any ideas?

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• Dry way to apply current class to navigation elements in django templates

  - by Lakshman Prasad

  Say there are navigations, nav1, nav2, nav3 and many subnavs within each of them. For the css effect, you need to apply class=current to the currently selected navs. What is the dry way to do that. For the subnavs, you can check to see if request.get_full_path is the same as the url that subnav refers to, in the base template. How can you make it dry for the navigation too.

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• in django admin, can we have a multiple select based on choices

  - by Rasiel

  http://docs.djangoproject.com/en/dev/ref/models/fields/#choices i've read through the documentation and this implies using a database table for dynamic data, however it states choices is meant for static data that doesn't change much, if ever. so what if i want to use choices, but have it select multiple because the data i'm using is quite static, e.g days of the week. is there anyway to achieve this without a database table?

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• SelectDateWidget in Django Admin?

  - by Maria

  Can i change default AdminDateWidget to SelectDateWidget in my models? How can i do this? I try: class RespondentAdmin(admin.ModelAdmin): formfield_overrides = { models.DateField: {'widget': SelectDateWidget}, } but it doesn't work

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• Django sys.path.append for project *and* app needed under WSGI

  - by GerardJP

  Hi all, Could somebody give me a pointer on why I need to add my project root path to the python path as well as the application itself in my WSGI file? Project base is called 'djapp', the application is called 'myapp'. sys.path.append(os.path.dirname(os.path.abspath(__file__)) + '/..') sys.path.append(os.path.dirname(os.path.abspath(__file__)) + '/../djapp') os.environ['DJANGO_SETTINGS_MODULE'] = 'djapp.settings' If I omit the line with "/../djapp/" the log tells my that 'myapp' can not be imported, even though 'djapp.settings' is. (validating 'djapp' was imported) It al runs properly with the ./manage.py command. there's a __init__ in the project folder. For testings sake, I see the same issue using addsitedir: site.addsitedir('/home/user/web/project/') site.addsitedir('/home/user/web/project/djapp') Thanx a lot. Gerard.

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• Django url tag multiple parameters

  - by Overdose

  I have two similar codes. The first one works as expected. urlpatterns = patterns('', (r'^(?P<n1>\d)/test/', test), (r'', test2), {% url testapp.views.test n1=5 %} But adding the second parameter makes the result return empty string. urlpatterns = patterns('', (r'^(?P<n1>\d)/test(?P<n2>\d)/', test), (r'', test2),) {% url testapp.views.test n1=5, n2=2 %} Views signature: def test(request, n1, n2=1):

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• Django save method

  - by Marijus

  So I have a model with a FileField for excel spreadsheet. What I need to do this add another column in this spreadsheet, in each row let user pick from a drop-down list then save it and display it in html. All the picking and uploading will happen through the admin interface. So I have figured out way how to display a spreadsheet in html, however I have no idea how to write this save method. I could really use some hints and tips..

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• how to display fixed dropdown in django admin?

  - by FurtiveFelon

  Hi all, I would like to display priority information in a drop down. Currently i am using a integer field to store the priority, but i would like to display high/medium/low instead of letting user type in a priority. A way to approximate this is to use a Priority database which stores 3 elements, 1:high, 2:medium, 3:low, but it seems like an overkill. Any easier way would be much appreciated! Jason

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• Django: How can I protect against concurrent modification of data base entries

  - by Ber

  If there a way to protect against concurrent modifications of the same data base entry by two or more users? It would be acceptable to show an error message to the user performing the second commit/save operation, but data should not be silently overwritten. I think locking the entry is not an option, as a user might use the "Back" button or simply close his browser, leaving the lock for ever.

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• Django: Serializing models in a nested data structure?

  - by Rosarch

  It's easy to serialize models in an iterable: def _toJSON(models): return serializers.serialize("json", models, ensure_ascii=False) What about when I have something more complicated: [ (Model_A_1, [Model_B_1, Model_B_2, Model_B_3]), (Model_A_2, [Model_B_3, Model_B_4, Model_B_5, Model_B_59]), (Model_A_3, [Model_B_6, Model_B_7]), ] I tried serializing each model as it was added to the structure, then serializing the whole thing with simplejson.dumps, but that causes the JSON defining each model to be escaped. Is there a better way to do this?

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• Django get_FOO_display and distinct()

  - by datakid

  I've seen answers to both halves of my question, but I can't work out how to marry the two. I have a book model, and a translatedBook model. The translatedBook has a langage set up as model choices in the usual way: LANGUAGES = ( (u'it', u'Italian'), (u'ja', u'Japanese'), (u'es', u'Spanish'), (u'zh-cn', u'Simplified Chinese'), (u'zh-tw', u'Traditional Chinese'), (u'fr', u'French'), (u'el', u'Greek'), (u'ar', u'Arabic'), (u'bg', u'Bulgarian'), (u'bn', u'Bengali'), etc I know that to get "Italian" I have to do translatedBook.get_language_display on a Book object. But how do I get a list of distinct languages in their long format? I've tried: lang_avail = TargetText.objects.values('language').distinct().order_by('language') lang_avail = TargetText.objects.distinct().order_by('language').values('language'). lang_avail = TargetText.objects.all().distinct('language').order_by('language') but I can't seem to get what I want - which is a list like: "English, Italian, Simplified Chinese, Spanish" The final lang_avail listed above didn't return the list of 5, it returned the list of 355 (ie, # of books) with multiple repeats....

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• Django Generating RSS feed with description

  - by Issy

  Hey Guys, I am trying to generate a full rss feed, however when loading the feed in Mail, it just shows the title, with a read more link at the bottom. I have tried several different options. But none seem to work. I would like to generate the feed with a combination of several feeds in my modl. Here is the code i have tried: class LatestEvents(Feed): description_template = "events_description.html" def title(self): return "%s Events" % SITE.name def link(self): return '/events/' def items(self): events = list(Event.objects.all().order_by('-published_date')[:5]) return events author_name = 'Latest Events' def item_pubdate(self, item): return item.published_date And in my template which is stored in TEMPLATE_ROOT/feeds/ {{ obj.description|safe }} <h1>Event Location Details</h1> {{ obj.location|safe }} Even if i hard code the description it does not work.

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• Django comments form validation, check if the comment is from authenticated user

  - by Headcrab

  I wanted to add reCaptcha to the comment form, but only for non-authenticated users. I've created my own comments app, using reCaptcha more or less according to this post, except that I don't have any (explicit) view function to post the comment, but instead I check captcha as a part of form validation process. Everything works fine so far. Now, how do I disable reCaptcha for a logged in user? Removing all reCaptcha-related fields in the template is no problem, but how do I validate the form, not checking captcha if the comment if from an authenticated user? Alternatively, is there a way to use two different comment form classes, depending on whether the user is authenticated or not?

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• django related_name for field clashes.

  - by Absolute0

  I am getting a field clash in my models: class Visit(models.Model): user = models.ForeignKey(User) visitor = models.ForeignKey(User) Error: One or more models did not validate: profiles.visit: Accessor for field 'user' clashes with related field 'User.visit_set'. Add a related_name argument to the definition for 'user'. profiles.visit: Accessor for field 'visitor' clashes with related field 'User.visit_set'. Add a related_name argument to the definition for 'visitor'. what would be a sensible 'related_field' to use on visitor field? This model basically represents the visits that take place to a particular user's profile. Also should I replace any of the ForeignKey's with a ManyToManyField? The logic is a bit confusing. Edit: This seems to fix it, but I am unsure if its what I want. :) class Visit(models.Model): user = models.ForeignKey(User) visitor = models.ForeignKey(User, related_name='visitors')

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• Django left join m2m field.

  - by duder

  Here's my Model: class User(models.Model): pass class Item(models.Model): pass class ItemVote(models.Model): user = models.ForeignKey(User) item = models.ForeignKey(Item) vote = models.BooleanField() I want to retrieve a list of Items, and I want to know if the current user has voted for each Item. How do I alter my query object so that it will generate sql similar to: SELECT ... FROM items LEFT OUTER JOIN item_votes ON (item_votes.user_id = ? AND item_votes.item_id = items.id)

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• Django nested formsets

  - by dragoon

  Hi, I have an edit object view that contains a formset(one or many if this matters), now I want to create a page that can display multiple edit object forms and submit it in a single form. What is the correct way to achieve this task?

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• How to cutomize a modelform widget in django 1.1?

  - by muudscope

  I'm trying to modify a django form to use a textarea instead of a normal input for the "address" field in my house form. The docs seem to imply this changed from django 1.1 (which I'm using) to 1.2. But neither approach is working for me. Here's what I've tried: class HouseForm(forms.ModelForm): address = forms.Textarea() # Should work with django 1.1, but doesn't class Meta: model = House #widgets = { 'address': forms.Textarea() } # 1.2 style - doesn't work either.

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• django admin site - filtering available objects for user

  - by JPG

  I have models that belong to some 'group' (Company class). I want to add users, who will also belong to a one group and should be able to edit/manage/add objects with membership in associated group. something like: class Company() class Something() company = ForeignKey(Company) user Microsoft_admin company = ForeignKey(Company) and this user should only see and edit objects belonging to associated Company in the Admin Interface. How to acomplish that?

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• Django Piston - how can I create custom methods?

  - by orokusaki

  I put my questions in the code comments for clarity: from piston.handler import AnonymousBaseHandler class AnonymousAPITest(AnonymousBaseHandler): fields = ('update_subscription',) def update_subscription(self, request, months): # Do some stuff here to update a subscription based on the # number of months provided. # How the heck can I call this method? return {'msg': 'Your subscription has been updated!'} def read(self, request): return { 'msg': 'Why would I need a read() method on a fully custom API?' }

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• How do I restrict foreign keys choices to related objects only in django

  - by Jeff Mc

  I have a two way foreign relation similar to the following class Parent(models.Model): name = models.CharField(max_length=255) favoritechild = models.ForeignKey("Child", blank=True, null=True) class Child(models.Model): name = models.CharField(max_length=255) myparent = models.ForeignKey(Parent) How do I restrict the choices for Parent.favoritechild to only children whose parent is itself? I tried class Parent(models.Model): name = models.CharField(max_length=255) favoritechild = models.ForeignKey("Child", blank=True, null=True, limit_choices_to = {"myparent": "self"}) but that causes the admin interface to not list any children.

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• Specifying different initial values for fields in inherited models (django)

  - by Shawn Chin

  Question : What is the recommended way to specify an initial value for fields if one uses model inheritance and each child model needs to have different default values when rendering a ModelForm? Take for example the following models where CompileCommand and TestCommand both need different initial values when rendered as ModelForm. # ------ models.py class ShellCommand(models.Model): command = models.Charfield(_("command"), max_length=100) arguments = models.Charfield(_("arguments"), max_length=100) class CompileCommand(ShellCommand): # ... default command should be "make" class TestCommand(ShellCommand): # ... default: command = "make", arguments = "test" I am aware that one can used the initial={...} argument when instantiating the form, however I would rather store the initial values within the context of the model (or at least within the associated ModelForm). My current approach What I'm doing at the moment is storing an initial value dict within Meta, and checking for it in my views. # ----- forms.py class CompileCommandForm(forms.ModelForm): class Meta: model = CompileCommand initial_values = {"command":"make"} class TestCommandForm(forms.ModelForm): class Meta: model = TestCommand initial_values = {"command":"make", "arguments":"test"} # ------ in views FORM_LOOKUP = { "compile": CompileCommandFomr, "test": TestCommandForm } CmdForm = FORM_LOOKUP.get(command_type, None) # ... initial = getattr(CmdForm, "initial_values", {}) form = CmdForm(initial=initial) This feels too much like a hack. I am eager for a more generic / better way to achieve this. Suggestions appreciated. Other attempts I have toyed around with overriding the constructor for the submodels: class CompileCommand(ShellCommand): def __init__(self, *args, **kwargs): kwargs.setdefault('command', "make") super(CompileCommand, self).__init__(*args, **kwargs) and this works when I try to create an object from the shell: >>> c = CompileCommand(name="xyz") >>> c.save() <CompileCommand: 123> >>> c.command 'make' However, this does not set the default value when the associated ModelForm is rendered, which unfortunately is what I'm trying to achieve. Update 2 (looks promising) I now have the following in forms.py which allow me to set Meta.default_initial_values without needing extra code in views. class ModelFormWithDefaults(forms.ModelForm): def __init__(self, *args, **kwargs): if hasattr(self.Meta, "default_initial_values"): kwargs.setdefault("initial", self.Meta.default_initial_values) super(ModelFormWithDefaults, self).__init__(*args, **kwargs) class TestCommandForm(ModelFormWithDefaults): class Meta: model = TestCommand default_initial_values = {"command":"make", "arguments":"test"}

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