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  • How to save link with tag e parameters in TextField

    - by xRobot
    I have this simple Post model: class Post(models.Model): title = models.CharField(_('title'), max_length=60, blank=True, null=True) body = models.TextField(_('body')) blog = models.ForeignKey(Blog, related_name="posts") user = models.ForeignKey(User) I want that when I insert in the form the links, the these links are saved in the body from this form: http://www.example.com or www.example.com to this form ( with tag and rel="nofollow" parameter ): <a href="http://www.example.com" rel="nofollow">www.example.com</a> How can I do this ? Thanks ^_^

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  • How to save links with tags and parameters in TextField

    - by xRobot
    I have this simple Post model: class Post(models.Model): title = models.CharField(_('title'), max_length=60, blank=True, null=True) body = models.TextField(_('body')) blog = models.ForeignKey(Blog, related_name="posts") user = models.ForeignKey(User) I want that when I insert in the form the links, then these links are saved in the body from this form: http://www.example.com or www.example.com to this form ( with tag and rel="nofollow" parameter ): <a href="http://www.example.com" rel="nofollow">www.example.com</a> How can I do this ? Thanks ^_^

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  • How can I access the "through" object of a Django ManyToManyField?

    - by Macha
    I have the following models in my Django app. How can I from the Team model find all the User objects who have accepted as True in the Membership model? I know I need to use Team.objects.filter(), but I'm not sure how to check the value of the accepted field. from django.contrib.auth.models import User class Team(models.Model): members = models.ManyToManyField(User, through="Membership") class Membership(models.Model): user = models.ForeignKey(User) team = models.ForeignKey(Team) accepted = models.BooleanField(default=False)

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  • Finding object count where a field is unique in Django

    - by Johnd
    I have a model that is something like this: class Input(models.Model): details = models.CharField(max_length=1000) user = models.ForeignKey(User) class Case(Input): title = models.CharField(max_length=200) views = models.IntegerField() class Argument(Input): case = models.ForeignKey(Case) side = models.BooleanField() A user can submit many arguments, per case. I want to be able to say how many users have submitted side=true arguments. I mean if 1 user had 10 arguments and another user had 2 arguments (both side=true) I'd want the count to be 2, not 12.

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  • Django. Invalid keyword argument for this function. ManyToMany

    - by sagem_tetra
    I have this error: 'people' is an invalid keyword argument for this function class Passage(models.Model): name= models.CharField(max_length = 255) who = models.ForeignKey(UserProfil) class UserPassage(models.Model): passage = models.ForeignKey(Passage) people = models.ManyToManyField(UserProfil, null=True) class UserProfil(models.Model): user = models.OneToOneField(User) name = models.CharField(max_length=50) I try: def join(request): user = request.user user_profil = UserProfil.objects.get(user=user) passage = Passage.objects.get(id=2) #line with error up = UserPassage.objects.create(people= user_profil, passage=passage) return render_to_response('thanks.html') How to do correctly? Thanks!

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  • Django one form for two models

    - by martinthenext
    Hello! I have a ForeignKey relationship between TextPage and Paragraph and my goal is to make front-end TextPage creating/editing form as if it was in ModelAdmin with 'inlines': several field for the TextPage and then a couple of Paragraph instances stacked inline. The problem is that i have no idea about how to validate and save that: @login_required def textpage_add(request): profile = request.user.profile_set.all()[0] if not (profile.is_admin() or profile.is_editor()): raise Http404 PageFormSet = inlineformset_factory(TextPage, Paragraph, extra=5) if request.POST: try: textpageform = TextPageForm(request.POST) # formset = PageFormSet(request.POST) except forms.ValidationError as error: textpageform = TextPageForm() formset = PageFormSet() return render_to_response('textpages/manage.html', { 'formset' : formset, 'textpageform' : textpageform, 'error' : str(error), }, context_instance=RequestContext(request)) # Saving data if textpageform.is_valid() and formset.is_valid(): textpageform.save() formset.save() return HttpResponseRedirect(reverse(consults)) else: textpageform = TextPageForm() formset = PageFormSet() return render_to_response('textpages/manage.html', { 'formset' : formset, 'textpageform' : textpageform, }, context_instance=RequestContext(request)) I know I't a kind of code-monkey style to post code that you don't even expect to work but I wanted to show what I'm trying to accomplish. Here is the relevant part of models.py: class TextPage(models.Model): title = models.CharField(max_length=100) page_sub_category = models.ForeignKey(PageSubCategory, blank=True, null=True) def __unicode__(self): return self.title class Paragraph(models.Model): article = models.ForeignKey(TextPage) title = models.CharField(max_length=100, blank=True, null=True) text = models.TextField(blank=True, null=True) def __unicode__(self): return self.title Any help would be appreciated. Thanks!

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  • Limit foreign key choices in select in an inline form in admin

    - by mightyhal
    Edited :-) Hopefully a bit clearer now. The logic is of the model is: A Building has many Rooms A Room may be inside another Room (a closet, for instance--ForeignKey on 'self') A Room can only in inside of another Room in the same building (this is the tricky part) Here's the code I have: #spaces/models.py from django.db import models class Building(models.Model): name=models.CharField(max_length=32) def __unicode__(self): return self.name class Room(models.Model): number=models.CharField(max_length=8) building=models.ForeignKey(Building) inside_room=models.ForeignKey('self',blank=True,null=True) def __unicode__(self): return self.number and: #spaces/admin.py from ex.spaces.models import Building, Room from django.contrib import admin class RoomAdmin(admin.ModelAdmin): pass class RoomInline(admin.TabularInline): model = Room extra = 2 class BuildingAdmin(admin.ModelAdmin): inlines=[RoomInline] admin.site.register(Building, BuildingAdmin) admin.site.register(Room) The inline will display only rooms in the current building (which is what I want). The problem, though, is that for the inside_room drop down, it displays all of the rooms in the Rooms table (including those in other buildings). In the inline of rooms, I need to limit the inside_room choices to only rooms which are in the current building being displayed by the main form. I can't figure out a way to do it with either a limit_choices_to in the model, nor can I figure out how exactly to override the admin's inline formset properly (I feel like I should be somehow create a custom inline form, pass the building_id of the main form to the custom inline, then limit the queryset for the field's choices based on that--but I just can't wrap my head around how to do it). Maybe this is too complex for the admin site, but it seems like something that would be generally useful... Thanks again for your help!

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  • Django Cannot set values on a ManyToManyField which specifies an intermediary model

    - by dana
    i am using a m2m and a through table, and when i was trying to save, my error was: Cannot set values on a ManyToManyField which specifies an intermediary model so, i've modified my code, so that when i save the form, to insert data into the 'through' table too.But now, i'm having another error. (i've bolded the lines where i think i am wrong) i have in models.py: class Classroom(models.Model): user = models.ForeignKey(User, related_name = 'classroom_creator') classname = models.CharField(max_length=140, unique = True) date = models.DateTimeField(auto_now=True) open_class = models.BooleanField(default=True) members = models.ManyToManyField(User,related_name="list of invited members", through = 'Membership') class Membership(models.Model): accept = models.BooleanField(User) date = models.DateTimeField(auto_now = True) classroom = models.ForeignKey(Classroom, related_name = 'classroom_membership') member = models.ForeignKey(User, related_name = 'user_membership') and in def save_classroom(request): if request.method == 'POST': form = ClassroomForm(request.POST, request.FILES, user = request.user) **classroom_instance = Classroom member_instance = Membership** if form.is_valid(): new_obj = form.save(commit=False) new_obj.user = request.user r = Relations.objects.filter(initiated_by = request.user) membership = Membership.objects.create(**classroom = classroom_instance, member = member_instance,date=datetime.datetime.now())** new_obj.save() form.save_m2m() return HttpResponseRedirect('/classroom/classroom_view/{{user}}/') else: form = ClassroomForm(user = request.user) return render_to_response('classroom/classroom_form.html', { 'form': form, }, context_instance=RequestContext(request)) but i don't seem to initialise okay the classroom_instance and menber_instance.My error os: Cannot assign "": "Membership.classroom" must be a "Classroom" instance. Thanks!

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  • Add data to Django form class using modelformset_factory

    - by dean
    I have a problem where I need to display a lot of forms for detail data for a hierarchical data set. I want to display some relational fields as labels for the forms and I'm struggling with a way to do this in a more robust way. Here is the code... class Category(models.Model): name = models.CharField(max_length=160) class Item(models.Model): category = models.ForeignKey('Category') name = models.CharField(max_length=160) weight = models.IntegerField(default=0) class Meta: ordering = ('category','weight','name') class BudgetValue(models.Model): value = models.IntegerField() plan = models.ForeignKey('Plan') item = models.ForeignKey('Item') I use the modelformset_factory to create a formset of budgetvalue forms for a particular plan. What I'd like is item name and category name for each BudgetValue. When I iterate through the forms each one will be labeled properly. class BudgetValueForm(forms.ModelForm): item = forms.ModelChoiceField(queryset=Item.objects.all(),widget=forms.HiddenInput()) plan = forms.ModelChoiceField(queryset=Plan.objects.all(),widget=forms.HiddenInput()) category = "" < assign dynamically on form creation > item = "" < assign dynamically on form creation > class Meta: model = BudgetValue fields = ('item','plan','value') What I started out with is just creating a dictionary of budgetvalue.item.category.name, budgetvalue.item.name, and the form for each budget value. This gets passed to the template and I render it as I intended. I'm assuming that the ordering of the forms in the formset and the querset used to genererate the formset keep the budgetvalues in the same order and the dictionary is created correctly. That is the budgetvalue.item.name is associated with the correct form. This scares me and I'm thinking there has to be a better way. Any help would be greatly appreciated.

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  • How make this special many2many fields validation using Django ORM?

    - by e-satis
    I have the folowing model: class Step(models.Model): order = models.IntegerField() latitude = models.FloatField() longitude = models.FloatField() date = DateField(blank=True, null=True) class Journey(models.Model): boat = models.ForeignKey(Boat) route = models.ManyToManyField(Step) departure = models.ForeignKey(Step, related_name="departure_of", null=True) arrival = models.ForeignKey(Step, related_name="arrival_of", null=True) I would like to implement the following check: # If a there is less than one step, raises ValidationError. routes = tuple(self.route.order_by("date")) if len(routes) <= 1: raise ValidationError("There must be at least two setps in the route") # save the first and the last step as departure and arrival self.departure = routes[0] self.arrival = routes[-1] # departure and arrival must at least have a date if not (self.departure.date or self.arrival.date): raise ValidationError("There must be an departure and an arrival date. " "Please set the date field for the first and last Step of the Journey") # departure must occurs before arrival if not (self.departure.date > self.arrival.date): raise ValidationError("Departure must take place the same day or any date before arrival. " "Please set accordingly the date field for the first and last Step of the Journey") I tried to do that by overloading save(). Unfortunately, Journey.route is empty in save(). What's more, Journey.id doesn't exists yet. I didn't try django.db.models.signals.post_save but suppose it will fail because Journey.route is empty as well (when does this get filled anyway?). I see a solution in django.db.models.signals.m2m_changed but there are a lot of steps (thousands), and I want to avoid to perform an operation for every single of them.

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  • Django m2m form appearing fields

    - by dana
    I have a classroom application,and a follow relation. Users can follow each other and can create classrooms.When a user creates a classroom, he can invite only the people that are following him. The Classroom model is a m2m to User table. i have in models. py: class Classroom(models.Model): creator = models.ForeignKey(User) classname = models.CharField(max_length=140, unique = True) date = models.DateTimeField(auto_now=True) open_class = models.BooleanField(default=True) members = models.ManyToManyField(User,related_name="list of invited members") and in models.py of the follow application: class Relations(models.Model): initiated_by = models.ForeignKey(User, editable=False) date_initiated = models.DateTimeField(auto_now=True, editable = False) follow = models.ForeignKey(User, editable = False, related_name = "follow") date_follow = models.DateTimeField(auto_now=True, editable = False) and in views.py of the classroom app: def save_classroom(request, username): if request.method == 'POST': u = User.objects.get(username=username) form = ClassroomForm(request.POST, request.FILES) if form.is_valid(): new_obj = form.save(commit=False) new_obj.creator = request.user r = Relations.objects.filter(initiated_by = request.user) # new_obj.members = new_obj.save() return HttpResponseRedirect('.') else: form = ClassroomForm() return render_to_response('classroom/classroom_form.html', { 'form': form, }, context_instance=RequestContext(request)) i'm using a ModelForm for the classroom form, and the default view, taking in consideration my many to many relation with User table, in the field Members, is a list of all Users in my database. But i only want in that list the users that are in a follow relationship with the logged in user - the one who creates the classroom. How can i do that? Thanks!

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  • How do I create self-relationships in polymorphic inheritance in Elixir and Pylons?

    - by Turukawa
    I am new to programming and am following the example in the Pylons documentation on creating a Wiki. The database I want to link to the wiki was created with Elixir so I rewrote the Wiki database schema and have continued from there. In the wiki there is a requirement for a Navigation table which is inherited by Pages and Sections. A section can have many pages, while a page can only have one section. In addition, each sibling node can be chain-referenced to each other. So: Nav has "section" (OneToMany) and "before" (OneToOne - to reference preceeding node) Page has "section" (ManyToOne - many pages in one section) and inherits "before" Section inherits all from Nav The code I've written looks like this: class Nav(Entity): using_options(inheritance='multi') name = Field(Unicode(30), default=u'Untitled Node') path = Field(Unicode(255), default=u'') section = OneToMany('Page', inverse='section') after = OneToOne('Nav', inverse='before') before = OneToMany('Nav', inverse='after') class Page(Nav): using_options(inheritance='multi') content = Field(UnicodeText, nullable=False) posted = Field(DateTime, default=now()) title = Field(Unicode(255), default=u'Untitled Page') heading = Field(Unicode(255)) tags = ManyToMany('Tag') comments = OneToMany('Comment') section = ManyToOne('Nav', inverse='section') class Section(Nav): using_options(inheritance='multi') Errors received on this: sqlalchemy.exc.OperationalError: (OperationalError) table nav has no column named aftr_id u'INSERT INTO nav (name, path, aftr_id, row_type) VALUES (?, ?, ?, ?)' I've also tried: before = ManyToMany('Nav', inverse='before') on Nav in the hopes this might break the problem, but also not. The original SQLAlchemy code from the tutorial for these declarations is as follows: nav_table = schema.Table('nav', meta.metadata, schema.Column('id', types.Integer(), schema.Sequence('nav_id_seq', optional=True), primary_key=True), schema.Column('name', types.Unicode(255), default=u'Untitled Node'), schema.Column('path', types.Unicode(255), default=u''), schema.Column('section', types.Integer(), schema.ForeignKey('nav.id')), schema.Column('before', types.Integer(), default=None), schema.Column('type', types.String(30), nullable=False) ) page_table = schema.Table('page', meta.metadata, schema.Column('id', types.Integer, schema.ForeignKey('nav.id'), primary_key=True), schema.Column('content', types.Text(), nullable=False), schema.Column('posted', types.DateTime(), default=now), schema.Column('title', types.Unicode(255), default=u'Untitled Page'), schema.Column('heading', types.Unicode(255)), ) section_table = sa.Table('section', meta.metadata, schema.Column('id', types.Integer, schema.ForeignKey('nav.id'), primary_key=True), ) orm.mapper(Nav, nav_table, polymorphic_on=nav_table.c.type, polymorphic_identity='nav') orm.mapper(Section, section_table, inherits=Nav, polymorphic_identity='section') orm.mapper(Page, page_table, inherits=Nav, polymorphic_identity='page', properties={ 'comments':orm.relation(Comment, backref='page', cascade='all'), 'tags':orm.relation(Tag, secondary=pagetag_table) }) Any help is much appreciated.

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  • sqlalchemy relation through another (declarative)

    - by clayg
    Is anyone familiar with ActiveRecord's "has_many :through" relations for models? I'm not really a Rails guy, but that's basically what I'm trying to do. As a contrived example consider Projects, Programmers, and Assignments: from sqlalchemy import create_engine from sqlalchemy.orm import sessionmaker from sqlalchemy import Column, ForeignKey from sqlalchemy.types import Integer, String, Text from sqlalchemy.orm import relation from sqlalchemy.ext.declarative import declarative_base Base = declarative_base() class Assignment(Base): __tablename__ = 'assignment' id = Column(Integer, primary_key=True) description = Column(Text) programmer_id = Column(Integer, ForeignKey('programmer.id')) project_id = Column(Integer, ForeignKey('project.id')) def __init__(self, description=description): self.description = description def __repr__(self): return '<Assignment("%s")>' % self.description class Programmer(Base): __tablename__ = 'programmer' id = Column(Integer, primary_key=True) name = Column(String(64)) assignments = relation("Assignment", backref='programmer') def __init__(self, name=name): self.name = name def __repr__(self): return '<Programmer("%s")>' % self.name class Project(Base): __tablename__ = 'project' id = Column(Integer, primary_key=True) name = Column(String(64)) description = Column(Text) assignments = relation("Assignment", backref='project') def __init__(self, name=name, description=description): self.name = name self.description = description def __repr__(self): return '<Project("%s", "%s...")>' % (self.name, self.description[:10]) engine = create_engine('sqlite://') Base.metadata.create_all(engine) Session = sessionmaker(bind=engine) session = Session() Projects have many Assignments. Programmers have many Assignments. (understatement?) But in my office at least, Programmers also have many Projects - I'd like this relationship to be inferred through the Assignments assigned to the Programmer. I'd like the Programmer model to have a attribute "projects" which will return a list of Projects associated to the Programmer through the Assignment model. me = session.query(Programmer).filter_by(name='clay').one() projects = session.query(Project).\ join(Project.assignments).\ join(Assignment.programmer).\ filter(Programmer.id==me.id).all() How can I describe this relationship clearly and simply using the sqlalchemy declarative syntax? Thanks!

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  • Creating self-referential tables with polymorphism in SQLALchemy

    - by Jace
    I'm trying to create a db structure in which I have many types of content entities, of which one, a Comment, can be attached to any other. Consider the following: from datetime import datetime from sqlalchemy import create_engine from sqlalchemy import Column, ForeignKey from sqlalchemy import Unicode, Integer, DateTime from sqlalchemy.orm import relation, backref from sqlalchemy.ext.declarative import declarative_base Base = declarative_base() class Entity(Base): __tablename__ = 'entities' id = Column(Integer, primary_key=True) created_at = Column(DateTime, default=datetime.utcnow, nullable=False) edited_at = Column(DateTime, default=datetime.utcnow, onupdate=datetime.utcnow, nullable=False) type = Column(Unicode(20), nullable=False) __mapper_args__ = {'polymorphic_on': type} # <...insert some models based on Entity...> class Comment(Entity): __tablename__ = 'comments' __mapper_args__ = {'polymorphic_identity': u'comment'} id = Column(None, ForeignKey('entities.id'), primary_key=True) _idref = relation(Entity, foreign_keys=id, primaryjoin=id == Entity.id) attached_to_id = Column(Integer, ForeignKey('entities.id'), nullable=False) #attached_to = relation(Entity, remote_side=[Entity.id]) attached_to = relation(Entity, foreign_keys=attached_to_id, primaryjoin=attached_to_id == Entity.id, backref=backref('comments', cascade="all, delete-orphan")) text = Column(Unicode(255), nullable=False) engine = create_engine('sqlite://', echo=True) Base.metadata.bind = engine Base.metadata.create_all(engine) This seems about right, except SQLAlchemy doesn't like having two foreign keys pointing to the same parent. It says ArgumentError: Can't determine join between 'entities' and 'comments'; tables have more than one foreign key constraint relationship between them. Please specify the 'onclause' of this join explicitly. How do I specify onclause?

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  • How to create instances of related models in Django

    - by sevennineteen
    I'm working on a CMSy app for which I've implemented a set of models which allow for creation of custom Template instances, made up of a number of Fields and tied to a specific Customer. The end-goal is that one or more templates with a set of custom fields can be defined through the Admin interface and associated to a customer, so that customer can then create content objects in the format prescribed by the template. I seem to have gotten this hooked up such that I can create any number of Template objects, but I'm struggling with how to create instances - actual content objects - in those templates. For example, I can define a template "Basic Page" for customer "Acme" which has the fields "Title" and "Body", but I haven't figured out how to create Basic Page instances where these fields can be filled in. Here are my (somewhat elided) models... class Customer(models.Model): ... class Field(models.Model): ... class Template(models.Model): label = models.CharField(max_length=255) clients = models.ManyToManyField(Customer, blank=True) fields = models.ManyToManyField(Field, blank=True) class ContentObject(models.Model): label = models.CharField(max_length=255) template = models.ForeignKey(Template) author = models.ForeignKey(User) customer = models.ForeignKey(Customer) mod_date = models.DateTimeField('Modified Date', editable=False) def __unicode__(self): return '%s (%s)' % (self.label, self.template) def save(self): self.mod_date = datetime.datetime.now() super(ContentObject, self).save() Thanks in advance for any advice!

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  • Setting up relations/mappings for a SQLAlchemy many-to-many database

    - by Brent Ramerth
    I'm new to SQLAlchemy and relational databases, and I'm trying to set up a model for an annotated lexicon. I want to support an arbitrary number of key-value annotations for the words which can be added or removed at runtime. Since there will be a lot of repetition in the names of the keys, I don't want to use this solution directly, although the code is similar. My design has word objects and property objects. The words and properties are stored in separate tables with a property_values table that links the two. Here's the code: from sqlalchemy import Column, Integer, String, Table, create_engine from sqlalchemy import MetaData, ForeignKey from sqlalchemy.orm import relation, mapper, sessionmaker from sqlalchemy.ext.declarative import declarative_base engine = create_engine('sqlite:///test.db', echo=True) meta = MetaData(bind=engine) property_values = Table('property_values', meta, Column('word_id', Integer, ForeignKey('words.id')), Column('property_id', Integer, ForeignKey('properties.id')), Column('value', String(20)) ) words = Table('words', meta, Column('id', Integer, primary_key=True), Column('name', String(20)), Column('freq', Integer) ) properties = Table('properties', meta, Column('id', Integer, primary_key=True), Column('name', String(20), nullable=False, unique=True) ) meta.create_all() class Word(object): def __init__(self, name, freq=1): self.name = name self.freq = freq class Property(object): def __init__(self, name): self.name = name mapper(Property, properties) Now I'd like to be able to do the following: Session = sessionmaker(bind=engine) s = Session() word = Word('foo', 42) word['bar'] = 'yes' # or word.bar = 'yes' ? s.add(word) s.commit() Ideally this should add 1|foo|42 to the words table, add 1|bar to the properties table, and add 1|1|yes to the property_values table. However, I don't have the right mappings and relations in place to make this happen. I get the sense from reading the documentation at http://www.sqlalchemy.org/docs/05/mappers.html#association-pattern that I want to use an association proxy or something of that sort here, but the syntax is unclear to me. I experimented with this: mapper(Word, words, properties={ 'properties': relation(Property, secondary=property_values) }) but this mapper only fills in the foreign key values, and I need to fill in the other value as well. Any assistance would be greatly appreciated.

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  • forms problem in django 1.1

    - by alexarsh
    I have the following form: class ModuleItemForm2(forms.ModelForm): class Meta: model = Module_item fields = ('title', 'media', 'thumb', 'desc', 'default', 'player_option') The model is: class Module_item(models.Model): title = models.CharField(max_length=100) layout = models.CharField(max_length=5, choices=LAYOUTS_CHOICE) media = models.CharField(help_text='Media url', max_length=500, blank=True, null=True) conserv = models.ForeignKey(Conserv, help_text= 'Redirect to Conserv', blank=True, null=True) conserve_section = models.CharField(max_length=100, help_text= 'Section within the redirected Conserv', blank=True, null=True) parent = models.ForeignKey('self', help_text='Upper menu.', blank=True, null=True) module = models.ForeignKey(Module, blank=True, null=True) thumb = models.FileField(upload_to='sms/module_items/thumbs', blank=True, null=True) desc = models.CharField(max_length=500, blank=True, null=True) auto_play = models.IntegerField(help_text='Auto start play (miliseconds)', blank=True, null=True) order = models.IntegerField(help_text='Display order', blank=True, null=True) depth = models.IntegerField(help_text='The layout depth', blank=True, null=True) flow_replace = models.IntegerField(blank=True, null=True) default = models.IntegerField(help_text='The selected sub item (Note: Starting from 0)', blank=True, null=True) player_options = models.CharField(max_length=1000, null=True, blank=True) In my view I build form: module_item_form2 = ModuleItemForm2() print module_item_form2 And I get the following error on the print line: 'NoneType' object has no attribute 'label' It works fine with django 1.0.2. I see the error only in django 1.1. Do you have an idea what am I doing wrong? Regards, Arshavski Alexander.

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  • How to save to Django Model that Have Mulitple Foreign Keys Fields

    - by Spikie
    I have Models for business Apps class staff_name(models.Model): TITLE_CHOICES = ( ('Mr', 'Mr'), ('Miss', 'Miss'), ( 'Mrs', 'Mrs'), ( 'chief', 'chief'), ) titlename = models.CharField(max_length=10,choices=TITLE_CHOICES) firstname = models.CharField(max_length=150) surname = models.CharField(max_length=150) date = models.DateTimeField(auto_now=True) class meta: ordering = ["date"] get_latest_by = "date" class inventory_transaction(models.Model): stock_in = models.DecimalField(blank=True, null=True,max_digits=8, decimal_places=2) stock_out = models.DecimalField(blank=True,null=True,max_digits=8, decimal_places=2) Number_container = models.ForeignKey(container_identity) staffs = models.ForeignKey(staff_name) goods_details = models.ForeignKey(departments) balance = models.DecimalField(max_digits=8, decimal_places=2) date = models.DateTimeField(auto_now=True) What i want to accomplish is check if the staff have made entry to the table before if yes add the value for the stock in plus (last) balance field and assign to balance if no just assign stock in value to balance field and save these are my codes These are my codes: try: s = staffname.staffs_set.all().order_by("-date").latest() # staffname is the instant of the class model staff_name e = s.staffs_set.create(stockin=vdataz,balance=s.balance + vdataz ) # e is the instant of the class model inventory_transaction e.save e.staffs.add(s) e.from_container.add(containersno) e.goods_details.add(department) except ObjectDoesNotExist: e = staff_name.objects.create(stockin=vdataz,balance=vdataz ) e.save e.staffs.add(staffname) e.from_container.add(containersno) e.goods_details.add(department) I will really appreciate a solution Thanks I hope it make more sense now. iam on online if you need more explanation just ask in the comment.Thank you for your help

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  • How to provide an inline model field with a queryset choices without losing field value for inline r

    - by Judith Boonstra
    The code displayed below is providing the choices I need for the app field, and the choices I need for the attr field when using Admin. I am having a problem with the attr field on the inline form for already saved records. The attr selected for these saved does show in small print above the field, but not within the field itself. # MODELS: Class Vocab(models.Model): entity = models.Charfield, max_length = 40, unique = True) Class App(models.Model): name = models.ForeignKey(Vocab, related_name = 'vocab_appname', unique = True) app = SelfForeignKey('self, verbose_name = 'parent', blank = True, null = True) attr = models.ManyToManyField(Vocab, related_name = 'vocab_appattr', through ='AppAttr' def parqs(self): a method that provides a queryset consisting of available apps from vocab, excluding self and any apps within the current app's dependent line. def attrqs(self): a method that provides a queryset consisting of available attr from vocab excluding those already selected by current app, 2) those already selected by any apps within the current app's parent line, and 3) those selected by any apps within the current app's dependent line. Class AppAttr(models.Model): app = models.ForeignKey(App) attr = models.ForeignKey(Vocab) # FORMS: from models import AppAttr def appattr_form_callback(instance, field, *args, **kwargs) if field.name = 'attr': if instance: return field.formfield(queryset = instance.attrqs(), *kwargs) return field.formfield(*kwargs) # ADMIN: necessary imports class AppAttrInline(admin.TabularInline): model = AppAttr def get_formset(self, request, obj = None, **kwargs): kwargs['formfield_callback'] = curry(appattr_form_callback, obj) return super(AppAttrInline, self).get_formset(request, obj, **kwargs) class AppForm(forms.ModelForm): class Meta: model = App def __init__(self, *args, **kwargs): super(AppForm, self).__init__(*args, **kwargs) if self.instance.id is None: working = App.objects.all() else: thisrec = App.objects.get(id = self.instance.id) working = thisrec.parqs() self.fields['par'].queryset = working class AppAdmin(admin.ModelAdmin): form = AppForm inlines = [AppAttrInline,] fieldsets = .......... necessary register statements

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  • Django ManyToMany Membership errors making associations

    - by jmitchel3
    I'm trying to have a "member admin" in which they have hundreds of members in the group. These members can be in several groups. Admins can remove access for the member ideally in the view. I'm having trouble just creating the group. I used a ManytoManyField to get started. Ideally, the "member admin" would be able to either select existing Users OR it would be able to Add/Invite new ones via email address. Here's what I have: #views.py def membership(request): group = Group.objects.all().filter(user=request.user) GroupFormSet = modelformset_factory(Group, form=MembershipForm) if request.method == 'POST': formset = GroupFormSet(request.POST, request.FILES, queryset=group) if formset.is_valid(): formset.save(commit=False) for form in formset: form.instance.user = request.user formset.save() return render_to_response('formset.html', locals(), context_instance=RequestContext(request)) else: formset= GroupFormSet(queryset=group) return render_to_response('formset.html', locals(), context_instance=RequestContext(request)) #models.py class Group(models.Model): name = models.CharField(max_length=128) members = models.ManyToManyField(User, related_name='community_members', through='Membership') user = models.ForeignKey(User, related_name='community_creator', null=True) def __unicode__(self): return self.name class Membership(models.Model): member = models.ForeignKey(User, related_name='user_membership', blank=True, null=True) group = models.ForeignKey(Group, related_name='community_membership', blank=True, null=True) date_joined = models.DateField(auto_now=True, blank=True, null=True) class Meta: unique_together = ('member', 'group') Any ideas? Thank you for your help.

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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) objects = OpenCvManager() 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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  • 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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  • Django Admin Running Same Query Thousands of Times for Model

    - by Tom
    Running into an odd . . . loop when trying to view a model in the Django admin. I have three related models (code trimmed for brevity, hopefully I didn't trim something I shouldn't have): class Association(models.Model): somecompany_entity_id = models.CharField(max_length=10, db_index=True) name = models.CharField(max_length=200) def __unicode__(self): return self.name class ResidentialUnit(models.Model): building = models.CharField(max_length=10) app_number = models.CharField(max_length=10) unit_number = models.CharField(max_length=10) unit_description = models.CharField(max_length=100, blank=True) association = models.ForeignKey(Association) created = models.DateTimeField(auto_now_add=True) updated = models.DateTimeField(auto_now=True) def __unicode__(self): return '%s: %s, Unit %s' % (self.association, self.building, self.unit_number) class Resident(models.Model): unit = models.ForeignKey(ResidentialUnit) type = models.CharField(max_length=20, blank=True, default='') lookup_key = models.CharField(max_length=200) jenark_id = models.CharField(max_length=20, blank=True) user = models.ForeignKey(User) is_association_admin = models.BooleanField(default=False, db_index=True) show_in_contact_list = models.BooleanField(default=False, db_index=True) created = models.DateTimeField(auto_now_add=True) updated = models.DateTimeField(auto_now=True) _phones = {} home_phone = None work_phone = None cell_phone = None app_number = None account_cache_key = None def __unicode__(self): return '%s' % self.user.get_full_name() It's the last model that's causing the problem. Trying to look at a Resident in the admin takes 10-20 seconds. If I take 'self.association' out of the __unicode__ method for ResidentialUnit, a resident page renders pretty quickly. Looking at it in the debug toolbar, without the association name in ResidentialUnit (which is a foreign key on Resident), the page runs 14 queries. With the association name put back in, it runs a far more impressive 4,872 queries. The strangest part is the extra queries all seem to be looking up the association name. They all come from the same line, the __unicode__ method for ResidentialUnit. Each one is the exact same thing, e.g., SELECT `residents_association`.`id`, `residents_association`.`jenark_entity_id`, `residents_association`.`name` FROM `residents_association` WHERE `residents_association`.`id` = 1096 ORDER BY `residents_association`.`name` ASC I assume I've managed to create a circular reference, but if it were truly circular, it would just die, not run 4000x and then return. Having trouble finding a good Google or StackOverflow result for this.

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  • Implementing a popularity algorithm in Django

    - by TheLizardKing
    I am creating a site similar to reddit and hacker news that has a database of links and votes. I am implementing hacker news' popularity algorithm and things are going pretty swimmingly until it comes to actually gathering up these links and displaying them. The algorithm is simple: Y Combinator's Hacker News: Popularity = (p - 1) / (t + 2)^1.5` Votes divided by age factor. Where` p : votes (points) from users. t : time since submission in hours. p is subtracted by 1 to negate submitter's vote. Age factor is (time since submission in hours plus two) to the power of 1.5.factor is (time since submission in hours plus two) to the power of 1.5. I asked a very similar question over yonder http://stackoverflow.com/questions/1964395/complex-ordering-in-django but instead of contemplating my options I choose one and tried to make it work because that's how I did it with PHP/MySQL but I now know Django does things a lot differently. My models look something (exactly) like this class Link(models.Model): category = models.ForeignKey(Category) user = models.ForeignKey(User) created = models.DateTimeField(auto_now_add = True) modified = models.DateTimeField(auto_now = True) fame = models.PositiveIntegerField(default = 1) title = models.CharField(max_length = 256) url = models.URLField(max_length = 2048) def __unicode__(self): return self.title class Vote(models.Model): link = models.ForeignKey(Link) user = models.ForeignKey(User) created = models.DateTimeField(auto_now_add = True) modified = models.DateTimeField(auto_now = True) karma_delta = models.SmallIntegerField() def __unicode__(self): return str(self.karma_delta) and my view: def index(request): popular_links = Link.objects.select_related().annotate(karma_total = Sum('vote__karma_delta')) return render_to_response('links/index.html', {'links': popular_links}) Now from my previous question, I am trying to implement the algorithm using the sorting function. An answer from that question seems to think I should put the algorithm in the select and sort then. I am going to paginate these results so I don't think I can do the sorting in python without grabbing everything. Any suggestions on how I could efficiently do this? EDIT This isn't working yet but I think it's a step in the right direction: from django.shortcuts import render_to_response from linkett.apps.links.models import * def index(request): popular_links = Link.objects.select_related() popular_links = popular_links.extra( select = { 'karma_total': 'SUM(vote.karma_delta)', 'popularity': '(karma_total - 1) / POW(2, 1.5)', }, order_by = ['-popularity'] ) return render_to_response('links/index.html', {'links': popular_links}) This errors out into: Caught an exception while rendering: column "karma_total" does not exist LINE 1: SELECT ((karma_total - 1) / POW(2, 1.5)) AS "popularity", (S... EDIT 2 Better error? TemplateSyntaxError: Caught an exception while rendering: missing FROM-clause entry for table "vote" LINE 1: SELECT ((vote.karma_total - 1) / POW(2, 1.5)) AS "popularity... My index.html is simply: {% block content %} {% for link in links %} karma-up {{ link.karma_total }} karma-down {{ link.title }} Posted by {{ link.user }} to {{ link.category }} at {{ link.created }} {% empty %} No Links {% endfor %} {% endblock content %} EDIT 3 So very close! Again, all these answers are great but I am concentrating on a particular one because I feel it works best for my situation. from django.db.models import Sum from django.shortcuts import render_to_response from linkett.apps.links.models import * def index(request): popular_links = Link.objects.select_related().extra( select = { 'popularity': '(SUM(links_vote.karma_delta) - 1) / POW(2, 1.5)', }, tables = ['links_link', 'links_vote'], order_by = ['-popularity'], ) return render_to_response('links/test.html', {'links': popular_links}) Running this I am presented with an error hating on my lack of group by values. Specifically: TemplateSyntaxError at / Caught an exception while rendering: column "links_link.id" must appear in the GROUP BY clause or be used in an aggregate function LINE 1: ...karma_delta) - 1) / POW(2, 1.5)) AS "popularity", "links_lin... Not sure why my links_link.id wouldn't be in my group by but I am not sure how to alter my group by, django usually does that.

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  • Kendo Grid: Foreign Key Dropdown does not update grid cell after update

    - by JookyDFW
    I have a Kendo MVC grid that contains a nullable property (short) that is bound as a foreign key and uses a dropdown list as an editor template. I am also using inline editing. When the property value is null, the dropdown list selected value does not get set into the grid cell after the update button is clicked. This works fine if incell editing is used. I am looking for a workaround that will solve my problem. I am including a stripped down version of my code below Everything works if the nullable value is set to a non-null value. GRID @(Html.Kendo().Grid<AssetViewModel>() .Name("DealAssets") .Columns(c => { c.Bound(x => x.Name); c.ForeignKey(x => x.AssetTypeID, (IEnumerable<SelectListItem>)ViewBag.AssetTypeList, "Value", "Text"); c.ForeignKey(x => x.SeniorityTypeID, seniorityTypeList, "Value", "Text").EditorTemplateName("GridNullableForeignKey"); c.ForeignKey(x => x.RateBaseID, rateBaseList, "Value", "Text").EditorTemplateName("GridNullableForeignKey"); ; c.Command(m => { m.Edit(); m.Destroy(); }); }) .ToolBar(toolbar => toolbar.Create().Text("Add New Asset")) .Editable(x => x.Mode(GridEditMode.InLine)) .DataSource(ds => ds .Ajax() .Model(model => model.Id(request => request.ID)) .Read(read => read.Action("ReadAssets", "Deal", new { id = Model.ID })) .Create(create => create.Action("CreateAsset", "Deal", new { currentDealID = Model.ID })) .Update(update => update.Action("UpdateAsset", "Deal")) .Destroy(destroy => destroy.Action("DeleteAsset", "Deal")) ) ) EDITOR TEMPLATE @model short? @{ var controlName = ViewData.TemplateInfo.GetFullHtmlFieldName(""); } @( Html.Kendo().DropDownListFor(m => m) .Name(controlName) .OptionLabel("- Please select -") .BindTo((SelectList)ViewData[ViewData.TemplateInfo.GetFullHtmlFieldName("") + "_Data"]) ) UPDATE ACTION public ActionResult UpdateAsset([DataSourceRequest] DataSourceRequest request, int ID) { var dealAsset = DataContext.DealAssets.SingleOrDefault(o => o.ID == ID); if (dealAsset != null) { if (TryUpdateModel(dealAsset.Asset, new[] {"Name","AssetTypeID","SeniorityTypeID","RateBaseID" })) { DataContext.SaveChanges(); } } return Json(new[] { new AssetViewModel(dealAsset) }.ToDataSourceResult(request, ModelState), JsonRequestBehavior.AllowGet); }

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