Search Results

Search found 3558 results on 143 pages for 'django templatetags'.

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

Page 50/143 | < Previous Page | 46 47 48 49 50 51 52 53 54 55 56 57  | Next Page >

  • How to extend models of other applications in Django?

    - by culebrón
    I have a small app with Category model and want to make a required foreign key referencing it from Photologue Gallery model. What's the right approach? I can make many-to-many field in Category, but this way it will not be required in Gallery. Use "register" and modify the Gallery model? Inherit it in my app?

    Read the article

  • Django says the "id may not be NULL" but why is it?

    - by Oli
    I'm going crazy today. I just tried to insert a new record and it threw back a "post_blogpost.id may not be NULL" error. Here's my model: class BlogPost(models.Model): title = models.CharField(max_length=100) slug = models.SlugField(max_length=100) who = models.ForeignKey(User, default=1) when = models.DateTimeField() intro = models.TextField(blank=True, null=True) content = models.TextField(blank=True, null=True) counter = models.PositiveIntegerField(default=0) published = models.BooleanField(default=False) css = models.TextField(blank=True, null=True) class Meta: ordering = ('-when', 'id') There are a number of functions beneath the model too but I won't include them in full here. Their names are: content_cache_key, clear_cache, __unicode__, reads, read, processed_content. I'm adding through the admin... And I'm running out of hair.

    Read the article

  • Django Querysets -- need a less expensive way to do this..

    - by rh0dium
    Hi all, I have a problem with some code and I believe it is because of the expense of the queryset. I am looking for a much less expensive (in terms of time) way to to this.. log.info("Getting Users") employees = Employee.objects.filter(is_active = True) log.info("Have Users") if opt.supervisor: if opt.hierarchical: people = getSubs(employees, " ".join(args)) else: people = employees.filter(supervisor__name__icontains = " ".join(args)) else: log.info("Filtering Users") people = employees.filter(name__icontains = " ".join(args)) | \ employees.filter(unix_accounts__username__icontains = " ".join(args)) log.info("Filtered Users") log.info("Processing data") np = [] for person in people: unix, p4, bugz = "No", "No", "No" if len(person.unix_accounts.all()): unix = "Yes" if len(person.perforce_accounts.all()): p4 = "Yes" if len(person.bugzilla_accounts.all()): bugz = "Yes" if person.cell_phone != "": exphone = fixphone(person.cell_phone) elif person.other_phone != "": exphone = fixphone(person.other_phone) else: exphone = "" np.append({ 'name':person.name, 'office_phone': fixphone(person.office_phone), 'position': person.position, 'location': person.location.description, 'email': person.email, 'functional_area': person.functional_area.name, 'department': person.department.name, 'supervisor': person.supervisor.name, 'unix': unix, 'perforce': p4, 'bugzilla':bugz, 'cell_phone': fixphone(exphone), 'fax': fixphone(person.fax), 'last_update': person.last_update.ctime() }) log.info("Have data") Now this results in a log which looks like this.. 19:00:55 INFO phone phone Getting Users 19:00:57 INFO phone phone Have Users 19:00:57 INFO phone phone Processing data 19:01:30 INFO phone phone Have data As you can see it's taking over 30 seconds to simply iterate over the data. That is way too expensive. Can someone clue me into a more efficient way to do this. I thought that if I did the first filter that would make things easier but seems to have no effect. I'm at a loss on this one. Thanks To be clear this is about 1500 employees -- Not too many!!

    Read the article

  • How to limit fields in django-admin depending on user?

    - by minder
    I suppose similar problem would have been discussed here, but I couldn't find it. Let's suppose I have an Editor and a Supervisor. I want the Editor to be able to add new content (eg. a news post) but before publication it has to be acknowledged by Supervisor. When Editor lists all items, I want to set some fields on the models (like an 'ack' field) as read-only (so he could know what had been ack'ed and what's still waiting approval) but the Supervisor should be able to change everything (list_editable would be perfect) What are the possible solutions to this problem?

    Read the article

  • Django admin breaking with non-default primary_key for model with a m2m relationship ?

    - by Gj
    I have a simple Post model with a m2m field to a Tag model. The Tag had for some reason to use a non default primary key. Inside the admin page for a Post, the labels for the multiple selection field for Tags appear, but not the input field itself. I also tried using the filter_horizontal for the tags, but still only the labels appear without the actual field. Any ideas why it breaks and/or workarounds? Thanks!

    Read the article

  • How do I store multiple copies of the same field in Django?

    - by Alistair
    I'm storing OLAC metadata which describes linguistic resources. Many of the elements of the metadata are repeatable -- for example, a resource can have two languages, three authors and four dates associated with it. Is there any way of storing this in one model? It seems like overkill to define a model for each repeatable metadata element -- especially since the models will only have one field: it's value.

    Read the article

  • Django many-to-many relationship to self with extra data, how do I select from a certain direction?

    - by Jake
    I have some hierarchical data where each Set can have many members and can belong to more than one Set(group) Here are the models: class Set(models.Model): ... groups = models.ManyToManyField('self', through='Membership', symmetrical=False) members = models.ManyToManyField('self', through='Membership', symmetrical=False) class Membership(models.Model): group = models.ForeignKey( Set, related_name='Members' ) member = models.ForeignKey( Set, related_name='Groups' ) order = models.IntegerField( default=-1 ) I want to know how to get all the members or all the groups for a Set instance. I think I can do it as follows, but it's not very logical, can anyone tell me what's going on and how I should be doing it? # This gives me a set of Sets # Which seems to be the groups this Set belongs to set_instance.set_set.all() # These give me a set of Memberships, not Sets set_instance.Members.all() set_instance.Groups.all() # These they both return a set of Sets # which seem to be the members of this one set_instance.members.all() set_instance.groups.all()

    Read the article

  • How can you dispatch on request method in Django URLpatterns?

    - by rcampbell
    It's clear how to create a URLPattern which dispatches from a URL regex: (r'^books/$', books), where books can further dispatch on request method: def books(request): if request.method == 'POST': ... else ... I'd like to know if there is an idiomatic way to include the request method inside the URLPattern, keeping all dispatch/route information in a single location, such as: (r'^books/$', GET, retrieve-book), (r'^books/$', POST, update-books), (r'^books/$', PUT, create-books),

    Read the article

  • Right way to return proxy model instance from a base model instance in Django ?

    - by sotangochips
    Say I have models: class Animal(models.Model): type = models.CharField(max_length=255) class Dog(Animal): def make_sound(self): print "Woof!" class Meta: proxy = True class Cat(Animal): def make_sound(self): print "Meow!" class Meta: proxy = True Let's say I want to do: animals = Animal.objects.all() for animal in animals: animal.make_sound() I want to get back a series of Woofs and Meows. Clearly, I could just define a make_sound in the original model that forks based on animal_type, but then every time I add a new animal type (imagine they're in different apps), I'd have to go in and edit that make_sound function. I'd rather just define proxy models and have them define the behavior themselves. From what I can tell, there's no way of returning mixed Cat or Dog instances, but I figured maybe I could define a "get_proxy_model" method on the main class that returns a cat or a dog model. Surely you could do this, and pass something like the primary key and then just do Cat.objects.get(pk = passed_in_primary_key). But that'd mean doing an extra query for data you already have which seems redundant. Is there any way to turn an animal into a cat or a dog instance in an efficient way? What's the right way to do what I want to achieve?

    Read the article

  • why isn't my id showing up in django admin list?

    - by FurtiveFelon
    Hi all, I have a class Task(models.Model), and i didn't define id field explicitly (since it defines automatically for you). I checked in the database, it exists for the Task. Now i would like to display it in the list via list_display property in admin.ModelAdmin. I have a bunch of things in there, only id is not showing up for any of the rows i have. Everything else works fine. Anyone know anything special i have to do to get id to display? Thanks a lot! Jason

    Read the article

  • How to add default value on django save form?

    - by Ignacio
    I have an object Task and a form that saves it. I want to automatically asign created_by field to the currently logged in user. So, my view is this: def new_task(request, task_id=None): message = None if task_id is not None: task = Task.objects.get(pk=task_id) message = 'TaskOK' submit = 'Update' else: task = Task(created_by = GPUser(user=request.user)) submit = 'Create' if request.method == 'POST': # If the form has been submitted... form = TaskForm(request.POST, instance=task) if form.is_valid(): task = form.save(commit=False); task.created_by = GPUser(user=request.user) task.save() if message == None: message = 'taskOK' return tasks(request, message) else: form = TaskForm(instance=task) return custom_render('user/new_task.html', {'form': form, 'submit': submit, 'task_id':task.id}, request) The problem is, you guessed, the created_by field doesn't get saved. Any ideas? Thanks

    Read the article

  • Is there a way to simplify this Django query?

    - by Mark
    accepted_bids = Bid.objects.filter(shipment__user=u, status='acc').select_related('shipment') completed_shipments = [] for b in accepted_bids: completed_shipments.append(b.shipment) vehicles_shipped = [] for s in completed_shipments: vehicles_shipped.extend(s.items.all()) In the end, I want a list of shipped vehicles. A vehicle is shipped if it's part of a shipment that's completed. A shipment is completed if it has an accepted bid. I'd prefer not to iterate over the querysets thereby forcing a hit to the DB before its necessary... isn't there a way to get all the associated shipments from a list of bids, for example?

    Read the article

  • How to get all objects with their children using django orm?

    - by kender
    Hi, I got very simple hierarchical structure: every object can have 0 or 1 parent. There's no limit on how many children each object can have. So in my application I got such a model: class O(Model): name = CharField(max_length = 20) parent = ForeignKey('O', related_name = 'children') Now I would like to be able to fetch all objects who have a particular one Object1 in their parent-tree (as in their parent or parent of their parents, etc). Should I use mptt or is there a simpler approach?

    Read the article

  • Django admin: how do I add an unrelated model field to a model change/add page?

    - by NP
    I have the following models: class Foo(models.Model): field1 = models.IntegerField() ... class Bar(models.Model): field1 = models.IntegerField() ... class Foo_bar(models.Model): foo = models.ForeignKey(Foo) bar = models.ForeignKey(Bar) ... In the admin, I want it so that in the Foo change/add page, you can specify a Bar object, and on save I want to create a Foo_bar object to represent the relationship. How can I do this through customizing the Admin site/ModelAdmins? Thanks.

    Read the article

  • How do I create an empty Django formset using modelformset_factory?

    - by nbolton
    I'm creating a formset, but it seems to populate it with all of the existing data in the table for that object. I can't figure out how to start with a blank formset; the only way seems to be to delete all of the data from the table, but clearly this isn't an option. I will post code if necessary (but there's lots of it, so knowing what is relevant is tricky).

    Read the article

  • Django: How can I add weight/ordering to a many to many relationship?

    - by Klaas van Schelven
    I'm having Pages with TextBlocks on them. Text Blocks may appear on different pages, pages may have several text blocks on them. Every page may have these blocks in an ordering of it's own. This can be solved by using a separate through parameter. Like so: class TextBlock(models.Model): title = models.CharField(max_length=255) text = models.TextField() class Page(models.Model): textblocks = models.ManyToManyField(TextBlock, through=OrderedTextBlock) class OrderedTextBlock(models.Model): text_block = models.ForeignKey(TextBlock) product_page = models.ForeignKey(ProductPage) weight = models.IntegerField() class Meta: ordering = ('weight',) But I'm not very enthousiastic about the violations of DRY for my app. (There's a lot of ordered ManyToMany relations). Is there a recommended way to go about this?

    Read the article

  • How to show raw_id value of a ManyToMany relation in the Django admin?

    - by luc
    Hello, I have an app using raw_id on both ForeignKeyField and ManyToManyField. The admin displays the value of the foreign key on the right of the edit box. Unfortunatey, it doesn't work with ManyToMany. I've checked the code and I think that it is the normal behavior. However I would like to know if someone has an easy tip to change this behavior? Thanks in advance Update: I've tried to subclass the ManyToManyRawIdWidget but I don't know how to say that the raw_id_fields should use my custom widget. formfield_overrides doesn't seem to work with raw_id fields

    Read the article

  • How can I have multiple navigation paths with Django, like a simplifies wizard path and a full path?

    - by Zeta
    Lets say I have an application with a structure such as: System set date set name set something Other set death ray target calibrate and I want to have "back" and "next" buttons on a page. The catch is, if you're going in via the "wizard", I want the nav path to be something like "set name" - "set death ray target" - "set name". If you go via the Advanced options menu, I want to just iterate options... "set date" - "set name" - "set something" - "set death ray target" - calibrate. So far, I'm thinking I have to use different URIs, but that's that. Any ideia how this could be done? Thanks.

    Read the article

  • In django, how can I include some default records in my models.py?

    - by kdt
    If I have a models.py like class WidgetType(models.Model): name = models.CharField(max_length=200) class Widget(models.Model): typeid = models.ForeignKey(WidgetType) data = models.CharField(max_length=200) How can I build in a set of built in constant values for WidgetType when I know I'm only going to have a certain few types of widget? Clearly I could fire up my admin interface and add them by hand, but I'd like to simplify configuration by having it built into the python.

    Read the article

  • Is there a way to get direct_to_template to pass RequestContext in django?

    - by BigJason
    I have found myself writing the same view over and over. It is basically this: def home_index(request): return render_to_response('home/index.html', RequestContext(request)) To keep with the dry principal, I would like to utilize a generic view. I have seen direct_to_template, but it passes an empty context. So how can I use a generic view and still get the power of RequestContext?

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

< Previous Page | 46 47 48 49 50 51 52 53 54 55 56 57  | Next Page >