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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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  • Rails rspec expects Admin::PostsController, which is there.

    - by berkes
    I have a file app/controllers/admin/posts_controller.rb class Admin::PostsController < ApplicationController layout 'admin' # GET /admin/posts def index @pposts = Post.paginate :page => params[:page], :order => 'created_at DESC' end # ...Many more standard CRUD/REST methods... end And an rspec test spec/controllers/admin/posts_controller_spec.rb require 'spec_helper' describe Admin::PostsController do describe "GET 'index'" do it "should be successful" do get 'index' response.should be_success end end #...many more test for all CRUD/REST methods end However, running that spec throws an error. I have no idea what that error means, nor how to start solving it. /home/...../active_support/dependencies.rb:492:in `load_missing_constant': Expected /home/...../app/controllers/admin/posts_controller.rb to define Admin::PostsController (LoadError) I may have it all set up wrong, or may be doing something really silly, but all I want is my CRUD actions on /admin, with separate before filters and a separate layout. And to test these controllers. EDIT ZOMG, made a terrible copy-paste error into this SO posting. The controller was PostsController, not the PagesController that I pasted into there. Problem still stands, as my code is correct, just the SO post, here was wrong.

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  • Django not recognizing django admin urls

    - by colorfulgrayscale
    I just registered my models my models with django admin. I navigate to the django admin at /admin. I log in sucessfully and I can see all my models. great so far. But now if I try to click one of the links, for Ex: 'users', django gives me a 404 saying The current URL, admin/auth/user/, didn't match any of these. Its really weird because in my urls.py I have it mapped correctly (r'^admin/', include(admin.site.urls)), I have all the required middleware enabled and have these in my installed apps 'django.contrib.auth', 'django.contrib.contenttypes', 'django.contrib.sessions', 'django.contrib.sites', 'django.contrib.messages', 'django.contrib.admin', anyone have any idea? Thanks.

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  • Cannot Extend Django 1.2.1 Admin Template

    - by jcady
    I am attempting to override/extend the header for the Django admin in version 1.2.1. However when I try to extend the admin template and simply change what I need documented here: http://docs.djangoproject.com/en/dev/ref/contrib/admin/#overriding-vs-replacing-an-admin-template), I run into a recursion problem. I have an index.html file in my project's templates/admin/ directory that starts with {% extends "admin/index.html" %} But it seems that this is referencing the local index file (a.k.a. itself) rather than the default Django copy. I want to extend the default Django template and simply change a few blocks. When I try this file, I get a recursion depth error. How can I extend parts of the admin? Thanks.

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  • django admin gives warning "Field 'X' doesn't have a default value"

    - by noam
    I have created two models out of an existing legacy DB , one for articles and one for tags that one can associate with articles: class Article(models.Model): article_id = models.AutoField(primary_key=True) text = models.CharField(max_length=400) class Meta: db_table = u'articles' class Tag(models.Model): tag_id = models.AutoField(primary_key=True) tag = models.CharField(max_length=20) article=models.ForeignKey(Article) class Meta: db_table = u'article_tags' I want to enable adding tags for an article from the admin interface, so my admin.py file looks like this: from models import Article,Tag from django.contrib import admin class TagInline(admin.StackedInline): model = Tag class ArticleAdmin(admin.ModelAdmin): inlines = [TagInline] admin.site.register(Article,ArticleAdmin) The interface looks fine, but when I try to save, I get: Warning at /admin/webserver/article/382/ Field 'tag_id' doesn't have a default value

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  • Django: Using 2 different AdminSite instances with different models registered

    - by omat
    Apart from the usual admin, I want to create a limited admin for non-staff users. This admin site will have different registered ModelAdmins. I created a folder /useradmin/ in my project directory and similar to contrib/admin/_init_.py I added an autodiscover() which will register models defined in useradmin.py modules instead of admin.py: # useradmin/__init__.py def autodiscover(): # Same as admin.autodiscover() but registers useradmin.py modules ... for app in settings.INSTALLED_APPS: mod = import_module(app) try: before_import_registry = copy.copy(site._registry) import_module('%s.useradmin' % app) except: site._registry = before_import_registry if module_has_submodule(mod, 'useradmin'): raise I also cretated sites.py under useradmin/ to override AdminSite similar to contrib/admin/sites: # useradmin/sites.py class UserAdminSite(AdminSite): def has_permission(self, request): # Don't care if the user is staff return request.user.is_active def login(self, request): # Do the login stuff but don't care if the user is staff if request.user.is_authenticated(): ... else: ... site = UserAdminSite(name='useradmin') In the project's URLs: # urls.py from django.contrib import admin import useradmin admin.autodiscover() useradmin.autodiscover() urlpatterns = patterns('', (r'^admin/', include(admin.site.urls)), (r'^useradmin/', include(useradmin.site.urls)), ) And I try to register different models in admin.py and useradmin.py modules under app directories: # products/useradmin.py import useradmin class ProductAdmin(useradmin.ModelAdmin): pass useradmin.site.register(Product, ProductAdmin) But when registering models in useradmin.py like useradmin.site.register(Product, ProductAdmin), I get 'module' object has no attribute 'ModelAdmin' exception. Though when I try this via shell; import useradmin from useradmin import ModelAdmin does not raise any exception. Any ideas what might be wrong? Edit: I tried going the @Luke way and arranged the code as follows as minimal as possible: (file paths are relative to the project root) # admin.py from django.contrib.admin import autodiscover from django.contrib.admin.sites import AdminSite user_site = AdminSite(name='useradmin') # urls.py (does not even have url patterns; just calls autodiscover()) import admin admin.autodiscover() # products/admin.py import admin from products.models import Product admin.user_site.register(Product) As a result I get an AttributeError: 'module' object has no attribute 'user_site' when admin.user_site.register(Product) in products/admin.py is called. Any ideas? Solution: I don't know if there are better ways but, renaming the admin.py in the project root to useradmin.py and updating the imports accordingly resolved the last case, which was a naming and import conflict.

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  • Error cloning gitosis-admin on new setup

    - by michaelmior
    I have the following in my gitosis.conf. (Created via gitsosis-init < id_rsa.pub with the key from my laptop) [gitosis] loglevel = DEBUG [group gitosis-admin] writable = gitosis-admin members = michael@laptop When I try git clone git@SERVER:gitsos-admin.git, I get the following errors: Initialized empty Git repository in /home/michael/gitsos-admin/.git/ DEBUG:gitosis.serve.main:Got command "git-upload-pack 'gitsos-admin.git'" DEBUG:gitosis.access.haveAccess:Access check for 'michael@laptop' as 'writable' on 'gitsos-admin.git'... DEBUG:gitosis.access.haveAccess:Stripping .git suffix from 'gitsos-admin.git', new value 'gitsos-admin' DEBUG:gitosis.group.getMembership:found 'michael@laptop' in 'gitosis-admin' DEBUG:gitosis.access.haveAccess:Access check for 'michael@laptop' as 'writeable' on 'gitsos-admin.git'... DEBUG:gitosis.access.haveAccess:Stripping .git suffix from 'gitsos-admin.git', new value 'gitsos-admin' DEBUG:gitosis.group.getMembership:found 'michael@laptop' in 'gitosis-admin' DEBUG:gitosis.access.haveAccess:Access check for 'michael@laptop' as 'readonly' on 'gitsos-admin.git'... DEBUG:gitosis.access.haveAccess:Stripping .git suffix from 'gitsos-admin.git', new value 'gitsos-admin' DEBUG:gitosis.group.getMembership:found 'michael@laptop' in 'gitosis-admin' ERROR:gitosis.serve.main:Repository read access denied fatal: The remote end hung up unexpectedly I know my key is being accepted because I have tried logging in via SSH and although a terminal won't be allocated, the authorization works.

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  • 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

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  • Help me find an appropriate ruby/python parser generator

    - by Geo
    The first parser generator I've worked with was Parse::RecDescent, and the guides/tutorials available for it were great, but the most useful feature it has was it's debugging tools, specifically the tracing capabilities ( activated by setting $RD_TRACE to 1 ). I am looking for a parser generator that can help you debug it's rules. The thing is, it has to be written in python or in ruby, and have a verbose mode/trace mode or very helpful debugging techniques. Does anyone know such a parser generator ? EDIT: when I said debugging, I wasn't referring to debugging python or ruby. I was referring to debugging the parser generator, see what it's doing at every step, see every char it's reading, rules it's trying to match. Hope you get the point. BOUNTY EDIT: to win the bounty, please show a parser generator framework, and illustrate some of it's debugging features. I repeat, I'm not interested in pdb, but in parser's debugging framework. Also, please don't mention treetop. I'm not interested in it.

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  • No access to Windows 2003 admin shares

    - by ARomo
    This is the environment: Several Win 2003 SP 2 servers and several Win XP SP2 & SP3 clients. All in the same LAN. Firewall is disabled everywhere. No recent Windows updates or configuration changes. This is the problem: Since last Thursday, I log on to any other server or workstation as any regular (non-admin) user and I fail to be able to open ADMIN SHARES ONLY (namely \\server1\c$, \\server1\e$ and \\server1\admin$). The error message is: "\server1\c$ is not accessible. You might not have permission to use this network resource. Contact the administrator of this server to find out if you have access permissions. Multiple connections to a server or shared resource by the same user, using more than one user name, are not allowed. Disconnect all previous connections to the server or shared resource and try again." I can, however, open the same shares if I use FQDN or IP address: \\server1.domain.local\c$ \\172.0.0.1\c$ Other shares do not have this issue and I can open them without any issue. Any ideas or suggestion would be truly appreciated. Thank you in advance.

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  • Django Admin: OneToOne Relation as an Inline?

    - by Jim Robert
    I am putting together the admin for a satchmo application. Satchmo uses OneToOne relations to extend the base Product model, and I'd like to edit it all on one page. It is possible to have a OneToOne relation as an Inline? If not, what is the best way to add a few fields to a given page of my admin that will eventually be saved into the OneToOne relation? for example: class Product(models.Model): name = models.CharField(max_length=100) ... class MyProduct(models.Model): product = models.OneToOne(Product) ... I tried this for my admin but it does not work, and seems to expect a Foreign Key: class ProductInline(admin.StackedInline): model = Product fields = ('name',) class MyProductAdmin(admin.ModelAdmin): inlines = (AlbumProductInline,) admin.site.register(MyProduct, MyProductAdmin) Which throws this error: <class 'satchmo.product.models.Product'> has no ForeignKey to <class 'my_app.models.MyProduct'> Is the only way to do this a Custom Form? edit: Just tried the following code to add the fields directly... also does not work: class AlbumAdmin(admin.ModelAdmin): fields = ('product__name',)

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  • Local User & Local Admin User Server 2008

    - by Ammo
    Hi I had a test recently and one of the questions was to create a file and local user and give the local user write permission to that file. I created the local user successfully however when I went to add permission to the file it would not find the local user when name was entered correctly, and idea what could have prevented this. Secondly I was asked to create a local admin account and give full permissions to the file, to my knowledge server 2008 has a built in admin account, and neither was the server a domain controller. Could you tell me what you would do in this situation? Many Thanks!

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  • Django Admin Actions missing

    - by Andrew C
    One of my Django sites is missing the Django Admin Action bar shown here: http://docs.djangoproject.com/en/dev/ref/contrib/admin/actions/#ref-contrib-admin-actions There is no checkbox next to each row and no Action select box near the top of the page. This is happening on every model. I have several sites running Django 1.1, and they all show the Admin Actions, so it feels like a local configuration issue. Anyone seen this before?

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  • Authlogic admin subsite

    - by MrThomas
    Following this tutorial getting the following errors: NameError in Admin/dashboardsController#show uninitialized constant Admin::DashboardsController NameError in Admin sessionController#new uninitialized constant Admin::AdminHelper not sure how to correct this!

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  • What do you think of this generator syntax?

    - by ChaosPandion
    I've been working on an ECMAScript dialect for quite some time now and have reached a point where I am comfortable adding new language features. I would love to hear some thoughts and suggestions on the syntax. Example generator { yield 1; yield 2; yield 3; if (true) { yield break; } yield continue generator { yield 4; yield 5; yield 6; }; } Syntax GeneratorExpression:     generator  {  GeneratorBody  } GeneratorBody:     GeneratorStatementsopt GeneratorStatements:     StatementListopt GeneratorStatement GeneratorStatementsopt GeneratorStatement:     YieldStatement     YieldBreakStatement     YieldContinueStatement YieldStatement:     yield  Expression  ; YieldBreakStatement:     yield  break  ; YieldContinueStatement:     yield  continue  Expression  ; Semantics The YieldBreakStatement allows you to end iteration early. This helps avoid deeply indented code. You'll be able to write something like this: generator { yield something1(); if (condition1 && condition2) yield break; yield something2(); if (condition3 && condition4) yield break; yield something3(); } instead of: generator { yield something1(); if (!condition1 && !condition2) { yield something2(); if (!condition3 && !condition4) { yield something3(); } } } The YieldContinueStatement allows you to combine generators: function generateNumbers(start) { return generator { yield 1 + start; yield 2 + start; yield 3 + start; if (start < 100) { yield continue generateNumbers(start + 1); } }; }

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  • How to customize a many-to-many inline model in django admin

    - by Jonathan
    I'm using the admin interface to view invoices and products. To make things easy, I've set the products as inline to invoices, so I will see the related products in the invoice's form. As you can see I'm using a many-to-many relationship. In models.py: class Product(models.Model): name = models.TextField() price = models.DecimalField(max_digits=10,decimal_places=2) class Invoice(models.Model): company = models.ForeignKey(Company) customer = models.ForeignKey(Customer) products = models.ManyToManyField(Product) In admin.py: class ProductInline(admin.StackedInline): model = Invoice.products.through class InvoiceAdmin(admin.ModelAdmin): inlines = [FilteredApartmentInline,] admin.site.register(Product, ProductAdmin) The problem is that django presents the products as a table of drop down menus (one per associated product). Each drop down contains all the products listed. So if I have 5000 products and 300 are associated with a certain invoice, django actually loads 300x5000 product names. Also the table is not aesthetic. How can I change it so that it'll just display the product's name in the inline table? Which form should I override, and how?

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  • ValueError with multi-table inheritance in Django Admin

    - by jorde
    I created two new classes which inherit model Entry: class Entry(models.Model): LANGUAGE_CHOICES = settings.LANGUAGES language = models.CharField(max_length=2, verbose_name=_('Comment language'), choices=LANGUAGE_CHOICES) user = models.ForeignKey(User) country = models.ForeignKey(Country, null=True, blank=True) created = models.DateTimeField(auto_now=True) class Comment(Entry): comment = models.CharField(max_length=2000, blank=True, verbose_name=_('Comment in English')) class Discount(Entry): discount = models.CharField(max_length=2000, blank=True, verbose_name=_('Comment in English')) coupon = models.CharField(max_length=2000, blank=True, verbose_name=_('Coupon code if needed')) After adding these new models to admin via admin.site.register I'm getting ValueError when trying to create a comment or a discount via admin. Adding entries works fine. Error msg: ValueError at /admin/reviews/discount/add/ Cannot assign "''": "Discount.discount" must be a "Discount" instance. Request Method: GET Request URL: http://127.0.0.1:8000/admin/reviews/discount/add/ Exception Type: ValueError Exception Value: Cannot assign "''": "Discount.discount" must be a "Discount" instance. Exception Location: /Library/Python/2.6/site-packages/django/db/models/fields/related.py in set, line 211 Python Executable: /usr/bin/python Python Version: 2.6.1

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  • Django admin.py missing field error

    - by user782400
    When I include 'caption', I get an error saying EntryAdmin.fieldsets[1][1]['fields']' refers to field 'caption' that is missing from the form In the admin.py; I have imported the classes from joe.models import Entry,Image Is that because my class from models.py is not getting imported properly ? Need help in resolving this issue. Thanks. models.py class Image(models.Model): image = models.ImageField(upload_to='joe') caption = models.CharField(max_length=200) imageSrc = models.URLField(max_length=200) user = models.CharField(max_length=20) class Entry(models.Model): image = models.ForeignKey(Image) mimeType = models.CharField(max_length=20) name = models.CharField(max_length=200) password = models.URLField(max_length=50) admin.py class EntryAdmin(admin.ModelAdmin): fieldsets = [ ('File info', {'fields': ['name','password']}), ('Upload image', {'fields': ['image','caption']})] list_display = ('name', 'mimeType', 'password') admin.site.register(Entry, EntryAdmin) admin.site.register(Image)

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  • How can I build a Truth Table Generator?

    - by KingNestor
    I'm looking to write a Truth Table Generator as a personal project. There are several web-based online ones here and here. (Example screenshot of an existing Truth Table Generator) I have the following questions: How should I go about parsing expressions like: ((P = Q) & (Q = R)) = (P = R) Should I use a parser generator like ANTLr or YACC, or use straight regular expressions? Once I have the expression parsed, how should I go about generating the truth table? Each section of the expression needs to be divided up into its smallest components and re-built from the left side of the table to the right. How would I evaluate something like that? Can anyone provide me with tips concerning the parsing of these arbitrary expressions and eventually evaluating the parsed expression?

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  • Dajano admin site foreign key fields

    - by user292652
    hi i have the following models setup class Player(models.Model): #slug = models.slugField(max_length=200) Player_Name = models.CharField(max_length=100) Nick = models.CharField(max_length=100, blank=True) Jersy_Number = models.IntegerField() Team_id = models.ForeignKey('Team') Postion_Choices = ( ('M', 'Manager'), ('P', 'Player'), ) Poistion = models.CharField(max_length=1, blank=True, choices =Postion_Choices) Red_card = models.IntegerField( blank=True, null=True) Yellow_card = models.IntegerField(blank=True, null=True) Points = models.IntegerField(blank=True, null=True) #Pic = models.ImageField(upload_to=path/for/upload, height_field=height, width_field=width, max_length=100) class PlayerAdmin(admin.ModelAdmin): list_display = ('Player_Name',) search_fields = ['Player_Name',] admin.site.register(Player, PlayerAdmin) class Team(models.Model): """Model docstring""" #slug = models.slugField(max_length=200) Team_Name = models.CharField(max_length=100,) College = models.CharField(max_length=100,) Win = models.IntegerField(blank=True, null=True) Loss = models.IntegerField(blank=True, null=True) Draw = models.IntegerField(blank=True, null=True) #logo = models.ImageField(upload_to=path/for/upload, height_field=height, width_field=width, max_length=100) class Meta: pass #def __unicode__(self): # return Team_Name #def save(self, force_insert=False, force_update=False): # pass @models.permalink def get_absolute_url(self): return ('view_or_url_name') class TeamAdmin(admin.ModelAdmin): list_display = ('Team_Name',) search_fields = ['Team_Name',] admin.site.register(Team, TeamAdmin) my question is how do i get to the admin site to show Team_name in the add player form Team_ID field currently it is only showing up as Team object in the combo box

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  • How to deal with multiple sub-type of one super-type in Django admin

    - by Henri
    What would be the best solution for adding/editing multiple sub-types. E.g a super-type class Contact with sub-type class Client and sub-type class Supplier. The way shown here works, but when you edit a Contact you get both inlines i.e. sub-type Client AND sub-type Supplier. So even if you only want to add a Client you also get the fields for Supplier of vice versa. If you add a third sub-type , you get three sub-type field groups, while you actually only want one sub-type group, in the mentioned example: Client. E.g.: class Contact(models.Model): contact_name = models.CharField(max_length=128) class Client(models.Model): contact = models.OneToOneField(Contact, primary_key=True) user_name = models.CharField(max_length=128) class Supplier(models.Model): contact.OneToOneField(Contact, primary_key=True) company_name = models.CharField(max_length=128) and in admin.py class ClientInline(admin.StackedInline): model = Client class SupplierInline(admin.StackedInline): model = Supplier class ContactAdmin(admin.ModelAdmin): inlines = (ClientInline, SupplierInline,) class ClientAdmin(admin.ModelAdmin): ... class SupplierAdmin(admin.ModelAdmin): ... Now when I want to add a Client, i.e. only a Client I edit Contact and I get the inlines for both Client and Supplier. And of course the same for Supplier. Is there a way to avoid this? When I want to add/edit a Client that I only see the Inline for Client and when I want to add/edit a Supplier that I only see the Inline for Supplier, when adding/editing a Contact? Or perhaps there is a different approach. Any help or suggestion will be greatly appreciated.

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