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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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  • Django admin - remove field if editing an object

    - by John McCollum
    I have a model which is accessible through the Django admin area, something like the following: # model class Foo(models.Model): field_a = models.CharField(max_length=100) field_b = models.CharField(max_length=100) # admin.py class FooAdmin(admin.ModelAdmin): pass Let's say that I want to show field_a and field_b if the user is adding an object, but only field_a if the user is editing an object. Is there a simple way to do this, perhaps using the fields attribute? If if comes to it, I could hack a JavaScript solution, but it doesn't feel right to do that at all!

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  • Django admin site populated combo box based on imput

    - by user292652
    hi i have to following model class Match(models.Model): Team_one = models.ForeignKey('Team', related_name='Team_one') Team_two = models.ForeignKey('Team', related_name='Team_two') Stadium = models.CharField(max_length=255, blank=True) Start_time = models.DateTimeField(auto_now_add=False, auto_now=False, blank=True, null=True) Rafree = models.CharField(max_length=255, blank=True) Judge = models.CharField(max_length=255, blank=True) Winner = models.ForeignKey('Team', related_name='winner', blank=True) updated = models.DateTimeField('update date', auto_now=True ) created = models.DateTimeField('creation date', auto_now_add=True ) def save(self, force_insert=False, force_update=False): pass @models.permalink def get_absolute_url(self): return ('view_or_url_name') class MatchAdmin(admin.ModelAdmin): list_display = ('Team_one','Team_two', 'Winner') search_fields = ['Team_one','Team_tow'] admin.site.register(Match, MatchAdmin) i was wondering is their a way to populated the winner combo box once the team one and team two is selected in admin site ?

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  • Objects with permissions assigned by django-guardian not visible in admin

    - by jul
    I'm using django-guardian in order to manage per object permission. For a given user I give permission all permission on one object: joe = User.objects.get(username="joe") mytask = Task.objects.get(pk=1) assign('add_task', joe, mytask) assign('change_task', joe, mytask) assign('delete_task', joe, mytask) and I get, as expected: In [57]: joe.has_perm("add_task", mytask) Out[57]: True In [58]: joe.has_perm("change_task", mytask) Out[58]: True In [59]: joe.has_perm("delete_task", mytask) Out[59]: True In admin.py I also make TaskAdmin inherit from GuardedModelAdmin instead of admin.ModelAdmin Now when I connect to my site with joe, on the admin I get: You don't have permission to edit anything Am I not supposed to be able to edit the object mytask? Do I have to set some permissions using the built-in model-based permission system? Am I missing anything? Thank you

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  • django-admin formfield_for_* change default value per/depending on instance

    - by Nick Ma.
    Hi, I'm trying to change the default value of a foreignkey-formfield to set a Value of an other model depending on the logged in user. But I'm racking my brain on it... This: Changing ForeignKey’s defaults in admin site would an option to change the empty_label, but I need the default_value. #Now I tried the following without errors but it didn't had the desired effect: class EmployeeAdmin(admin.ModelAdmin): ... def formfield_for_foreignkey(self, db_field, request=None, **kwargs): formfields= super(EmployeeAdmin, self).formfield_for_foreignkey(db_field, request, **kwargs) if request.user.is_superuser: return formfields if db_field.name == "company": #This is the RELEVANT LINE kwargs["initial"] = request.user.default_company return db_field.formfield(**kwargs) admin.site.register(Employee, EmployeeAdmin) ################################################################## # REMAINING Setups if someone would like to know it but i think # irrelevant concerning the problem ################################################################## from django.contrib.auth.models import User, UserManager class CompanyUser(User): ... objects = UserManager() company = models.ManyToManyField(Company) default_company= models.ForeignKey(Company, related_name='default_company') #I registered the CompanyUser instead of the standard User, # thats all up and working ... class Employee(models.Model): company = models.ForeignKey(Company) ... Hint: kwargs["default"] ... doesn't exist. Thanks in advance, Nick

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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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  • Django admin site auto populate combo box based on input

    - by user292652
    hi i have to following model class Match(models.Model): Team_one = models.ForeignKey('Team', related_name='Team_one') Team_two = models.ForeignKey('Team', related_name='Team_two') Stadium = models.CharField(max_length=255, blank=True) Start_time = models.DateTimeField(auto_now_add=False, auto_now=False, blank=True, null=True) Rafree = models.CharField(max_length=255, blank=True) Judge = models.CharField(max_length=255, blank=True) Winner = models.ForeignKey('Team', related_name='winner', blank=True) updated = models.DateTimeField('update date', auto_now=True ) created = models.DateTimeField('creation date', auto_now_add=True ) def save(self, force_insert=False, force_update=False): pass @models.permalink def get_absolute_url(self): return ('view_or_url_name') class MatchAdmin(admin.ModelAdmin): list_display = ('Team_one','Team_two', 'Winner') search_fields = ['Team_one','Team_tow'] admin.site.register(Match, MatchAdmin) i was wondering is their a way to populated the winner combo box once the team one and team two is selected in admin site ?

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  • Django Admin drop down combobox and assigned values

    - by Daniel Garcia
    I have several question for the Django Admin feature. Im kind of new in Django so im not sure how to do it. Basically what Im looking to do is when Im adding information on the model. Some of the fields i want them to be drop-downs and maybe combo-boxes with AutoCompleteMode. Also looking for some fields to have the same information, for example if i have a datatime field I want that information to feed the fields day, month and year from hoti.hotiapp.models import Occurrence from django.contrib import admin class MyModelAdmin(admin.ModelAdmin): exclude = ['reference',] admin.site.register(Occurrence, MyModelAdmin) Anything helps Thanks in advance

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  • django 1.1 beta issue

    - by ha22109
    Hello all, I m using django 1.1 beta.I m facing porblem in case of list_editable.First it was throughing exception saying need ordering in case of list_editable" then i added ordering in model but know it is giving me error.The code is working fine with django1.1 final. here is my code model.py class User(models.Model): advertiser = models.ForeignKey(WapUser,primary_key=True) status = models.CharField(max_length=20,choices=ADVERTISER_INVITE_STATUS,default='invited') tos_version = models.CharField(max_length=5) contact_email = models.EmailField(max_length=80) contact_phone = models.CharField(max_length=15) contact_mobile = models.CharField(max_length=15) contact_person = models.CharField(max_length=80) feedback=models.BooleanField(choices=boolean_choices,default=0) def __unicode__(self): return self.user.login class Meta: db_table = u'roi_advertiser_info' managed=False ordering=['feedback',] admin.py class UserAdmin(ReadOnlyAdminFields, admin.ModelAdmin): list_per_page = 15 fields = ['advertiser','contact_email','contact_phone','contact_mobile','contact_person'] list_display = ['advertiser','contact_email','contact_phone','contact_mobile','contact_person','status','feedback'] list_editable=['feedback'] readonly = ('advertiser',) search_fields = ['advertiser__login_id'] radio_fields={'approve_auto': admin.HORIZONTAL} list_filter=['status','feedback'] admin.site.register(User,UserADmin)

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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 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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  • Django as S3 proxy

    - by schneck
    Hi there, I extended a ModelAdmin with a custom field "Download file", which is a link to a URL in my Django project, like: http://www.myproject.com/downloads/1 There, I want to serve a file which is stored in a S3-bucket. The files in the bucket are not public readable, and the user may not have direct access to it. Now I want to avoid that the file has to be loaded in the server memory (these are multi-gb-files) avoid to have temp files on the server The ideal solution would be to let django act as a proxy that streams S3-chunks directly to the user. I use boto, but did not find a possibility to stream the chunks. Any ideas? Thanks.

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  • How do I flag only one of the formsets in django admin?

    - by azuer88
    I have these (simplified) models: class Question(models.Model): question = models.CharField(max_length=60) class Choices(models.Model): question = models.ForeignKey(Question) text = models.CharField(max_length=60) is_correct = models.BooleanField(default=False) I've made Choices as an inline of Question (in admin). Is there a way to make sure that only one Choice will have is_correct = True? Ideally, is_correct will be displayed as a radio button when it is displayed in the admin formset (TabularInline). my admin.py has: from django.contrib import admin class OptionInline(admin.TabularInline): model = Option extra = 5 max_num = 5 class QuestionAdmin(admin.ModelAdmin): inlines = [OptionInline, ] admin.site.register(QType) admin.site.register(Question, QuestionAdmin)

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  • list editabale error

    - by ha22109
    Hello all, I m using django 1.1 beta.I m facing porblem in case of list_editable.First it was throughing exception saying need ordering in case of list_editable" then i added ordering in model but know it is giving me error.The code is working fine with django1.1 final. here is my code model.py class User(models.Model): advertiser = models.ForeignKey(WapUser,primary_key=True) status = models.CharField(max_length=20,choices=ADVERTISER_INVITE_STATUS,default='invited') tos_version = models.CharField(max_length=5) contact_email = models.EmailField(max_length=80) contact_phone = models.CharField(max_length=15) contact_mobile = models.CharField(max_length=15) contact_person = models.CharField(max_length=80) feedback=models.BooleanField(choices=boolean_choices,default=0) def __unicode__(self): return self.user.login class Meta: db_table = u'roi_advertiser_info' managed=False ordering=['feedback',] admin.py class UserAdmin(ReadOnlyAdminFields, admin.ModelAdmin): list_per_page = 15 fields = ['advertiser','contact_email','contact_phone','contact_mobile','contact_person'] list_display = ['advertiser','contact_email','contact_phone','contact_mobile','contact_person','status','feedback'] list_editable=['feedback'] readonly = ('advertiser',) search_fields = ['advertiser__login_id'] radio_fields={'approve_auto': admin.HORIZONTAL} list_filter=['status','feedback'] admin.site.register(User,UserADmin)

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  • Why does sorl.thumbnail ImageField fail in the admin?

    - by Mark0978
    I have code that looks like this: from sorl.thumbnail import ImageField class Gallery(models.Model): pass class GalleryImage(models.Model): image = ImageField(upload_to='galleries') In the admin: class GalleryImageInline(admin.TabularInline): model = GalleryImage class GalleryAdmin(admin.ModelAdmin): inlines = (GalleryImageInline,) If I use the sorl.thumbnail as above, it is impossible to add images in the admin. I get the validation error Enter a list of values. If I replace the sorl.thumbnail.ImageField with a plain django ImageField, everything works. If I want sorl.thumbnail to clean up the cache thumbnails, I need to use it in the model, but if I use it in the model, I can't seem to add any images to need thumbnails. Anyone else found and fixed this problem yet?

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  • django admin how to limit selectbox values

    - by SledgehammerPL
    model: class Store(models.Model): name = models.CharField(max_length = 20) class Admin: pass def __unicode__(self): return self.name class Stock(Store): products = models.ManyToManyField(Product) class Admin: pass def __unicode__(self): return self.name class Product(models.Model): name = models.CharField(max_length = 128, unique = True) parent = models.ForeignKey('self', null = True, blank = True, related_name='children') (...) def __unicode__(self): return self.name mptt.register(Product, order_insertion_by = ['name']) admin.py: from bar.drinkstore.models import Store, Stock from django.contrib import admin admin.site.register(Store) admin.site.register(Stock) Now when I look at admin site I can select any product from the list. But I'd like to have a limited choice - only leaves. In mptt class there's function: is_leaf_node() -- returns True if the model instance is a leaf node (it has no children), False otherwise. But I have no idea how to connect it I'm trying to make a subclass: in admin.py: from bar.drinkstore.models import Store, Stock from django.contrib import admin admin.site.register(Store) class StockAdmin(admin.ModelAdmin): def queryset(self, request): qs = super(StockAdmin, self).queryset(request).filter(ihavenoideawhatfilter) admin.site.register(Stock, StockAdmin) but I'm not sure if it's right way, and what filter set.

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  • How to accept localized date format (e.g dd/mm/yy) in a DateField on an admin form ?

    - by tomjerry
    Is it possible to customize a django application to have accept localized date format (e.g dd/mm/yy) in a DateField on an admin form ? I have a model class : class MyModel(models.Model): date = models.DateField("Date") And associated admin class class MyModelAdmin(admin.ModelAdmin): pass On django administration interface, I would like to be able to input a date in following format : dd/mm/yyyy. However, the date field in the admin form expects yyyy-mm-dd. How can I customize things ? Nota bene : I have already specified my custom language code (fr-FR) in settings.py, but it seems to have no effect on this date input matter. Thanks in advance for your answer

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  • Dropdown sorting in django-admin

    - by Andrey
    I'd like to know how can I sort values in the Django admin dropdowns. For example, I have a model called Article with a foreign key pointing to the Users model, smth like: class Article(models.Model): title = models.CharField(_('Title'), max_length=200) slug = models.SlugField(_('Slug'), unique_for_date='publish') author = models.ForeignKey(User) body = models.TextField(_('Body')) status = models.IntegerField(_('Status')) categories = models.ManyToManyField(Category, blank=True) publish = models.DateTimeField(_('Publish date')) I edit this model in django admin: class ArticleAdmin(admin.ModelAdmin): list_display = ('title', 'publish', 'status') list_filter = ('publish', 'categories', 'status') search_fields = ('title', 'body') prepopulated_fields = {'slug': ('title',)} admin.site.register(Article, ArticleAdmin) and of course it makes the nice user select dropdown for me, but it's not sorted and it takes a lot of time to find a user by username.

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  • Inlines in Django Admin

    - by Oli
    I have two models, Order and UserProfile. Each Order has a ForeignKey to UserProfile, to associate it with that user. On the django admin page for each Order, I'd like to display the UserProfile associated with it, for easy processing of information. I have tried inlines: class UserInline(admin.TabularInline): model = UserProfile class ValuationRequestAdmin(admin.ModelAdmin): list_display = ('address1', 'address2', 'town', 'date_added') list_filter = ('town', 'date_added') ordering = ('-date_updated',) inlines = [ UserInline, ] But it complains that UserProfile "has no ForeignKey to" Order - which it doesn't, it's the other way around. Is there a way to do what I want?

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  • verbose_name for a model's method

    - by mawimawi
    How can I set a verbose_name for a model's method, so that it might be displayed in the admin's change_view form? example: class Article(models.Model): title = models.CharField(max_length=64) created_date = models.DateTimeField(....) def created_weekday(self): return self.created_date.strftime("%A") in admin.py: class ArticleAdmin(admin.ModelAdmin): readonly_fields = ('created_weekday',) fields = ('title', 'created_weekday') Now the label for created_weekday is "Created Weekday", but I'd like it to have a different label which should be i18nable using ugettext_lazy as well. I've tried created_weekday.verbose_name=... after the method, but that did not show any result. Is there a decorator or something I can use, so I could make my own "verbose_name" / "label" / whateverthename is?

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  • Django adminsite customize search_fields query

    - by dArignac
    Howdy! In the django admin you can set the search_fields for the ModelAdmin to be able to search over the properties given there. My model class has a property that is not a real model property, means it is not within the database table. The property relates to another database table that is not tied to the current model through relations. But I want to be able to search over it, so I have to somehow customize the query the admin site creates to do the filtering when the search field was filled - is this possible and if, how? I can query the database table of my custom property and it then returns the ids of the model classes fitting the search. This then, as I said, has to flow into the admin site search query. Thanks!

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  • Overriding the admin Media class

    - by shacker
    Given an admin media class that sets up a rich text editor, like: class TutorialAdmin(admin.ModelAdmin): fields... class Media: js = ['/paths/to/tinymce.js',] I would like the ability to selectively override js depending on a field value in the model it references. I've added a "use_editor" boolean to the Tutorial model. The question is, how can I detect whether the current instance has that bool set? I'd like to end up with something like: class Media: if self.use_editor: js = ['/path/to/tinymce.js',] else: js = '' Ideas? Thanks.

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  • 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

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  • How can I copy a queryset to a new model in django admin?

    - by user3806832
    I'm trying to write an action that allows the user to select the queryset and copy it to a new table. So: John, Mark, James, Tyler and Joe are in a table 1( called round 1) The user selects the action that say to "move to next round" and those same instances that were chosen are now also in the table for "round 2". I started trying with an action but don't really know where to go from here: def Round_2(modeladmin, request, queryset): For X in queryset: X.pk = None perform.short_description = "Move to Round 2" How can I copy them to the next table with all of their information (pk doesn't have to be the same)? Thanks

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  • Django: Extending User Model - Inline User fields in UserProfile

    - by Jack Sparrow
    Is there a way to display User fields under a form that adds/edits a UserProfile model? I am extending default Django User model like this: class UserProfile(models.Model): user = models.OneToOneField(User, unique=True) about = models.TextField(blank=True) I know that it is possible to make a: class UserProfileInlineAdmin(admin.TabularInline): and then inline this in User ModelAdmin but I want to achieve the opposite effect, something like inverse inlining, displaying the fields of the model pointed by the OneToOne Relationship (User) in the page of the model defining the relationship (UserProfile). I don't care if it would be in the admin or in a custom view/template. I just need to know how to achieve this. I've been struggling with ModelForms and Formsets, I know the answer is somewhere there, but my little experience in Django doesn't allow me to come up with the solution yet. A little example would be really helpful!

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