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  • Django admin panel doesn't work after modify default user model.

    - by damienix
    I was trying to extend user profile. I founded a few solutions, but the most recommended was to create new user class containing foreign key to original django.contrib.auth.models.User class. I did it with this so i have in models.py: class UserProfile(models.Model): user = models.ForeignKey(User, unique=True) website_url = models.URLField(verify_exists=False) and in my admin.py from django.contrib import admin from someapp.models import * from django.contrib.auth.admin import UserAdmin # Define an inline admin descriptor for UserProfile model class UserProfileInline(admin.TabularInline): model = UserProfile fk_name = 'user' max_num = 1 # Define a new UserAdmin class class MyUserAdmin(UserAdmin): inlines = [UserProfileInline, ] # Re-register UserAdmin admin.site.unregister(User) admin.site.register(User, MyUserAdmin) And now when I'm trying to create/edit user in admin panel i have an error: "Unknown column 'content_userprofile.id' in 'field list'" where content is my appname. I was trying to add line AUTH_PROFILE_MODULE = 'content.UserProfile' to my settings.py but with no effect. How to tell panel admin to know how to correctly display fields in user form?

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  • Custom Django admin URL + changelist view for custom list filter by Tags

    - by Botondus
    In django admin I wanted to set up a custom filter by tags (tags are introduced with django-tagging) I've made the ModelAdmin for this and it used to work fine, by appending custom urlconf and modifying the changelist view. It should work with URLs like: http://127.0.0.1:8000/admin/reviews/review/only-tagged-vista/ But now I get 'invalid literal for int() with base 10: 'only-tagged-vista', error which means it keeps matching the review edit page instead of the custom filter page, and I cannot figure out why since it used to work and I can't find what change might have affected this. Any help appreciated. Relevant code: class ReviewAdmin(VersionAdmin): def changelist_view(self, request, extra_context=None, **kwargs): from django.contrib.admin.views.main import ChangeList cl = ChangeList(request, self.model, list(self.list_display), self.list_display_links, self.list_filter, self.date_hierarchy, self.search_fields, self.list_select_related, self.list_per_page, self.list_editable, self) cl.formset = None if extra_context is None: extra_context = {} if kwargs.get('only_tagged'): tag = kwargs.get('tag') cl.result_list = cl.result_list.filter(tags__icontains=tag) extra_context['extra_filter'] = "Only tagged %s" % tag extra_context['cl'] = cl return super(ReviewAdmin, self).changelist_view(request, extra_context=extra_context) def get_urls(self): from django.conf.urls.defaults import patterns, url urls = super(ReviewAdmin, self).get_urls() def wrap(view): def wrapper(*args, **kwargs): return self.admin_site.admin_view(view)(*args, **kwargs) return update_wrapper(wrapper, view) info = self.model._meta.app_label, self.model._meta.module_name my_urls = patterns('', # make edit work from tagged filter list view # redirect to normal edit view url(r'^only-tagged-\w+/(?P<id>.+)/$', redirect_to, {'url': "/admin/"+self.model._meta.app_label+"/"+self.model._meta.module_name+"/%(id)s"} ), # tagged filter list view url(r'^only-tagged-(P<tag>\w+)/$', self.admin_site.admin_view(self.changelist_view), {'only_tagged':True}, name="changelist_view"), ) return my_urls + urls Edit: Original issue fixed. I now receive 'Cannot filter a query once a slice has been taken.' for line: cl.result_list = cl.result_list.filter(tags__icontains=tag) I'm not sure where this result list is sliced, before tag filter is applied. Edit2: It's because of the self.list_per_page in ChangeList declaration. However didn't find a proper solution yet. Temp fix: if kwargs.get('only_tagged'): list_per_page = 1000000 else: list_per_page = self.list_per_page cl = ChangeList(request, self.model, list(self.list_display), self.list_display_links, self.list_filter, self.date_hierarchy, self.search_fields, self.list_select_related, list_per_page, self.list_editable, self)

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  • Django - Override admin site's login form

    - by TrojanCentaur
    I'm currently trying to override the default form used in Django 1.4 when logging in to the admin site (my site uses an additional 'token' field required for users who opt in to Two Factor Authentication, and is mandatory for site staff). Django's default form does not support what I need. Currently, I've got a file in my templates/ directory called templates/admin/login.html, which seems to be correctly overriding the template used with the one I use throughout the rest of my site. The contents of the file are simply as below: # admin/login.html: {% extends "login.html" %} The actual login form is as below: # login.html: {% load url from future %}<!DOCTYPE html> <html> <head> <title>Please log in</title> </head> <body> <div id="loginform"> <form method="post" action="{% url 'id.views.auth' %}"> {% csrf_token %} <input type="hidden" name="next" value="{{ next }}" /> {{ form.username.label_tag }}<br/> {{ form.username }}<br/> {{ form.password.label_tag }}<br/> {{ form.password }}<br/> {{ form.token.label_tag }}<br/> {{ form.token }}<br/> <input type="submit" value="Log In" /> </form> </div> </body> </html> My issue is that the form provided works perfectly fine when accessed using my normal login URLs because I supply my own AuthenticationForm as the form to display, but through the Django Admin login route, Django likes to supply it's own form to this template and thus only the username and password fields render. Is there any way I can make this work, or is this something I am just better off 'hard coding' the HTML fields into the form for?

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  • Django internationalization for admin pages - translate model name and attributes

    - by geekQ
    Django's internationalization is very nice (gettext based, LocaleMiddleware), but what is the proper way to translate the model name and the attributes for admin pages? I did not find anything about this in the documentation: http://docs.djangoproject.com/en/dev/topics/i18n/internationalization/ http://www.djangobook.com/en/2.0/chapter19/ I would like to have "???????? ????? ??? ?????????" instead of "???????? order ??? ?????????". Note, the 'order' is not translated. First, I defined a model, activated USE_I18N = True in settings.py, run django-admin makemessages -l ru. No entries are created by default for model names and attributes. Grepping in the Django source code I found: $ ack "Select %s to change" contrib/admin/views/main.py 70: self.title = (self.is_popup and ugettext('Select %s') % force_unicode(self.opts.verbose_name) or ugettext('Select %s to change') % force_unicode(self.opts.verbose_name)) So the verbose_name meta property seems to play some role here. Tried to use it: class Order(models.Model): subject = models.CharField(max_length=150) description = models.TextField() class Meta: verbose_name = _('order') Now the updated po file contains msgid 'order' that can be translated. So I put the translation in. Unfortunately running the admin pages show the same mix of "???????? order ??? ?????????". I'm currently using Django 1.1.1. Could somebody point me to the relevant documentation? Because google can not. ;-) In the mean time I'll dig deeper into the django source code...

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  • Question about rights for svn

    - by diadiora
    I have a web application written in asp.net mvc. I have in MyApp.Web assembly the list of views and and the content files(images, scripts, css, and so on). In MyApp.WebBase I have the rest of fonctionality(Controllers, domain(entities, repositories, services)). Now the question is the following: I want to give to third party html coder access only to MyApp.Web source code in order he to be able to compile the application locally and see the results. By other hand the developer team shoul have access to full source code. The problem is that in order the html coder to be able to compile the application locally he need in his project the references to the MyApp.WebBase.dll Can anyone help me? Thanks.

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  • how do you set the admin password on openldap 2.4

    - by dingfelder
    I am getting started with openLdap 2.4 and am having a bit of trouble, all the examples I see seem to refer to previous versions which used the text config file slapd.conf but from what I see on discussions about v2.4, this has been deprecated. I thought prehaps I needed to add a user, and log in as them but when I try and run an ldapadd command, I get a prompt to enter a password: Enter LDAP Password: ldap_bind: Invalid credentials (49) Notes: I installed openldap server via yum (in fedora 15), and have installed phpldapadminbut also can try things on the command line if anyone has suggestions. After installing and starting I get the following response from a search: # ldapsearch -x -b '' -s base '(objectclass=*)' namingContexts # extended LDIF # LDAPv3 # base <> with scope baseObject # filter: (objectclass=*) # requesting: namingContexts dn: namingContexts: dc=my-domain,dc=com # search result search: 2 result: 0 Success # numResponses: 2 # numEntries: 1 I am glad to remove and reinstall the server if that helps, can anyone provide a link to tips that works for version 2.4 for a new setup?

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  • Fed up of Server Admin - Want career change guidance

    - by JB04
    Hi All, I am SA in top level MNC and what I liked turned out to be my most disliked. I feel that I am capable of doing more than what I am doing at present. This 1 hour , 2 hour SLA is not my kind. I wanna get a better life.. The rotational shift is also something I am hating these days. Awkward shifts and too many process to follow. I have 3 years of Experience. I dont wanna waste this 3 yrs of experience I wanna get into OS developer or kind of so that this three years of experience is not wasted !! Please help me out

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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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  • Django: UserProfile with Unique Foreign Key in Django Admin

    - by lazerscience
    Hi, I have extended Django's User Model using a custom user profile called UserExtension. It is related to User through a unique ForeignKey Relationship, which enables me to edit it in the admin in an inline form! I'm using a signal to create a new profile for every new user: def create_user_profile(sender, instance, created, **kwargs): if created: try: profile, created = UserExtension.objects.get_or_create(user=instance) except: pass post_save.connect(create_user_profile, sender=User) (as described here for example: http://stackoverflow.com/questions/44109/extending-the-user-model-with-custom-fields-in-django) The problem is, that, if I create a new user through the admin, I get an IntegritiyError on saving "column user_id is not unique". It doesnt seem that the signal is called twice, but i guess the admin is trying to save the profile AFTERWARDS? But I need the creation through signal if I create a new user in other parts of the system!

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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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  • Django Admin interface with pickled set

    - by Rosarch
    I have a model that has a pickled set of strings. (It has to be pickled, because Django has no built in set field, right?) class Foo(models.Model): __bar = models.TextField(default=lambda: cPickle.dumps(set()), primary_key=True) def get_bar(self): return cPickle.loads(str(self.__bar)) def set_bar(self, values): self.__bar = cPickle.dumps(values) bar = property(get_bar, set_bar) I would like the set to be editable in the admin interface. Obviously the user won't be working with the pickled string directly. Also, the interface would need a widget for adding/removing strings from a set. What is the best way to go about doing this? I'm not super familiar with Django's admin system. Do I need to build a custom admin widget or something? Update: If I do need a custom widget, this looks helpful: http://www.fictitiousnonsense.com/archives/22

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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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  • Using Django Admin for a custom database solution

    - by Prashanth Ellina
    A client wants to have a simple intranet application to manage his process. He runs a Quarry and wishes to track number of loads delivered per day and associated activities. Since I knew about Django's excellent Admin interface, I figured I could define the "Schema" in models.py and have Django Admin generate the forms. I did exactly that and the result is not bad at all. I've been able to customize the look and feel to suit the client's taste. Some questions: Is Django Admin the right choice for such a use-case? Will I run to problems in the future due to flexibility of the framework? Is there a better framework out there specifically designed for this use-case (general Database management for small businesses)? I prefer ones written in Python since I can hack it up to customize. Thanks!

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  • django admin app error (Model with property field): global name 'full_name' is not defined

    - by rxin
    This is my model: class Author(models.Model): first_name = models.CharField(max_length=200) last_name = models.CharField(max_length=200) middle_name = models.CharField(max_length=200, blank=True) def __unicode__(self): return full_name def _get_full_name(self): "Returns the person's full name." if self.middle_name == '': return "%s %s" % (self.first_name, self.last_name) else: return "%s %s %s" % (self.first_name, self.middle_name, self.last_name) full_name = property(_get_full_name) Everything is fine except when I go into admin interface, I see TemplateSyntaxError at /bibbase2/admin/bibbase2/author/ Caught an exception while rendering: global name 'full_name' is not defined It seems like the built-in admin app doesn't work with a property field. Is there something wrong with my code?

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  • group inlines in django admin

    - by pablo
    Hi I have a two models, Model1 and Model2. Model2 has a FK to Model1 and FK to iteself. In the admin I show Model2 as inlines in Model1 change_form. I want to modify the way the inlines are shown in the admin. I need to group all the instances that have the same parent_model2 and display them as a readonly field with a string of 'childs' in the parent Model2 instance. I know how to use itertools.groupby (or the django version) but don't know how to do it in the admin. What should I override to be able to iterate over all the Model2 instances, group them by parent, add children to the parent and remove children from the inlines? class Model1(models.Model): name = models.CharField() class Model2(models.Model): name = models.CharField() fk_model1 = models.ForeignKey('self', blank=True, null=True) parent_model2 = models.ForeignKey('self', blank=True, null=True) Thanks

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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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  • Filtering foreign keys with AJAX in Django admin

    - by cnobile
    I have most of this figured out already. I have AJAX returning the region/state/province when a country is selected. The correct foreign key is saved to the database, however, when the record is viewed afterwards the selected state is not shown in the select nor are any states for the selected country. I understand why this is happening as, the admin view is not aware of the relation between the state and the country. So here is the question. Is there a hook in the admin view that will allow me to load the correct states for the country and set the selected attribute on the option in the select tag? Or how can I override the admin view for any forms that require the country and region/state/province set? I am using jQuery and Djando-1.1. Thanks

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  • Django Admin: Many-to-Many listbox doesn't show up with a through parameter

    - by NP
    Hi All, I have the following models: class Message(models.Model): date = models.DateTimeField() user = models.ForeignKey(User) thread = models.ForeignKey('self', blank=True, null=True) ... class Forum(models.Model): name = models.CharField(max_length=24) messages = models.ManyToManyField(Message, through="Message_forum", blank=True, null=True) ... class Message_forum(models.Model): message = models.ForeignKey(Message) forum = models.ForeignKey(Forum) status = models.IntegerField() position = models.IntegerField(blank=True, null=True) tags = models.ManyToManyField(Tag, blank=True, null=True) In the admin site, when I go to add/change a forum, I don't see the messages listbox as you'd expect. However, it shows up if I remove the 'through' parameter in the ManyToManyField declaration. What's up with that? I've registered all three models (plus Tag) to the admin site in admin.py. TIA

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  • Can not login Magento admin page after move to my localhost

    - by Xinrui Ma
    I just move my Magento store to my localhost environment for testing use, I also using Git to maintain code, but after I move all the files to my local environment, I can't login my admin page, but I can still see my frontend pages, and the git, the database, seems works well. When I type a wrong admin/password to my admin page, it still gives me "Invalid password". But when I enter the right one, it just refresh the page and stay at the login page, nothing happens. Does anyone has met this problem before? Has any ideas? Thanks in advance!

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  • Sorting related objects in the Django Admin form interface

    - by Carver
    I am looking to sort the related objects that show up when editing an object using the admin form. So for example, I would like to take the following object: class Person(models.Model): first_name = models.CharField( ... ) last_name = models.CharField( ... ) hero = models.ForeignKey( 'self', null=True, blank=True ) and edit the first name, last name and hero using the admin interface. I want to sort the objects as they show up in the drop down by last name, first name (ascending). How do I do that? Context I'm using Django v1.1. I started by looking for help in the django admin docs, but didn't find the solution As you can see in the example, the foreign key is pointing to itself, but I expect it would be the same as pointing to a different model object. Bonus points for being able to filter the related objects, too (eg~ only allow selecting a hero with the same first name)

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  • Django admin, hide a model

    - by Hellnar
    Hello At the root page of the admin site where registered models appear, I want to hide several models that are registered to the Django admin. If I directly unregister those, I am not able to add new records as the add new symbol "+" dissapears. How can this be done ?

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  • Edit/show Primary Key in Django Admin

    - by emcee
    It appears Django hides fields that are flagged Primary Key from being displayed/edited in the Django admin interface. Let's say I'd like to input data in which I may or may not want to specify a primary key. How would I go about displaying primary keys in Django-admin, and how could I make specifying it optional? Many thanks in advance, beloved hive-mind.

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  • django admin site make CharField a PasswordInput

    - by Paul
    I have a Django site in which the site admin inputs their Twitter Username/Password in order to use the Twitter API. The Model is set up like this: class TwitterUser(models.Model): screen_name = models.CharField(max_length=100) password = models.CharField(max_length=255) def __unicode__(self): return self.screen_name I need the Admin site to display the password field as a password input, but can't seem to figure out how to do it. I have tried using a ModelAdmin class, a ModelAdmin with a ModelForm, but can't seem to figure out how to make django display that form as a password input...

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  • Using ckEditor on selective text areas in django admin forms

    - by Rahul
    Hi, I want to apply ckeditor on specific textarea in django admin form not on all the text areas. Like snippet below will apply ckeditor on every textarea present on django form: class ProjectAdmin(admin.ModelAdmin): formfield_overrides = {models.TextField: {'widget': forms.Textarea(attrs={'class':'ckeditor'})}, } class Media: js = ('ckeditor/ckeditor.js',) but i want it on a specific textarea not on every textarea.

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