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• Making only a part of model field available in Django

  - by Hellnar

  Hello I have a such model: GENDER_CHOICES = ( ('M', 'Male'), ('F', 'Female') ) class Profile(models.Model): user = models.ForeignKey(User) gender = models.CharField(max_length=1, choices=GENDER_CHOICES) class FrontPage(models.Model): female = models.ForeignKey(User,related_name="female") male = models.ForeignKey(User,related_name="male") Once I attempt to add a new FrontPage object via the Admin page, I can select "Female" profiles for the male field of FrontPage, how can I restrict that? Thanks

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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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• How to display total record count against models in django admin

  - by Rog

  Is there a neat way to make the record/object count for a model appear on the main model list in the admin module? I have found techniques for showing counts of related objects within sets in the list_display page (and I can see the total in the pagination section at the bottom of the same), but haven't come across a neat way to show the record count at the model list level.

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• Alternate datasource for django model?

  - by slypete

  I'm trying to seamlessly integrate some legacy data into a django application. I would like to know if it's possible to use an alternate datasource for a django model. For example, can I contact a server to populate a list of a model? The server would not be SQL based at all. Instead it uses some proprietary tcp based protocol. Copying the data is not an option, as the legacy application will continue to be used for some time. Would a custom manager allow me to do this? This model should behave just like any other django model. It should even pluggable to the admin interface. What do you think? Thanks, Pete

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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 Custom Template Tages: Inclusion Tags

  - by Harry

  Hello world! Im trieng to build my own template tags Im have no idea why I get the errors I get, im following the django doc's. this is my file structure of my app: pollquiz/ __init__.py show_pollquiz.html showpollquiz.py This is showpollquiz.py: from django import template from pollquiz.models import PollQuiz, Choice register = template.Library() @register.inclusion_tag('show_pollquiz.html') def show_poll(): poll = Choice.objects.all() return { 'poll' : poll } html file: <ul> {% for poll in poll <li>{{ poll.pollquiz }}</li> {% endfor </ul> in my base.html file im am including like this {% load showpollquiz %} and {% poll_quiz %} Bu then I get the the error: Exception Value: Caught an exception while rendering: show_pollquiz.html I have no idea why this happens. Any ideas? Please keep in mind Im still new to Django

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• UnicodeEncodeError when uploading files in Django admin

  - by Samuel Linde

  Note: I asked this question on StackOverflow, but I realize this might be a more proper place to ask this kind of question. I'm trying to upload a file called 'Testaråäö.txt' via the Django admin app. I'm running Django 1.3.1 with Gunicorn 0.13.4 and Nginx 0.7.6.7 on a Debian 6 server. Database is PostgreSQL 8.4.9. Other Unicode data is saved to the database with no problem, so I guess the problem must be with the filesystem somehow. I've set http { charset utf-8; } in my nginx.conf. LC_ALL and LANG is set to 'sv_SE.UTF-8'. Running 'locale' verifies this. I even tried setting LC_ALL and LANG in my nginx init script just to make sure locale is set properly. Here's the traceback: Traceback (most recent call last): File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/core/handlers/base.py", line 111, in get_response response = callback(request, *callback_args, **callback_kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/contrib/admin/options.py", line 307, in wrapper return self.admin_site.admin_view(view)(*args, **kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/utils/decorators.py", line 93, in _wrapped_view response = view_func(request, *args, **kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/views/decorators/cache.py", line 79, in _wrapped_view_func response = view_func(request, *args, **kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/contrib/admin/sites.py", line 197, in inner return view(request, *args, **kwargs) File "/srv/django/letebo/app/cms/admin.py", line 81, in change_view return super(PageAdmin, self).change_view(request, obj_id) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/utils/decorators.py", line 28, in _wrapper return bound_func(*args, **kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/utils/decorators.py", line 93, in _wrapped_view response = view_func(request, *args, **kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/utils/decorators.py", line 24, in bound_func return func(self, *args2, **kwargs2) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/db/transaction.py", line 217, in inner res = func(*args, **kwargs) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/contrib/admin/options.py", line 985, in change_view self.save_formset(request, form, formset, change=True) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/contrib/admin/options.py", line 677, in save_formset formset.save() File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/forms/models.py", line 482, in save return self.save_existing_objects(commit) + self.save_new_objects(commit) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/forms/models.py", line 613, in save_new_objects self.new_objects.append(self.save_new(form, commit=commit)) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/forms/models.py", line 717, in save_new obj.save() File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/db/models/base.py", line 460, in save self.save_base(using=using, force_insert=force_insert, force_update=force_update) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/db/models/base.py", line 504, in save_base self.save_base(cls=parent, origin=org, using=using) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/db/models/base.py", line 543, in save_base for f in meta.local_fields if not isinstance(f, AutoField)] File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/db/models/fields/files.py", line 255, in pre_save file.save(file.name, file, save=False) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/db/models/fields/files.py", line 92, in save self.name = self.storage.save(name, content) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/core/files/storage.py", line 48, in save name = self.get_available_name(name) File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/core/files/storage.py", line 74, in get_available_name while self.exists(name): File "/srv/.virtualenvs/letebo/lib/python2.6/site-packages/django/core/files/storage.py", line 218, in exists return os.path.exists(self.path(name)) File "/srv/.virtualenvs/letebo/lib/python2.6/genericpath.py", line 18, in exists st = os.stat(path) UnicodeEncodeError: 'ascii' codec can't encode characters in position 52-54: ordinal not in range(128) I tried running Gunicorn with debugging turned on, and the file uploads without any problem at all. I suppose this must mean that the issue is with Nginx. Still beats me where to look, though. Here are the raw response headers from Gunicorn and Nginx, if it makes any sense: Gunicorn: HTTP/1.1 302 FOUND Server: gunicorn/0.13.4 Date: Thu, 09 Feb 2012 14:50:27 GMT Connection: close Transfer-Encoding: chunked Expires: Thu, 09 Feb 2012 14:50:27 GMT Vary: Cookie Last-Modified: Thu, 09 Feb 2012 14:50:27 GMT Location: http://my-server.se:8000/admin/cms/page/15/ Cache-Control: max-age=0 Content-Type: text/html; charset=utf-8 Set-Cookie: messages="yada yada yada"; Path=/ Nginx: HTTP/1.1 500 INTERNAL SERVER ERROR Server: nginx/0.7.67 Date: Thu, 09 Feb 2012 14:50:57 GMT Content-Type: text/html; charset=utf-8 Transfer-Encoding: chunked Connection: close Vary: Cookie 500 UPDATE: Both locale.getpreferredencoding() and sys.getfilesystemencoding() outputs 'UTF-8'. locale.getdefaultlocale() outputs ('sv_SE', 'UTF8'). This seem correct to me, so I'm still not sure why I keep getting these errors.

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• Django: How do I add arbitrary html attributes to input fields on a form?

  - by User

  I have an input field that is rendered with a template like so: <div class="field"> {{ form.city }} </div> Which is rendered as: <div class="field"> <input id="id_city" type="text" name="city" maxlength="100" /> </div> Now suppose I want to add an autocomplete="off" attribute to the input element that is rendered, how would I do that? Or onclick="xyz()" or class="my-special-css-class"?

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• Trying to Nullify Django model fields with method where model and fields are parameters

  - by Johnny4000

  I'm trying to write a method like the below where a list of fields (a subset of all the fields) is passed in as a parameter and has their column values set to null. I would be happy of I could get a method with just the fields as a parameter like below, but having the model as a parameter would be even better. from my_project.my_app.models import MyModel def nullify_columns (self, null_fields): field_names = MyModel._meta.get_all_field_names() for field in field_names: if field in null_fields: # The below line does not work because I'm not sure how to # dynamically assign the field name. MyModel.objects.all().update( (MyModel.get_field(field).column) = None) Right now I have something like if 'column1' in list_of_fields: MyModel.objects.all().update(column1 = None) if 'column2' in list_of_fields: MyModel.objects.all().update(column2 = None) etc. which is horrible, but works.

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• Django's forms.Form vs forms.ModelForm

  - by Viet

  Could anyone please explain to me similarities and differences of Django's forms.Form & forms.ModelForm? I've used Django for 4 days, so excuse me if I don't know basic things. Thanks in advance!

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• Insert django form into template dynamically using javascript??

  - by qulzam

  I want to add same django form instance on same template. i already add one before and other add dynamically using javascript. for example 'form" is a django form: newcell.innerHTML = {{ form.firstname }}; The problem is that when i submit the form, in view the request object has only one value (that is not add using javascript). how can i get the values of other form elements values that is added dynamically runtime.

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• Django : debugging templatetags

  - by interstar

  How on earth do people debug Django templatetags? I created one, based on a working example, my new tag looks the same to me as the existing one. But I just get a 'my_lib' is not a valid tag library: Could not load template library from django.templatetags.my_lib, No module named my_lib I know that this is probably because of something failing when defining the lib. But how do I see what's going on? What do you use to debug this situation?

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• Django models query

  - by Hulk

  Code: class criteria(models.Model): details = models.CharField(max_length = 512) Headerid = models.ForeignKey(Header) def __unicode__(self): return self.id() the details corresponds to a textarea in the UI and a validation is done for 512 characters but when this is saved. /home/project/django/django/core/handlers/base.py in get_response, line 109 Is this any thing related with schema or number of characters entered from UI

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• Tying in to Django Admin's Model History

  - by akdom

  The Setup: I'm working on a Django application which allows users to create an object in the database and then go back and edit it as much as they desire. Django's admin site keeps a history of the changes made to objects through the admin site. The Question: How do I hook my application in to the admin site's change history so that I can see the history of changes users make to their "content" ?

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• Django: Generating a queryset from a GET request

  - by Nimmy Lebby

  I have a Django form setup using GET method. Each value corresponds to attributes of a Django model. What would be the most elegant way to generate the query? Currently this is what I do in the view: def search_items(request): if 'search_name' in request.GET: query_attributes = {} query_attributes['color'] = request.GET.get('color', '') if not query_attributes['color']: del query_attributes['color'] query_attributes['shape'] = request.GET.get('shape', '') if not query_attributes['shape']: del query_attributes['shape'] items = Items.objects.filter(**query_attributes) But I'm pretty sure there's a better way to go about it.

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• Django attribute error: 'module' object has no attribute 'is_usable'

  - by Robert A Henru

  Hi, I got the following error when calling the url in Django. It's working before, I guess it's related with some accidental changes I made, but I have no idea what they are. Thanks before for the help, Robert Environment: Request Method: GET Request URL: http://localhost:8000/time/ Django Version: 1.2 Python Version: 2.6.1 Installed Applications: ['django.contrib.auth', 'django.contrib.contenttypes', 'django.contrib.sessions', 'django.contrib.sites', 'django.contrib.messages', 'django.contrib.admin', 'djlearn.books'] Installed Middleware: ('django.middleware.common.CommonMiddleware', 'django.contrib.sessions.middleware.SessionMiddleware', 'django.middleware.csrf.CsrfViewMiddleware', 'django.contrib.auth.middleware.AuthenticationMiddleware', 'django.contrib.messages.middleware.MessageMiddleware') Traceback: File "/Library/Python/2.6/site-packages/django/core/handlers/base.py" in get_response 100. response = callback(request, *callback_args, **callback_kwargs) File "/Users/rhenru/Workspace/django/djlearn/src/djlearn/../djlearn/views.py" in current_datetime 16. return render_to_response('current_datetime.html',{'current_date':now,}) File "/Library/Python/2.6/site-packages/django/shortcuts/__init__.py" in render_to_response 20. return HttpResponse(loader.render_to_string(*args, **kwargs), **httpresponse_kwargs) File "/Library/Python/2.6/site-packages/django/template/loader.py" in render_to_string 181. t = get_template(template_name) File "/Library/Python/2.6/site-packages/django/template/loader.py" in get_template 157. template, origin = find_template(template_name) File "/Library/Python/2.6/site-packages/django/template/loader.py" in find_template 128. loader = find_template_loader(loader_name) File "/Library/Python/2.6/site-packages/django/template/loader.py" in find_template_loader 111. if not func.is_usable: Exception Type: AttributeError at /time/ Exception Value: 'module' object has no attribute 'is_usable'

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• Embedded Model Designing -- top down or bottom up?

  - by Jeff

  I am trying to learn RoR and develop a webapp. I have a few models I have thought of for this app, and they are fairly embedded. For example (please excuse my lack of RoR syntax): Model: textbook title:string type:string has_many: chapters Model: chapter content:text has_one: review_section Model: review_section title:string has_many: questions has_many: answers , through :questions Model: questions ... Model: answers ... My question is, with the example I gave, should I start at the top model (textbook) or the bottom most (answers)?

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• django crispy-forms inline forms

  - by abolotnov

  I'm trying to adopt crispy-forms and bootstrap and use as much of their functionality as possible instead of inventing something over and over again. Is there a way to have inline forms functionality with crispy-forms/bootstrap like django-admin forms have? Here is an example: class NewProjectForm(forms.Form): name = forms.CharField(required=True, label=_(u'???????? ???????'), widget=forms.TextInput(attrs={'class':'input-block-level'})) group = forms.ModelChoiceField(required=False, queryset=Group.objects.all(), label=_(u'?????? ????????'), widget=forms.Select(attrs={'class':'input-block-level'})) description = forms.CharField(required=False, label=_(u'???????? ???????'), widget=forms.Textarea(attrs={'class':'input-block-level'})) class Meta: model = Project fields = ('name','description','group') def __init__(self, *args, **kwargs): self.helper = FormHelper() self.helper.form_class = 'horizontal-form' self.helper.form_action = 'submit_new_project' self.helper.layout = Layout( Field('name', css_class='input-block-level'), Field('group', css_class='input-block-level'), Field('description',css_class='input-block-level'), ) self.helper.add_input(Submit('submit',_(u'??????? ??????'))) self.helper.add_input(Submit('cancel',_(u'? ?????????'))) super(NewProjectForm, self).__init__(*args, **kwargs) it will display a decent form: How do I go about adding a form that basically represents this model: class Link(models.Model): name = models.CharField(max_length=255, blank=False, null=False, verbose_name=_(u'????????')) url = models.URLField(blank=False, null=False, verbose_name=_(u'??????')) project = models.ForeignKey('Project') So there will be a project and name/url links and way to add many, like same thing is done in django-admin where you are able to add extra 'rows' with data related to your main model. On the sreenshot below you are able to fill out data for 'Question' object and below that you are able to add data for QuestionOption objects -you are able to click the '+' icon to add as many QuestionOptions as you want. I'm not looking for a way to get the forms auto-generated from models (that's nice but not the most important) - is there a way to construct a form that will let you add 'rows' of data like django-admin does?

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• ProgrammingError when aggregating over an annotated & grouped Django ORM query

  - by ento

  I'm trying to construct a query to get the "average, maximum, minimum number of items purchased by a single user". The data source is this simple sales record table: class SalesRecord(models.Model): id = models.IntegerField(primary_key=True) user_id = models.IntegerField() product_code = models.CharField() price = models.IntegerField() created_at = models.DateTimeField() A new record is inserted into this table for every item purchased by a user. Here's my attempt at building the query: q = SalesRecord.objects.all() q = q.values('user_id').annotate( # group by user and count the # of records count=Count('id'), # (= # of items) ).order_by() result = q.aggregate(Max('count'), Min('count'), Avg('count')) When I try to execute the code, a ProgrammingError is raised at the last line: (1064, "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FROM (SELECT sales_records.user_id AS user_id, COUNT(sales_records.`' at line 1") Django's error screen shows that the SQL is SELECT FROM (SELECT `sales_records`.`player_id` AS `player_id`, COUNT(`sales_records`.`id`) AS `count` FROM `sales_records` WHERE (`sales_records`.`created_at` >= %s AND `sales_records`.`created_at` <= %s ) GROUP BY `sales_records`.`player_id` ORDER BY NULL) subquery It's not selecting anything! Can someone please show me the right way to do this? Hacking Django I've found that clearing the cache of selected fields in django.db.models.sql.BaseQuery.get_aggregation() seems to solve the problem. Though I'm not really sure this is a fix or a workaround. @@ -327,10 +327,13 @@ # Remove any aggregates marked for reduction from the subquery # and move them to the outer AggregateQuery. + self._aggregate_select_cache = None + self.aggregate_select_mask = None for alias, aggregate in self.aggregate_select.items(): if aggregate.is_summary: query.aggregate_select[alias] = aggregate - del obj.aggregate_select[alias] + if alias in obj.aggregate_select: + del obj.aggregate_select[alias] ... yields result: {'count__max': 267, 'count__avg': 26.2563, 'count__min': 1}

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• Django - provide additional information in template

  - by Ninefingers

  Hi all, I am building an app to learn Django and have started with a Contact system that currently stores Contacts and Addresses. C's are a many to many relationship with A's, but rather than use Django's models.ManyToManyField() I've created my own link-table providing additional information about the link, such as what the address type is to the that contact (home, work etc). What I'm trying to do is pass this information out to a view, so in my full view of a contact I can do this: def contact_view_full(request, contact_id): c = get_object_or_404(Contact, id=contact_id) a = [] links = ContactAddressLink.objects.filter(ContactID=c.id) for link in links: b = Address.objects.get(id=link.AddressID_id) a.append(b) return render_to_response('contact_full.html', {'contact_item': c, 'addresses' : a }, context_instance=RequestContext(request)) And so I can do the equivalent of c.Addresses.all() or however the ManyToManyField works. What I'm interested to know is how can I pass out information about the link in the link object with the 'addresses' : a information, so that when my template does this: {% for address in addresses %} <!-- ... --> {% endfor %} and properly associate the correct link object data with the address. So what's the best way to achieve this? I'm thinking a union of two objects might be an idea but I haven't enough experience with Django to know if that's considered the best way of doing it. Suggestions? Thanks in advance. Nf

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• What is causing this OverflowError in Django?

  - by orokusaki

  I'm using a normal ModelForm.save() to create an object, and this exception comes up. It worked fine before until I added commit_manually, transaction.rollback() and transaction.commit() to my view. Has anyone else ran into this? Is this because of sqlite3? OverflowError: long too big to convert C:\Python26\Lib\site-packages\django-trunk\django\db\backends\sqlite3\base.py in execute, line 197 params: (203866156270872165269663274649746494334L,) query: u'SELECT (1) AS "a", "auth_user"."id", "auth_user"."username", "auth_user"."first_name", "auth_user"."last_name", "auth_user"."email", "auth_user"."password", "auth_user"."is_staff", "auth_user"."is_active", "auth_user"."is_superuser", "auth_user"."last_login", "auth_user"."date_joined" FROM "auth_user" WHERE "auth_user"."id" = ? LIMIT 1' self <django.db.backends.sqlite3.base.SQLiteCursorWrapper object at 0x015D5A98> Why would that L param be passed in, and

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• Asynchronous daemon processing / ORM interaction with Django

  - by perrierism

  I'm looking for a way to do asynchronous data processing with a daemon that uses Django ORM. However, the ORM isn't thread-safe; it's not thread-safe to try to retrieve / modify django objects from within threads. So I'm wondering what the correct way to achieve asynchrony is? Basically what I need to accomplish is taking a list of users in the db, querying a third party api and then making updates to user-profile rows for those users. As a daemon or background process. Doing this in series per user is easy, but it takes too long to be at all scalable. If the daemon is retrieving and updating the users through the ORM, how do I achieve processing 10-20 users at a time? I would use a standard threading / queue system for this but you can't thread interactions like models.User.objects.get(id=foo) ... Django itself is an asynchronous processing system which makes asynchronous ORM calls(?) for each request, so there should be a way to do it? I haven't found anything in the documentation so far. Cheers

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• django-admin: creating,saving and relating a m2m model

  - by pastylegs

  I have two models: class Production(models.Model): gallery = models.ManyToManyField(Gallery) class Gallery(models.Model): name = models.CharField() I have the m2m relationship in my productions admin, but I want that functionality that when I create a new Production, a default gallery is created and the relationship is registered between the two. So far I can create the default gallery by overwriting the productions save: def save(self, force_insert=False, force_update=False): if not ( Gallery.objects.filter(name__exact="foo").exists() ): g = Gallery(name="foo") g.save() self.gallery.add(g) This creates and saves the model instance (if it doesn't already exist), but I don't know how to register the relationship between the two?

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• Django and ImageField Question

  - by Hellnar

  Hello I have a such model: Foo (models.Model): slug = models.SlugField(unique=True) image = models.ImageField(upload_to='uploads/') I want to do two things with this: First of all, I want my image to be forced to resize to a specific width and height after the upload. I have tried this reading the documentation but seems to getting error: image = models.ImageField(upload_to='uploads/', height_field=258, width_field=425) Secondly, when adding an item via admin panel, I want my image's file name to be renamed as same as slug, if any issue arises (like if such named image already exists, add "_" to the end as it used to do. IE: My slug is i-love-you-guys , uploaded image such have i-love-you-guys.png at the end.

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• In Django-pagination Paginate does not working...

  - by mosg

  Hello. Python 2.6.2 django-pagination 1.0.5 Question: How to force pagination work correctly? The problem is that {% paginate %} does not work, but other {% load pagination_tags %} and {% autopaginate object_list 10 %} works! Error message appeared, when I add {% paginate %} into html page: TemplateSyntaxError at /logging Caught an exception while rendering: pagination/pagination.html What I have done: Install django-pagination without any problems. When I do in python import pagination, it's work well. Added pagination to INSTALLED_APP in settings.py: INSTALLED_APPS = ( # ..., 'pagination', ) Added in settings.py: TEMPLATE_CONTEXT_PROCESSORS = ( "django.core.context_processors.auth", "django.core.context_processors.debug", "django.core.context_processors.i18n", "django.core.context_processors.media", "django.core.context_processors.request" ) Also add to settings.py middleware: MIDDLEWARE_CLASSES = ( # ... 'pagination.middleware.PaginationMiddleware', ) Add to top in views.py: from django.template import RequestContext And finally add to my HTML template page lines: {% load pagination_tags %} ... {% autopaginate item_list 50 %} {% for item in item_list %} ... {% endfor %} {% paginate %} Thanks. PS: some edits required, because I can't django code style work well here :)

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