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• Advanced Django query with subselects and custom JOINS

  - by Bryan Ward

  I have been investigating this number theoretic function (found in the Height model) and I need to query for things based on the prime factorization of the primary key, or id. I have created a model for Factors of the id which maintains all of the prime factors. class Height(models.Model): b = models.IntegerField(null=True, blank=True) c = models.IntegerField(null=True, blank=True) d = models.FloatField(null=True, blank=True) class Factors(models.Model): height = models.ForeignKey(Height, null=True, blank=True) factor = models.IntegerField(null=True, blank=True) degree = models.IntegerField(null=True, blank=True) prime_id = models.IntegerField(null=True, blank=True) For example, if id=24, then the associated entries in the factors table would be height_id=24,factor=2,degree=3,prime_id=0 height_id=24,factor=3,degree=1,prime_id=1 the prime_id keep track of the relative order of the primes. Now let p < q < r < s all be prime numbers and a,b,c,d be positive integers. Then I want to be able to query for all Heights of the form id=(p**a)*(q**b)*(r**c)*(s**d). Now this is simple in the case that all of p,q,r,s,a,b,c,d are known in that I can just run Height.objects.get(id=(p**a)*(q**b)*(r**c)*(s**d)) But I need to be able to query for something like (2**a)*(3**2)*(r**c)*(s**d) where r,s,a,d are unknown and all Heights of such form will be returned. Furthermore, not all of the rows in Height will have exactly four prime factors, so I need to make sure that I am not matching rows of the form id=(p**a)*(q**b)*(r**c)*(s**d)*(t**e)... From what I can tell, the following MySQL query accomplishes this, but I would like to do it through the Django ORM. I also don't know if this MySQL query is the proper way to go about doing things. SELECT h.*,count(f.height_id) AS factorsCount FROM height AS h LEFT JOIN factors AS f ON ( f.height_id = h.id AND f.height_id IN (SELECT height_id FROM factors where prime_id=1 AND factor=2 AND degree=1) AND f.height_id IN (SELECT height_id FROM factors where prime_id=2 AND factor=3 AND degree=2) AND f.height_id IN (SELECT height_id FROM factors where prime_id=3 AND factor=5 AND degree=1) AND f.height_id IN (SELECT height_id FROM factors where prime_id=4 AND factor=7 ANd degree=1) ) GROUP BY h.id HAVING factorsCount=4 ORDER BY h.id; Any ideas or suggestions for things to try?

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• How to filter queryset in changelist_view in django admin?

  - by minder

  Let's say I have a site where Users can add Entries through admin panel. Each User has his own Category he is responsible for (each Category has an Editor assigned through ForeingKey/ManyToManyField). When User adds Entry, I limit the choices by using EntryAdmin like this: class EntryAdmin(admin.ModelAdmin): (...) def formfield_for_foreignkey(self, db_field, request, **kwargs): if db_field.name == 'category': if request.user.is_superuser: kwargs['queryset'] = Category.objects.all() else: kwargs['queryset'] = Category.objects.filter(editors=request.user) return db_field.formfield(**kwargs) return super(EntryAdmin, self).formfield_for_foreignkey(db_field, request, **kwargs) This way I can limit the categories to which a User can add Entry and it works perfect. Now the tricky part: On the Entry changelist/action page I want to show only those Entries which belong to current User's Category. I tried to do this using this method: def changelist_view(self, request, extra_context=None): if not request.user.is_superuser: self.queryset = self.queryset.filter(editors=request.user) But I get this error: AttributeError: 'function' object has no attribute 'filter' This is strange, because I thought it should be a typical QuerySet. Basically such methods are not well documented and digging through tons of Django code is not my favourite sport. Any ideas how can I achieve my goal?

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• How do you change a model name when it shows up in Django admin?

  - by Rhubarb

  I have a model name called StoreEntry. Django admin changes it to look like 'Store Entrys'. This is weird. If anything it should be Store entries. Any idea what's going on here and how to change it?

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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 version in GAE

  - by Alex

  I'm tring to use Django 1.1 in GAE, But when I uncomment use_library('django', '1.1') in this script import os os.environ['DJANGO_SETTINGS_MODULE'] = 'settings' from google.appengine.dist import use_library #use_library('django', '1.1') # Google App Engine imports. from google.appengine.ext.webapp import util # Force Django to reload its settings. from django.conf import settings settings._target = None import django.core.handlers.wsgi import django.core.signals import django.db import django.dispatch.dispatcher # Unregister the rollback event handler. django.dispatch.dispatcher.disconnect( django.db._rollback_on_exception, django.core.signals.got_request_exception) def main(): # Create a Django application for WSGI. application = django.core.handlers.wsgi.WSGIHandler() # Run the WSGI CGI handler with that application. util.run_wsgi_app(application) if __name__ == "__main__": main() I receives AttributeError: 'module' object has no attribute 'disconnect' What is going on?

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• django newbie question : cant start a new project

  - by Moayyad Yaghi

  hello . I'm totally new to django . and I'm using its documentation to get help on how to use it but seems like something is missing. i installed django using setup.py install command and i added the ( django/bin ) to system path variable but. i still cant start a new project i use the following syntax to start a project : django-admin.py startproject myNewProject but it says Type 'django-admin.py help' for usage. 1 do i miss anything ? thank u

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• Django 1.4 dependency when packaging a Precise application

  - by Caustic

  I am trying to package a program I wrote that depends on Django 1.4.1 in Ubuntu 12.04. As Django 1.4.1 isn't available in Precise I am wondering if it is best to: Package up Django 1.4.1 and drop it in my ppa OR write a script that wgets Django at build time and installs. OR Something better that I haven't thought of. I am still inexperienced with packaging and would appreciate some advice Thanks

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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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• Adding links to full change forms for inline items in django admin?

  - by David Eyk

  I have a standard admin change form for an object, with the usual StackedInline forms for a ForeignKey relationship. I would like to be able to link each inline item to its corresponding full-sized change form, as the inline item has inlined items of its own, and I can't nest them. I've tried everything from custom widgets to custom templates, and can't make anything work. So far, the "solutions" I've seen in the form of snippets just plain don't seem to work for inlines. I'm getting ready to try some DOM hacking with jQuery just to get it working and move on. I hope I must be missing something very simple, as this seems like such a simple task! Using Django 1.2.

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• unittest import error with virtualenv + google-app-engine-django

  - by Ray Yun

  I'm working with google-app-engine-django + zipped django. Just running "python manage.py test" succeeded without error. But with virtualenv, test was failed with "import unittest error". same error with Django 1.1. - OSX 10.5.6 - google-app-engine-django (r101 via svn) : r100 was failed with launcher 1.3.0 - GoogleAppLauncher 1.3.0 - Django 1.1 & 1.1.1 (zipped) : both failed - virtualenv 1.4.5 - virtualenvwrapper 1.24 Error Message: (django_appengine)Reiot:warclouds Reiot$ python manage.py test WARNING:root:Could not read datastore data from /var/folders/UZ/UZ1vQeLFH2ShHk4kIiLcFk+++TI/-Tmp-/django_google-app-engine-django.datastore INFO:root:zipimporter('/Volumes/data/Documents/warclouds/django.zip', 'django/core/serializers/') .WARNING:root:Can't open zipfile /Users/Reiot/.virtualenvs/django_appengine/lib/python2.5/site-packages/setuptools-0.6c11-py2.5.egg: IOError: [Errno 13] file not accessible: '/Users/Reiot/.virtualenvs/django_appengine/lib/python2.5/site-packages/setuptools-0.6c11-py2.5.egg' WARNING:root:Can't open zipfile /Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/site-packages/setuptools-0.6c9-py2.5.egg: IOError: [Errno 13] file not accessible: '/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/site-packages/setuptools-0.6c9-py2.5.egg' ERROR:root:Exception encountered handling request Traceback (most recent call last): File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 3177, in _HandleRequest self._Dispatch(dispatcher, self.rfile, outfile, env_dict) File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 3120, in _Dispatch base_env_dict=env_dict) File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 515, in Dispatch base_env_dict=base_env_dict) File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 2379, in Dispatch self._module_dict) File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 2289, in ExecuteCGI reset_modules = exec_script(handler_path, cgi_path, hook) File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 2185, in ExecuteOrImportScript exec module_code in script_module.__dict__ File "/Volumes/data/Documents/warclouds/main.py", line 28, in <module> from appengine_django import InstallAppengineHelperForDjango File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 1264, in Decorate return func(self, *args, **kwargs) File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 1914, in load_module return self.FindAndLoadModule(submodule, fullname, search_path) File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 1264, in Decorate return func(self, *args, **kwargs) File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 1816, in FindAndLoadModule description) File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 1264, in Decorate return func(self, *args, **kwargs) File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/tools/dev_appserver.py", line 1767, in LoadModuleRestricted description) File "/Volumes/data/Documents/warclouds/appengine_django/__init__.py", line 44, in <module> import unittest ImportError: No module named unittest INFO:root:"GET / HTTP/1.1" 500 - INFO:root:zipimporter('/Users/Reiot/.virtualenvs/django_appengine/lib/python2.5/site-packages/setuptools-0.6c11-py2.5.egg', '') INFO:root:zipimporter('/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/site-packages/setuptools-0.6c9-py2.5.egg', '') F........................................................... ====================================================================== FAIL: a request to the default page works in the dev_appserver ---------------------------------------------------------------------- Traceback (most recent call last): File "/Volumes/data/Documents/warclouds/appengine_django/tests/integration_test.py", line 176, in testBasic self.assertEquals(rv.status_code, 200) AssertionError: 500 != 200 I also tried with console import but it was ok. > which python /Users/Reiot/.virtualenvs/django_appengine/bin/python > python >>> import unittest Here is my environments: $ mkvirtualenv --no-site-packages no-django $ mkvirtualenv --no-site-packages django-1.1 $ mkvirtualenv --no-site-packages django-1.1.1 (django-1.1)$ easy_install Django-1.1.tar (django-1.1.1)$ easy_install Django-1.1.1.tar $ mkdir google-app-engine-django-svn $ cp -r google-app-engine-django-svn google-app-engine-django-svn-django-1.1 // copy appropriate django.zip $ cp -r google-app-engine-django-svn google-app-engine-django-svn-django-1.1.1 // copy appropriate django.zip

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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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• displaying list of registered user in django-admin

  - by theactiveactor

  My Book model has an author attribute which today is simply a CharField. The value for author should be one of the registered users of my Django site. When creating a new Book object in Django admin, I would like author to be displayed as a combo box showing all registered users. How would I go about achieving this?

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• Need "starting point" hints about adding "tabbed" interface to Django admin

  - by Edwin

  Hi, I'm new to the web development world - that means I'm new to javaScript/CSS. Now I'm building a web system with Python Django. I'm wondering would you like to give me some hints as the starting point for adding "tabbed" interface to Django admin? For example, there are 3 detail table for a master table, and I want to use 3 different tabs for editing that 3 detail tables in the 'edit' page for the master table. Thank you in advance!

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• Django admins disappearing from the admin index

  - by btol45

  We're running a Django website with rough 45 install Django admin classes. The handler is mod_fastcgi. Every once in a while about half the admins disappear from /admin/ screen. Touching the production.fcgi file restores everything to normal, but we have yet to determine the underling cause. Any thoughts on what the underlying issue might be?

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• Bringing admin to front-end in django

  - by martinthenext

  I'm trying to bring Django admin to front-end. The problem is in most cases I just have to write trivial model forms. Is there some way to use ModelAdmin forms outside of Django Admin?

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• Dynamically customize django admin columns ?

  - by tomjerry

  Is it possible to let the users choose / change dynamically the columns displayed in a object list in Django administration ? Things can surely be implemented "from scratch" by modifying the 'change_list.html' template but I was wondering if somebody has already had the same problem and/or if any django-pluggin can do that. Thanks in advance,

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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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• 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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• What's the straightforward way to implement one to many editing in list_editable in django admin?

  - by Nate Pinchot

  Given the following models: class Store(models.Model): name = models.CharField(max_length=150) class ItemGroup(models.Model): group = models.CharField(max_length=100) code = models.CharField(max_length=20) class ItemType(models.Model): store = models.ForeignKey(Store, on_delete=models.CASCADE, related_name="item_types") item_group = models.ForeignKey(ItemGroup) type = models.CharField(max_length=100) Inline's handle adding multiple item_types to a Store nicely when viewing a single Store. The content admin team would like to be able to edit stores and their types in bulk. Is there a simple way to implement Store.item_types in list_editable which also allows adding new records, similar to horizontal_filter? If not, is there a straightforward guide that shows how to implement a custom list_editable template? I've been Googling but haven't been able to come up with anything. Also, if there is a simpler or better way to set up these models that would make this easier to implement, feel free to comment.

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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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• Overwrite clean method in Django Custom Forms

  - by John

  Hi I have wrote a custom widget class AutoCompleteWidget(widgets.TextInput): """ widget to show an autocomplete box which returns a list on nodes available to be tagged """ def render(self, name, value, attrs=None): final_attrs = self.build_attrs(attrs, name=name) if not self.attrs.has_key('id'): final_attrs['id'] = 'id_%s' % name if not value: value = '[]' jquery = u""" <script type="text/javascript"> $("#%s").tokenInput('%s', { hintText: "Enter the word", noResultsText: "No results", prePopulate: %s, searchingText: "Searching..." }); $("body").focus(); </script> """ % (final_attrs['id'], reverse('ajax_autocomplete'), value) output = super(AutoTagWidget, self).render(name, "", attrs) return output + mark_safe(jquery) class MyForm(forms.Form): AutoComplete = forms.CharField(widget=AutoCompleteWidget) this widget uses a jquery function which autocompletes a word based on entries from the database. You can preset its initial values by setting prePopulate to a json string in the form ['name': 'some name', 'id': 'some id'] I do this by setting the inital value of the form field to this json string jquery_string = ['name': 'some name', 'id': 'some id'] form = MyForm(initial={'AutoComplete':jquery_string}) When submitting the form the the value of AutoComplete is returned as a comma seperated list of the selected ids e.g. 12,45,43,66 which if what I want. However if there is an error in the form, for example a required field has not been entered the value of the AutoComplete field is now 12,45,43,66 and not the json string which it requires. What is the best way to solve this. I was thinking about overwriting the clean method in the form class but I'm not sure how to find out if any other element has returned an error. e.g. if forms.errors form.cleaned_date['autocomplete'] = json string return form.cleaned_data Thanks

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• int() error in django views

  - by Hulk

  def displaydata(request): response_dict = {} offset = int(request.GET.get('iDisplayStart')) There is an error as, int() argument must be a string or a number at the above said line (i.e,`request.GET.get('iDisplayStart')) And in the template code, $(document).ready(function() { $.ajaxSetup({ cache: false }); oTable = $('#qp_table').dataTable( { "aoColumns": [ {"sWidth": "5%" }, {"sWidth": "35%" }, {"sWidth": "27%" }, {"sWidth": "15%"}, { "bSortable": false, "sWidth": "0%"}, {"bSortable": false, "sWidth": "0%"} ], "aaSorting": [[0, 'asc']], "bProcessing": true, "bServerSide": true, "sAjaxSource": "/diaplaydata/", "bJQueryUI": true, "sPaginationType": "full_numbers", "bFilter": false, "oLanguage" : { "sZeroRecords": "No data found", "sProcessing" : "Fetching Data" } });

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• django templates array assignment

  - by Hulk

  The following is in views: rows=query.evaluation_set.all() row_arr = [] for row in rows: row_arr.append(row.row_details) dict.update({'row_arr' : row_arr ,'col_arr' : col_arr}) return render_to_response('valuemart/show.html',context_instance=RequestContext(request,{'dict': dict})) How to extract the row_Arr array in the templates in javascript and list out all its values.row_Arr contains data of a column <script> var row_arr = '{{dict.row_arr}}'; //extract values here </script> Thanks..

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