Search Results

Search found 9098 results on 364 pages for 'django admin'.

Page 4/364 | < Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >

• Django Formset validation with an optional ForeignKey field

  - by Camilo Díaz

  Having a ModelFormSet built with modelformset_factory and using a model with an optional ForeignKey, how can I make empty (null) associations to validate on that form? Here is a sample code: ### model class Prueba(models.Model): cliente = models.ForeignKey(Cliente, null = True) valor = models.CharField(max_length = 20) ### view def test(request): PruebaFormSet = modelformset_factory(model = Prueba, extra = 1) if request.method == 'GET': formset = PruebaFormSet() return render_to_response('tpls/test.html', {'formset' : formset}, context_instance = RequestContext(request)) else: formset = PruebaFormSet(request.POST) # dumb tests, just to know if validating if formset.is_valid(): return HttpResponse('0') else: return HttpResponse('1') In my template, i'm just calling the {{ form.cliente }} method which renders the combo field, however, I want to be able to choose an empty (labeled "------") value, as the FK is optional... but when the form gets submitted it doesn't validate. Is this normal behaviour? How can i make this field to skip required validation?

  Read the article

• Django MultiWidget Phone Number Field

  - by Birdman

  I want to create a field for phone number input that has 2 text fields (size 3, 3, and 4 respectively) with the common "(" ")" "-" delimiters. Below is my code for the field and the widget, I'm getting the following error when trying to iterate the fields in my form during initial rendering (it happens when the for loop gets to my phone number field): Caught an exception while rendering: 'NoneType' object is unsubscriptable class PhoneNumberWidget(forms.MultiWidget): def __init__(self,attrs=None): wigs = (forms.TextInput(attrs={'size':'3','maxlength':'3'}),\ forms.TextInput(attrs={'size':'3','maxlength':'3'}),\ forms.TextInput(attrs={'size':'4','maxlength':'4'})) super(PhoneNumberWidget, self).__init__(wigs, attrs) def decompress(self, value): return value or None def format_output(self, rendered_widgets): return '('+rendered_widgets[0]+')'+rendered_widgets[1]+'-'+rendered_widgets[2] class PhoneNumberField(forms.MultiValueField): widget = PhoneNumberWidget def __init__(self, *args, **kwargs): fields=(forms.CharField(max_length=3), forms.CharField(max_length=3), forms.CharField(max_length=4)) super(PhoneNumberField, self).__init__(fields, *args, **kwargs) def compress(self, data_list): if data_list[0] in fields.EMPTY_VALUES or data_list[1] in fields.EMPTY_VALUES or data_list[2] in fields.EMPTY_VALUES: raise fields.ValidateError(u'Enter valid phone number') return data_list[0]+data_list[1]+data_list[2] class AdvertiserSumbissionForm(ModelForm): business_phone_number = PhoneNumberField(required=True)

  Read the article

• Django foreign key question

  - by Hulk

  All, i have the following model defined, class header(models.Model): title = models.CharField(max_length = 255) created_by = models.CharField(max_length = 255) def __unicode__(self): return self.id() class criteria(models.Model): details = models.CharField(max_length = 255) headerid = models.ForeignKey(header) def __unicode__(self): return self.id() class options(models.Model): opt_details = models.CharField(max_length = 255) headerid = models.ForeignKey(header) def __unicode__(self): return self.id() AND IN MY VIEWS I HAVE p= header(title=name,created_by=id) p.save() Now the data will be saved to header table .My question is that for this id generated in header table how will save the data to criteria and options table..Please let me know.. Thanks..

  Read the article

• How to manually render a Django template for an inlineformset_factory with can_delete = True / False

  - by chefsmart

  I have an inlineformset with a custom Modelform. So it looks something like this: MyInlineFormSet = inlineformset_factory(MyMainModel, MyInlineModel, form=MyCustomInlineModelForm) I am rendering this inlineformset manually in a template so that I have more control over widgets and javascript. So I go in a loop like {% for form in myformset.forms %} and then manually render each field as described on this page http://docs.djangoproject.com/en/dev/topics/forms/#customizing-the-form-template The formset has can_delete = True or can_delete = False depending on whether the user is creating new objects or editing existing ones. Question is, how do I manually render the can_delete checkbox?

  Read the article

• Django updating db for selected ids

  - by Hulk

  In the following, New row values in DB are 6,8.They are the ids of a field I want to update these some other fields in the table based on these values row_newid=request.POST.get('row_updated_id') //Array row_newdata=request.POST.get('row_updated_data') //Array for newrow in row_newid: //how to update row_newdata for newrow values No for all the ids in row_newid how do i update row_newdata. row_newdata has the values 'a' and 'b' for example. thanks....

  Read the article

• Django: Prefill a ManytoManyField

  - by Emile Petrone

  I have a ManyToManyField on a settings page that isn't rendering. The data was filled when the user registered, and I am trying to prefill that data when the user tries to change it. Thanks in advance for the help! The HTML: {{form.types.label}} {% if add %} {{form.types}} {% else %} {% for type in form.types.all %} {{type.description}} {% endfor %} {% endif %} The View: @csrf_protect @login_required def edit_host(request, host_id, template_name="host/newhost.html"): host = get_object_or_404(Host, id=host_id) if request.user != host.user: return HttpResponseForbidden() form = HostForm(request.POST) if form.is_valid(): if request.method == 'POST': if form.cleaned_data.get('about') is not None: host.about = form.cleaned_data.get('about') if form.cleaned_data.get('types') is not None: host.types = form.cleaned_data.get('types') host.save() form.save_m2m() return HttpResponseRedirect('/users/%d/' % host.user.id) else: form = HostForm(initial={ "about":host.about, "types":host.types, }) data = { "host":host, "form":form } return render_to_response(template_name, data, context_instance=RequestContext(request)) Form: class HostForm(forms.ModelForm): class Meta: model = Host fields = ('types', 'about', ) types = forms.ModelMultipleChoiceField( widget=forms.CheckboxSelectMultiple, queryset=Type.objects.all(), required=True) about = forms.CharField( widget=forms.Textarea(), required=True) def __init__(self, *args, **kwargs): super(HostForm, self).__init__(*args, **kwargs) self.fields['about'].widget.attrs = { 'placeholder':'Hello!'}

  Read the article

• Select a subset of foreign key elements in inlineformset_factory in Django

  - by Enis Afgan

  Hello, I have a model with two foreign keys: class Model1(models.Model): model_a = models.ForeignKey(ModelA) model_b = models.ForeignKey(ModelB) value = models.IntegerField() Then, I create an inline formset class, like so: an_inline_formset = inlineformset_factory(ModelA, Model1, fk_name="model_a") and then instantiate it, like so: a_formset = an_inline_formset(request.POST, instance=model_A_object) Once this formset gets rendered in a template/page, there is ChoiceField associated with the model_b field. The problem I'm having is that the elements in the resulting drop down menu include all of the elements found in ModelB table. I need to select a subset of those based on some criteria from ModelB. At the same time, I need to keep the reference to the instance of model_A_object when instantiating inlineformset_factory and, therefore, I can't just use this example. Any suggestions?

  Read the article

• Django models avaoid duplicates

  - by Hulk

  In models, class Getdata(models.Model): title = models.CharField(max_length=255) state = models.CharField(max_length=2, choices=STATE, default="0") name = models.ForeignKey(School) created_by = models.ForeignKey(profile) def __unicode__(self): return self.id() In templates <form> <input type="submit" save the data/> </form> If the user clicks on the save button and the above data is saved in the table how to avoid the duplicates,i.e, if the user again clicks on the same submit button there should not be another entry for the same values.Or is it some this that has to be handeled in views Thanks..

  Read the article

• Django models avoid duplicates

  - by Hulk

  In models, class Getdata(models.Model): title = models.CharField(max_length=255) state = models.CharField(max_length=2, choices=STATE, default="0") name = models.ForeignKey(School) created_by = models.ForeignKey(profile) def __unicode__(self): return self.id() In templates <form> <input type="submit" save the data/> </form> If the user clicks on the save button and the above data is saved in the table how to avoid the duplicates,i.e, if the user again clicks on the same submit button there should not be another entry for the same values.Or is it some this that has to be handeled in views Thanks..

  Read the article

• Django Admin Interface : New changes to models don't appear in admin interface even after syncdb !!

  - by Rohit

  Except admin other apps using those models work fine. Tables are created in database. What can be the problem with admin app? How do I fix it?

  Read the article

• Multiple model forms with some pre-populated fields

  - by jimbocooper

  Hi! Hope somebody can help me, since I've been stuck for a while with this... I switched to another task, but now back to the fight I still can't figure out how to come out from the black hole xD The thing is as follows: Let's say I've got a product model, and then a set of Clients which have rights to submit data for the products they've been subscribed (Many to Many from Client to Product). Whenever my client is going to submit data, I need to create as many forms as products he's subscribed, and pre-populate each one of them with the "product" field as long as perform a quite simple validation (some optional fields have to be completed if it's client's first submission). I would like one form "step" for each product submission, so I've tried formWizards... but the problem is you can't pre-assign values to the forms... this can be solved afterwards when submitting, though... but not the problem that it doesn't allow validation either, so at the end of each step I can check some data before rendering next step. Then I've tried model formsets, but then there's no way to pre-populate the needed fields. I came across some django plugins, but I'm not confident yet if any of them will make it.... Did anybody has a similar problem so he can give me a ray of light? Thanks a lot in advance!! :) edit: The code I used in the formsets way is as follows: prods = Products.objects.filter(Q(start_date__lte=today) & Q(end_date__gte=today), requester=client) num = len(prods) PriceSubmissionFormSet = modelformset_factory(PriceSubmission, extra=num) formset = PriceSubmissionFormSet(queryset=PriceSubmission.objects.none())

  Read the article

• CodeIgniter & Datamapper as frontend, Django Admin as backend, database tables inconsistent

  - by Rasiel

  I created a database for a site i'm doing using Django as the admin backend. However because the server where the site is hosted on, won't be able to support Python, I find myself needing to do the front end in PHP and as such i've decided to use CodeIgniter along with Datamapper to map the models/relationship. However DataMapper requires the tables to be in a specific format for it to work, and Django maps its tables differently, using the App name as the prefix in the table. I've tried using the prefix & join_prefix vars in datamapper but still doesn't map them correctly. Has anyone used a combination of this? and if so how have the fixed the issue of db table names being inconsistent? Is there anything out there that i can use to make them work together?

  Read the article

• Django admin clear button for form

  - by Daniel Garcia

  I was wondering if there is a way to put a "Clear Form" button on the admin forms of the apps in django?

  Read the article

• 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)

  Read the article

• django: cannot import settings, cannot login to admin, cannot change admin password

  - by xpanta

  Hi, It seems that I am completely lost here. Yesterday I noticed that I cannot login to the admin panel (don't use it much, so it's been some weeks since last login). I thought that I might have changed the admin password and now I can't remember it (though I doubt it). I tried django-admin.py changepassword (using django 1.2.1) but it said that 'changepassword' is unknown command (I have all the necessary imports in my settings.py. Admin interface used to work ok). Then I gave a django-admin.py validate. Then the hell begun. django-admin.py validate gave me this error: Error: Settings cannot be imported, because environment variable DJANGO_SETTINGS_MODULE is undefined. I then gave a set DJANGO_SETTINGS_MODULE=myproject.settings and then again a django-admin.py validate This is what I get now: Error: Could not import settings 'myproject.settings' (Is it on sys.path? Does it have syntax errors?): No module named myproject.settings and now I am lost. I tried django console and sys.path.append('c:\workspace') or sys.append('c:\workspace\myproject') but still get the same errors. I use windows 7 and my project dir is c:\workspace. I don't use a PYTHONPATH variable (although I tried setting it temporarily to C:\workspace but I still get the same error). I don't use Apache, just the django development server. What am I doing wrong? My web page works fine. I think that the fact that I can't login as admin is related to the previous import error, no? PS: I also tried this: http://coderseye.com/2007/howto-reset-the-admin-password-in-django.html but still I couldn't change admin password for some reason. Although I could create another admin user (with which I couldn't login).

  Read the article

• Problem with messages framework in Django 1.2

  - by Konstantin

  Hello! I'm running Django 1.2 beta and trying out the new feature: message framework. http://docs.djangoproject.com/en/dev/ref/contrib/messages/ Everything seems to work, but when I try to output the messages, I get nothing. Seems that messages variable is empty. I double checked all the settings, they seem to be just like in the manual. What could be wrong? settings.py MIDDLEWARE_CLASSES = ( 'django.middleware.common.CommonMiddleware', 'django.contrib.sessions.middleware.SessionMiddleware', 'django.contrib.messages.middleware.MessageMiddleware', #send messages to users 'django.middleware.locale.LocaleMiddleware', 'django.contrib.auth.middleware.AuthenticationMiddleware', #debug tool 'debug_toolbar.middleware.DebugToolbarMiddleware', ) TEMPLATE_CONTEXT_PROCESSORS = ( 'django.contrib.messages.context_processors.messages', #send messages to users 'django.core.context_processors.auth', ) #Store messages in sessions MESSAGE_STORAGE = 'django.contrib.messages.storage.session.SessionStorage'; INSTALLED_APPS = ( 'django.contrib.auth', 'django.contrib.contenttypes', 'django.contrib.sessions', #'django.contrib.sites', 'django.contrib.admin', 'django.contrib.messages', 'debug_toolbar', #my apps #... ) views.py def myview(request): from django.contrib import messages messages.error(request, 'error test'); messages.success(request, 'success test'); return render_to_response('mytemplate.html', locals()); mytemplate.html {% for message in messages %} {{ message }}<br /> {% endfor %} In template nothing is outputted.

  Read the article

• Django 1.4.1 error loading MySQLdb module when attempting 'python manage.py shell'

  - by Paul

  I am trying to set up MySQL, and can't seem to be able to enter the Django manage.py shell interpreter. Getting the output below: rrdhcp-10-32-44-126:django pavelfage$ python manage.py shell Traceback (most recent call last): File "manage.py", line 10, in <module> execute_from_command_line(sys.argv) File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/core/management/__init__.py", line 443, in execute_from_command_line utility.execute() File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/core/management/__init__.py", line 382, in execute self.fetch_command(subcommand).run_from_argv(self.argv) File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/core/management/base.py", line 196, in run_from_argv self.execute(*args, **options.__dict__) File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/core/management/base.py", line 232, in execute output = self.handle(*args, **options) File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/core/management/base.py", line 371, in handle return self.handle_noargs(**options) File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/core/management/commands/shell.py", line 45, in handle_noargs from django.db.models.loading import get_models File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/db/__init__.py", line 40, in <module> backend = load_backend(connection.settings_dict['ENGINE']) File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/db/__init__.py", line 34, in __getattr__ return getattr(connections[DEFAULT_DB_ALIAS], item) File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/db/utils.py", line 92, in __getitem__ backend = load_backend(db['ENGINE']) File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/db/utils.py", line 24, in load_backend return import_module('.base', backend_name) File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/utils/importlib.py", line 35, in import_module __import__(name) File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/db/backends/mysql/base.py", line 16, in <module> raise ImproperlyConfigured("Error loading MySQLdb module: %s" % e) django.core.exceptions.ImproperlyConfigured: Error loading MySQLdb module: No module named _mysql Any suggestions really appreciated.

  Read the article

• django admin: Add a "remove file" field for Image- or FileFields

  - by w-

  I was hunting around the net for a way to easily allow users to blank out imagefield/filefields they have set in the admin. I found this http://www.djangosnippets.org/snippets/894/ What was really interesting to me here was the code posted in the comment by rfugger remove_the_file = forms.BooleanField(required=False) def save(self, *args, **kwargs): object = super(self.__class__, self).save(*args, **kwargs) if self.cleaned_data.get('remove_the_file'): object.the_file = '' return object When i try to use this in my own form I basically added this to my admin.py which already had a BlahAdmin class BlahModelForm(forms.ModelForm): class Meta: model = Blah remove_img01 = forms.BooleanField(required=False) def save(self, *args, **kwargs): object = super(self.__class__, self).save(*args, **kwargs) if self.cleaned_data.get('remove_img01'): object.img01 = '' return object when i run it I get this error maximum recursion depth exceeded while calling a Python object at this line object = super(self.__class__, self).save(*args, **kwargs) When i think about it for a bit, it seems obvious that it is just infinitely calling itself causing the error. My problem is i can't figure out what is the correct way i should be doing this. Any suggestions? thanks

  Read the article

• Django admin causes high load for one model...

  - by Joe

  In my Django admin, when I try to view/edit objects from one particular model class the memory usage and CPU rockets up and I have to restart the server. I can view the list of objects fine, but the problem comes when I click on one of the objects. Other models are fine. Working with the object in code (i.e. creating and displaying) is ok, the problem only arises when I try to view an object with the admin interface. The class isn't even particularly exotic: class Comment(models.Model): user = models.ForeignKey(User) thing = models.ForeignKey(Thing) date = models.DateTimeField(auto_now_add=True) content = models.TextField(blank=True, null=True) approved = models.BooleanField(default=True) class Meta: ordering = ['-date'] Any ideas? I'm stumped. The only reason I could think of might be that the thing is quite a large object (a few kb), but as I understand it, it wouldn't get loaded until it was needed (correct?).

  Read the article

• How to resolve user registration and activation email error in Django registration?

  - by user2476295

  So I was just trying to setup a basic user authentication in Django and downloaded a django registration app with templates. Now when I run the server at 127.0.0.1:8000/accounts/register/ I get a basic registration page, I fill in the details and when I click submit I get this error "NoReverseMatch at /accounts/register/" Error during template rendering In template Users/sudhasinha/mysite/mysite/registration/templates/registration/activation_email.txt, error at line 4 'url' requires a non-empty first argument. The syntax changed in Django 1.5, see the docs. 1 {% load i18n %} 2 {% trans "Activate account at" %} {{ site.name }}: 3 4 http://{{ site.domain }}{**% url registration_activate activation_key %**} 5 6 {% blocktrans %}Link is valid for {{ expiration_days }} days.{% endblocktrans %} 7 This is what my activation_email.txt looks like: {% load i18n %} {% trans "Activate account at" %} {{ site.name }}: http://{{ site.domain }}{% url registration_activate activation_key %} {% blocktrans %}Link is valid for {{ expiration_days }} days.{% endblocktrans %} And this is what my registration_form.html looks like: {% extends "base.html" %} {% load i18n %} {% block content %} <form method="post" action="."> {{ form.as_p }} <input type="submit" value="{% trans 'Submit' %}" /> </form> {% endblock %} I have very minimal experience with Django and would appreciate some help to resolve this error. My urls seem to be setup correctly but I will post it if needed. Also pardon my horrible formatting

  Read the article

• Django - no module named app

  - by Koran

  Hi, I have been trying to get an application written in django working - but it is not working at all. I have been working on for some time too - and it is working on dev-server perfectly. But I am unable to put in the production env (apahce). My project name is apstat and the app name is basic. I try to access it as following Blockquote http://hostname/apstat But it shows the following error: MOD_PYTHON ERROR ProcessId: 6002 Interpreter: 'domU-12-31-39-06-DD-F4.compute-1.internal' ServerName: 'domU-12-31-39-06-DD-F4.compute-1.internal' DocumentRoot: '/home/ubuntu/server/' URI: '/apstat/' Location: '/apstat' Directory: None Filename: '/home/ubuntu/server/apstat/' PathInfo: '' Phase: 'PythonHandler' Handler: 'django.core.handlers.modpython' Traceback (most recent call last): File "/usr/lib/python2.6/dist-packages/mod_python/importer.py", line 1537, in HandlerDispatch default=default_handler, arg=req, silent=hlist.silent) File "/usr/lib/python2.6/dist-packages/mod_python/importer.py", line 1229, in _process_target result = _execute_target(config, req, object, arg) File "/usr/lib/python2.6/dist-packages/mod_python/importer.py", line 1128, in _execute_target result = object(arg) File "/usr/lib/pymodules/python2.6/django/core/handlers/modpython.py", line 228, in handler return ModPythonHandler()(req) File "/usr/lib/pymodules/python2.6/django/core/handlers/modpython.py", line 201, in __call__ response = self.get_response(request) File "/usr/lib/pymodules/python2.6/django/core/handlers/base.py", line 134, in get_response return self.handle_uncaught_exception(request, resolver, exc_info) File "/usr/lib/pymodules/python2.6/django/core/handlers/base.py", line 154, in handle_uncaught_exception return debug.technical_500_response(request, *exc_info) File "/usr/lib/pymodules/python2.6/django/views/debug.py", line 40, in technical_500_response html = reporter.get_traceback_html() File "/usr/lib/pymodules/python2.6/django/views/debug.py", line 114, in get_traceback_html return t.render(c) File "/usr/lib/pymodules/python2.6/django/template/__init__.py", line 178, in render return self.nodelist.render(context) File "/usr/lib/pymodules/python2.6/django/template/__init__.py", line 779, in render bits.append(self.render_node(node, context)) File "/usr/lib/pymodules/python2.6/django/template/debug.py", line 81, in render_node raise wrapped TemplateSyntaxError: Caught an exception while rendering: No module named basic Original Traceback (most recent call last): File "/usr/lib/pymodules/python2.6/django/template/debug.py", line 71, in render_node result = node.render(context) File "/usr/lib/pymodules/python2.6/django/template/debug.py", line 87, in render output = force_unicode(self.filter_expression.resolve(context)) File "/usr/lib/pymodules/python2.6/django/template/__init__.py", line 572, in resolve new_obj = func(obj, *arg_vals) File "/usr/lib/pymodules/python2.6/django/template/defaultfilters.py", line 687, in date return format(value, arg) File "/usr/lib/pymodules/python2.6/django/utils/dateformat.py", line 269, in format return df.format(format_string) File "/usr/lib/pymodules/python2.6/django/utils/dateformat.py", line 30, in format pieces.append(force_unicode(getattr(self, piece)())) File "/usr/lib/pymodules/python2.6/django/utils/dateformat.py", line 175, in r return self.format('D, j M Y H:i:s O') File "/usr/lib/pymodules/python2.6/django/utils/dateformat.py", line 30, in format pieces.append(force_unicode(getattr(self, piece)())) File "/usr/lib/pymodules/python2.6/django/utils/encoding.py", line 71, in force_unicode s = unicode(s) File "/usr/lib/pymodules/python2.6/django/utils/functional.py", line 201, in __unicode_cast return self.__func(*self.__args, **self.__kw) File "/usr/lib/pymodules/python2.6/django/utils/translation/__init__.py", line 62, in ugettext return real_ugettext(message) File "/usr/lib/pymodules/python2.6/django/utils/translation/trans_real.py", line 286, in ugettext return do_translate(message, 'ugettext') File "/usr/lib/pymodules/python2.6/django/utils/translation/trans_real.py", line 276, in do_translate _default = translation(settings.LANGUAGE_CODE) File "/usr/lib/pymodules/python2.6/django/utils/translation/trans_real.py", line 194, in translation default_translation = _fetch(settings.LANGUAGE_CODE) File "/usr/lib/pymodules/python2.6/django/utils/translation/trans_real.py", line 180, in _fetch app = import_module(appname) File "/usr/lib/pymodules/python2.6/django/utils/importlib.py", line 35, in import_module __import__(name) ImportError: No module named basic My settings.py is as follows: INSTALLED_APPS = ( 'django.contrib.auth', 'django.contrib.contenttypes', 'django.contrib.sessions', 'django.contrib.sites', 'apstat.basic', 'django.contrib.admin', ) If I remove the apstat.basic, it goes through, but that is not a solution. Is it something I am doing in apache? My apache - settings are - <VirtualHost *:80> ServerAdmin webmaster@localhost DocumentRoot /home/ubuntu/server/ <Directory /> Options None AllowOverride None </Directory> <Directory /home/ubuntu/server/apstat> AllowOverride None Order allow,deny allow from all </Directory> <Location "/apstat"> SetHandler python-program PythonHandler django.core.handlers.modpython SetEnv DJANGO_SETTINGS_MODULE apstat.settings PythonOption django.root /home/ubuntu/server/ PythonDebug On PythonPath "['/home/ubuntu/server/'] + sys.path" </Location> </VirtualHost> I have now sat for more than a day on this. If someone can help me out, it would be very nice.

  Read the article

• Formatting inline many-to-many related models presented in django admin

  - by Jonathan

  I've got two django models (simplified): class Product(models.Model): name = models.TextField() price = models.IntegerField() class Invoice(models.Model): company = models.TextField() customer = models.TextField() products = models.ManyToManyField(Product) I would like to see the relevant products as a nice table (of product fields) in an Invoice page in admin and be able to link to the individual respective Product pages. My first thought was using the admin's inline - but django used a select box widget per related Product. This isn't linked to the Product pages, and also as I have thousands of products, and each select box independently downloads all the product names, it quickly becomes unreasonably slow. So I turned to using ModelAdmin.filter_horizontal as suggested here, which used a single instance of a different widget, where you have a list of all Products and another list of related Products and you can add\remove products in the later from the former. This solved the slowness, but it still doesn't show the relevant Product fields, and it ain't linkable. So, what should I do? tweak views? override ModelForms? I Googled around and couldn't find any example of such code...

  Read the article

• Error after installing Django (supposed PATH or PYTHONPATH "error")

  - by illuminated

  Hi all, I guess this is a PATH/PYTHONPATH error, but my attempts failed so far to make django working. System is Ubuntu 10.04, 64bit: mx:~/webapps$ cat /etc/lsb-release DISTRIB_ID=Ubuntu DISTRIB_RELEASE=10.04 DISTRIB_CODENAME=lucid DISTRIB_DESCRIPTION="Ubuntu 10.04 LTS" Python version: 2.6.5: @mx:~/webapps$ python -V Python 2.6.5 When I run django-admin.py, the following happens: mx:~/webapps$ django-admin.py Traceback (most recent call last): File "/usr/local/bin/django-admin.py", line 2, in <module> from django.core import management ImportError: No module named django.core Similar when I import django in python shell: mx:~/webapps$ python Python 2.6.5 (r265:79063, Apr 16 2010, 13:09:56) [GCC 4.4.3] on linux2 Type "help", "copyright", "credits" or "license" for more information. >>> import django Traceback (most recent call last): File "<stdin>", line 1, in <module> ImportError: No module named django >>> quit() More details: mx:~/webapps$ python -c "from distutils.sysconfig import get_python_lib; print get_python_lib()" /usr/lib/python2.6/dist-packages Within python shell: Python 2.6.5 (r265:79063, Apr 16 2010, 13:09:56) [GCC 4.4.3] on linux2 Type "help", "copyright", "credits" or "license" for more information. >>> import sys >>> print sys.path ['', '/usr/lib/python2.6/dist-packages/django', '/usr/local/lib/python2.6/dist-packages/django/bin', '/usr/local/lib/python2.6/dist-packages/django', '/home/petra/webapps', '/usr/lib/python2.6', '/usr/lib/python2.6/plat-linux2', '/usr/lib/python2.6/lib-tk', '/usr/lib/python2.6/lib-old', '/usr/lib/python2.6/lib-dynload', '/usr/lib/python2.6/dist-packages', '/usr/lib/python2.6/dist-packages/PIL', '/usr/lib/pymodules/python2.6'] django-admin.py can be found here: mx:~/webapps$ locate django-admin.py ~/install/sources/Django-1.2.1/build/lib.linux-i686-2.6/django/bin/django-admin.py ~/install/sources/Django-1.2.1/build/scripts-2.6/django-admin.py ~/install/sources/Django-1.2.1/django/bin/django-admin.py /usr/local/bin/django-admin.py /usr/local/lib/python2.6/dist-packages/django/bin/django-admin.py /usr/local/lib/python2.6/dist-packages/django/bin/django-admin.pyc and in the end this doesn't help: export PYTHONPATH="/usr/lib/python2.6/dist-packages/django:$PYTHONPATH" nor this: export PYTHONPATH="/usr/local/lib/python2.6/dist-packages/django:$PYTHONPATH" How to solve this !? Thanks all in advance! :)

  Read the article

• 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)

  Read the article

• Django: Overriding the save() method: how do I call the delete() method of a child class

  - by Patti

  The setup = I have this class, Transcript: class Transcript(models.Model): body = models.TextField('Body') doPagination = models.BooleanField('Paginate') numPages = models.PositiveIntegerField('Number of Pages') and this class, TranscriptPages(models.Model): class TranscriptPages(models.Model): transcript = models.ForeignKey(Transcript) order = models.PositiveIntegerField('Order') content = models.TextField('Page Content', null=True, blank=True) The Admin behavior I’m trying to create is to let a user populate Transcript.body with the entire contents of a long document and, if they set Transcript.doPagination = True and save the Transcript admin, I will automatically split the body into n Transcript pages. In the admin, TranscriptPages is a StackedInline of the Transcript Admin. To do this I’m overridding Transcript’s save method: def save(self): if self.doPagination: #do stuff super(Transcript, self).save() else: super(Transcript, self).save() The problem = When Transcript.doPagination is True, I want to manually delete all of the TranscriptPages that reference this Transcript so I can then create them again from scratch. So, I thought this would work: #do stuff TranscriptPages.objects.filter(transcript__id=self.id).delete() super(Transcript, self).save() but when I try I get this error: Exception Type: ValidationError Exception Value: [u'Select a valid choice. That choice is not one of the available choices.'] ... and this is the last thing in the stack trace before the exception is raised: .../django/forms/models.py in save_existing_objects pk_value = form.fields[pk_name].clean(raw_pk_value) Other attempts to fix: t = self.transcriptpages_set.all().delete() (where self = Transcript from the save() method) looping over t (above) and deleting each item individually making a post_save signal on TranscriptPages that calls the delete method Any ideas? How does the Admin do it? UPDATE: Every once in a while as I'm playing around with the code I can get a different error (below), but then it just goes away and I can't replicate it again... until the next random time. Exception Type: MultiValueDictKeyError Exception Value: "Key 'transcriptpages_set-0-id' not found in " Exception Location: .../django/utils/datastructures.py in getitem, line 203 and the last lines from the trace: .../django/forms/models.py in _construct_form form = super(BaseInlineFormSet, self)._construct_form(i, **kwargs) .../django/utils/datastructures.py in getitem pk = self.data[pk_key]

  Read the article

< Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >