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  • Django "comment_was_flagged" signal

    - by tsoporan
    Hello, This is my first time working with django signals and I would like to hook the "comment_was_flagged" signal provided by the comments app to notify me when a comment is flagged. This is my code, but it doesn't seem to work, am I missing something? from django.contrib.comments.signals import comment_was_flagged from django.core.mail import send_mail def comment_flagged_notification(sender, **kwargs): send_mail('testing moderation', 'testing', 'test@localhost', ['[email protected]',]) comment_was_flagged.connect(comment_flagged_notification) (I am just testing the email for now, but I have assured the email is sending properly.) Thanks!

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  • Apps not showing in Django admin site

    - by jack
    I have a Django project with about 10 apps in it. But the admin interface only shows Auth and Site models which are part of Django distribution. Yes, the admin interface is up and working but none of my self-written apps shows there. INSTALLED_APPS INSTALLED_APPS = ( 'django.contrib.auth', 'django.contrib.sites', 'django.contrib.contenttypes', 'django.contrib.humanize', 'django.contrib.sessions', 'django.contrib.admin', 'django.contrib.admindocs', 'project.app1', ... app1/admin.py from django.contrib import admin from project.app1.models import * admin.site.register(model1) admin.site.register(model2) admin.site.register(model3) What could be wrong in this case? Looks like everything is configured as what document says. Thank you in advance.

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  • Django Foreign key queries

    - by Hulk
    In the following model: 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() If there is a row in the database for table header as Id=1, title=value-mart , createdby=CEO How do i query criteria and options tables to get all the values related to header table id=1 Also can some one please suggest a good link for queries examples, Thanks..

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  • django-multilingual and switching between languages on template side

    - by israkir
    I am trying to use django-multilingual and setup it properly. But what I found is that everything is clear for django-multilingual except a template usage example. I just started to use django and I don't know, maybe because of this reason, I cannot figure out how to switch between languages on template side. Is there any example that you can give or any 'more' clear source/documentation about this?

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  • Edit/show Primary Key in Django Admin

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

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  • Django User model, adding function

    - by Hellnar
    Hello, I want to add a new function to the default User model of Django for retrieveing a related list of Model type. Such Foo model: class Foo(models.Model): owner = models.ForeignKey(User, related_name="owner") likes = models.ForeignKey(User, related_name="likes") ........ #at some view user = request.user foos= user.get_related_foo_models() Hwo can this be achieved ?

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  • Django Admin Actions missing

    - by Andrew C
    One of my Django sites is missing the Django Admin Action bar shown here: http://docs.djangoproject.com/en/dev/ref/contrib/admin/actions/#ref-contrib-admin-actions There is no checkbox next to each row and no Action select box near the top of the page. This is happening on every model. I have several sites running Django 1.1, and they all show the Admin Actions, so it feels like a local configuration issue. Anyone seen this before?

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  • Django QuerySet filter method returns multiple entries for one record

    - by Yaroslav
    Trying to retrieve blogs (see model description below) that contain entries satisfying some criteria: Blog.objects.filter(entries__title__contains='entry') The results is: [<Blog: blog1>, <Blog: blog1>] The same blog object is retrieved twice because of JOIN performed to filter objects on related model. What is the right syntax for filtering only unique objects? Data model: class Blog(models.Model): name = models.CharField(max_length=100) def __unicode__(self): return self.name class Entry(models.Model): title = models.CharField(max_length=100) blog = models.ForeignKey(Blog, related_name='entries') def __unicode__(self): return self.title Sample data: b1 = Blog.objects.create(name='blog1') e1 = Entry.objects.create(title='entry 1', blog=b1) e1 = Entry.objects.create(title='entry 2', blog=b1)

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  • New Communications Industry Data Model with "Factory Installed" Predictive Analytics using Oracle Da

    - by charlie.berger
    Oracle Introduces Oracle Communications Data Model to Provide Actionable Insight for Communications Service Providers   We've integrated pre-installed analytical methodologies with the new Oracle Communications Data Model to deliver automated, simple, yet powerful predictive analytics solutions for customers.  Churn, sentiment analysis, identifying customer segments - all things that can be anticipated and hence, preconcieved and implemented inside an applications.  Read on for more information! TM Forum Management World, Nice, France - 18 May 2010 News Facts To help communications service providers (CSPs) manage and analyze rapidly growing data volumes cost effectively, Oracle today introduced the Oracle Communications Data Model. With the Oracle Communications Data Model, CSPs can achieve rapid time to value by quickly implementing a standards-based enterprise data warehouse that features communications industry-specific reporting, analytics and data mining. The combination of the Oracle Communications Data Model, Oracle Exadata and the Oracle Business Intelligence (BI) Foundation represents the most comprehensive data warehouse and BI solution for the communications industry. Also announced today, Hong Kong Broadband Network enhanced their data warehouse system, going live on Oracle Communications Data Model in three months. The leading provider increased its subscriber base by 37 percent in six months and reduced customer churn to less than one percent. Product Details Oracle Communications Data Model provides industry-specific schema and embedded analytics that address key areas such as customer management, marketing segmentation, product development and network health. CSPs can efficiently capture and monitor critical data and transform it into actionable information to support development and delivery of next-generation services using: More than 1,300 industry-specific measurements and key performance indicators (KPIs) such as network reliability statistics, provisioning metrics and customer churn propensity. Embedded OLAP cubes for extremely fast dimensional analysis of business information. Embedded data mining models for sophisticated trending and predictive analysis. Support for multiple lines of business, such as cable, mobile, wireline and Internet, which can be easily extended to support future requirements. With Oracle Communications Data Model, CSPs can jump start the implementation of a communications data warehouse in line with communications-industry standards including the TM Forum Information Framework (SID), formerly known as the Shared Information Model. Oracle Communications Data Model is optimized for any Oracle Database 11g platform, including Oracle Exadata, which can improve call data record query performance by 10x or more. Supporting Quotes "Oracle Communications Data Model covers a wide range of business areas that are relevant to modern communications service providers and is a comprehensive solution - with its data model and pre-packaged templates including BI dashboards, KPIs, OLAP cubes and mining models. It helps us save a great deal of time in building and implementing a customized data warehouse and enables us to leverage the advanced analytics quickly and more effectively," said Yasuki Hayashi, executive manager, NTT Comware Corporation. "Data volumes will only continue to grow as communications service providers expand next-generation networks, deploy new services and adopt new business models. They will increasingly need efficient, reliable data warehouses to capture key insights on data such as customer value, network value and churn probability. With the Oracle Communications Data Model, Oracle has demonstrated its commitment to meeting these needs by delivering data warehouse tools designed to fill communications industry-specific needs," said Elisabeth Rainge, program director, Network Software, IDC. "The TM Forum Conformance Mark provides reassurance to customers seeking standards-based, and therefore, cost-effective and flexible solutions. TM Forum is extremely pleased to work with Oracle to certify its Oracle Communications Data Model solution. Upon successful completion, this certification will represent the broadest and most complete implementation of the TM Forum Information Framework to date, with more than 130 aggregate business entities," said Keith Willetts, chairman and chief executive officer, TM Forum. Supporting Resources Oracle Communications Oracle Communications Data Model Data Sheet Oracle Communications Data Model Podcast Oracle Data Warehousing Oracle Communications on YouTube Oracle Communications on Delicious Oracle Communications on Facebook Oracle Communications on Twitter Oracle Communications on LinkedIn Oracle Database on Twitter The Data Warehouse Insider Blog

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  • Django authentication in django nonrel on GAE

    - by tooba
    I'm using the Django nonrel project on a google app engine project running locally in development. I've created my own models and these are fine when they are saved and retrieved in the datastore. I'm hoping to use django.contrib.auth to provide the user functionality. I can use the shell to create users and these get assigned an ID. When I create one of my own models which references User I have to pass in a user ID as it quite rightly fails otherwise. However, checking via the gae admin interface I can't see the User model in the datastore for the users I've created via the shell. Nor can I retreive the user details from one of my models which references them. Calls to mymodel.user.username return nothing. Nor can I log into admin using the username and password I've set up. I can see saved versions of the models I've made in the gae admin app. I get the impression that users are being created somewhere other than the datastore. Is there something else I need to do to use the standard contrib.auth users with django-nonrel and gae?

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  • Django model operating on a queryset

    - by jmoz
    I'm new to Django and somewhat to Python as well. I'm trying to find the idiomatic way to loop over a queryset and set a variable on each model. Basically my model depends on a value from an api, and a model method must multiply one of it's attribs by this api value to get an up-to-date correct value. At the moment I am doing it in the view and it works, but I'm not sure it's the correct way to achieve what I want. I have to replicate this looping elsewhere. Is there a way I can encapsulate the looping logic into a queryset method so it can be used in multiple places? I have this atm (I am using django-rest-framework): class FooViewSet(viewsets.ModelViewSet): model = Foo serializer_class = FooSerializer bar = # some call to an api def get_queryset(self): # Dynamically set the bar variable on each instance! foos = Foo.objects.filter(baz__pk=1).order_by('date') for item in foos: item.needs_bar = self.bar return items I would think something like so would be better: def get_queryset(self): bar = # some call to an api # Dynamically set the bar variable on each instance! return Foo.objects.filter(baz__pk=1).order_by('date').set_bar(bar) I'm thinking the api hit should be in the controller and then injected to instances of the model, but I'm not sure how you do this. I've been looking around querysets and managers but still can't figure it out nor decided if it's the best method to achieve what I want. Can anyone suggest the correct way to model this with django? Thanks.

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  • Model Driven Architecture Approach in programming / modelling

    - by yak
    I know the basics of the model driven architecture: it is all about model the system which I want to create and create the core code afterwards. I used CORBA a while ago. First thing that I needed to do was to create an abstract interface (some kind of model of the system I want to build) and generate core code later. But I have a different question: is model driven architecture a broad approach or not? I mean, let's say, that I have the language (modelling language) in which I want to model EXISTING system (opposite to the system I want to CREATE), and then analyze the model of the created system and different facts about that modeled abstraction. In this case, can the process I described above be considered the model driven architecture approach? I mean, I have the model, but this is the model of the existing system, not the system to be created.

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  • How to make a model instance read-only after saving it once?

    - by Ryszard Szopa
    One of the functionalities in a Django project I am writing is sending a newsletter. I have a model, Newsletter and a function, send_newsletter, which I have registered to listen to Newsletter's post_save signal. When the newsletter object is saved via the admin interface, send_newsletter checks if created is True, and if yes it actually sends the mail. However, it doesn't make much sense to edit a newsletter that has already been sent, for the obvious reasons. Is there a way of making the Newsletter object read-only once it has been saved? Edit: I know I can override the save method of the object to raise an error or do nothin if the object existed. However, I don't see the point of doing that. As for the former, I don't know where to catch that error and how to communicate the user the fact that the object wasn't saved. As for the latter, giving the user false feedback (the admin interface saying that the save succeded) doesn't seem like a Good Thing. What I really want is allow the user to use the Admin interface to write the newsletter and send it, and then browse the newsletters that have already been sent. I would like the admin interface to show the data for sent newsletters in an non-editable input box, without the "Save" button. Alternatively I would like the "Save" button to be inactive.

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  • PyDev and Django: PyDev breaking Django shell?

    - by Rosarch
    I've set up a new project, and populated it with simple models. (Essentially I'm following the tut.) When I run python manage.py shell on the command line, it works fine: >python manage.py shell Python 2.6.4 (r264:75708, Oct 26 2009, 08:23:19) [MSC v.1500 32 bit (Intel)] on win32 Type "help", "copyright", "credits" or "license" for more information. (InteractiveConsole) >>> from mysite.myapp.models import School >>> School.objects.all() [] Works great. Then, I try to do the same thing in Eclipse (using a Django project that is composed of the same files.) Right click on mysite project Django Shell with Django environment This is the output from the PyDev Console: >>> import sys; print('%s %s' % (sys.executable or sys.platform, sys.version)) C:\Python26\python.exe 2.6.4 (r264:75708, Oct 26 2009, 08:23:19) [MSC v.1500 32 bit (Intel)] >>> >>> from django.core import management;import mysite.settings as settings;management.setup_environ(settings) 'path\\to\\mysite' >>> from mysite.myapp.models import School >>> School.objects.all() Traceback (most recent call last): File "<console>", line 1, in <module> File "C:\Python26\lib\site-packages\django\db\models\query.py", line 68, in __repr__ data = list(self[:REPR_OUTPUT_SIZE + 1]) File "C:\Python26\lib\site-packages\django\db\models\query.py", line 83, in __len__ self._result_cache.extend(list(self._iter)) File "C:\Python26\lib\site-packages\django\db\models\query.py", line 238, in iterator for row in self.query.results_iter(): File "C:\Python26\lib\site-packages\django\db\models\sql\query.py", line 287, in results_iter for rows in self.execute_sql(MULTI): File "C:\Python26\lib\site-packages\django\db\models\sql\query.py", line 2368, in execute_sql cursor = self.connection.cursor() File "C:\Python26\lib\site-packages\django\db\backends\__init__.py", line 81, in cursor cursor = self._cursor() File "C:\Python26\lib\site-packages\django\db\backends\sqlite3\base.py", line 170, in _cursor self.connection = Database.connect(**kwargs) OperationalError: unable to open database file What am I doing wrong here?

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  • Django: UserProfile with Unique Foreign Key in Django Admin

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

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  • How to open python scripts directly by typing in their name in terminal (Mac OS X)

    - by Haffi112
    I'm working on installing django and running it on my system. I have a problem though, in this tutorial creating a project is explained by running the command django-admin.py startproject mysite My issue is that this doesn't work. I changed to the directory where django-admin.py is located and ran the command chmod +x django-admin.py with no results. I tried adding the directory with the file to my path without results. I ended up fixing my problem with this command python /location/of/django-admin.py startproject mysite which yielded the outcome I expected. My problem is: What do I need to change/configure such that command django-admin.py startproject mysite would be sufficient? Here are some experiments: 21:09~/Desktop/HI/NSN/Polls > django-admin.py startproject mysite -bash: django-admin.py: command not found 21:09~/Desktop/HI/NSN/Polls > ./django-admin.py startproject mysite -bash: ./django-admin.py: No such file or directory 21:09~/Desktop/HI/NSN/Polls > python django-admin.py startproject mysite python: can't open file 'django-admin.py': [Errno 2] No such file or directory 21:09~/Desktop/HI/NSN/Polls > /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py startproject prufa1 -bash: /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py: /opt/local/bin: bad interpreter: Permission denied 21:09~/Desktop/HI/NSN/Polls > sudo /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py startproject prufa1Password: sudo: unable to execute /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py: Permission denied 21:09~/Desktop/HI/NSN/Polls > sudo /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py startproject prufa1sudo: unable to execute /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py: Permission denied 21:09~/Desktop/HI/NSN/Polls > python /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py startproject prufa1 21:09~/Desktop/HI/NSN/Polls > ls mysite prufa1 Final edit: The problem is solved, see Ian C's answer for the right solution. Thank you everyone for helping my out, this was very fast!

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  • Django: Validation error in Admin

    - by tomwolber
    NEWBIE ALERT! background: For the first time, I am writing a model that needs to be validated. I cannot have two Items that have overlapping "date ranges". I have everything working, except when I raise forms.ValidationError, I get the yellow screen of death (debug=true) or a 500 page (debug=false). My question: How can I have an error message show up in the Admin (like when you leave a required filed blank)? Sorry for my inexperience, please let me know if I can clarify the question better. Models.py from django.db import models from django import forms from django.forms import ModelForm from django.db.models import Q class Item(models.Model): name = models.CharField(max_length=500) slug = models.SlugField(unique=True) startDate = models.DateField("Start Date", unique="true") endDate = models.DateField("End Date") def save(self, *args, **kwargs): try: Item.objects.get(Q(startDate__range=(self.startDate,self.endDate))|Q(endDate__range=(self.startDate,self.endDate))|Q(startDate__lt=self.startDate,endDate__gt=self.endDate)) #check for validation, which may raise an Item.DoesNotExist error, excepted below #if the validation fails, raise this error: raise forms.ValidationError('Someone has already got that date, or somesuch error message') except Item.DoesNotExist: super(Item,self).save(*args,**kwargs) def __unicode__(self): return self.name def get_absolute_url(self): return "/adtest/%s/" % self.slug

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  • Django - Override admin site's login form

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

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  • Django: 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]

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  • Configuration problems with django and mod_wsgi

    - by Jimbo
    Hi, I've got problems on getting django to work on apache 2.2 with mod_wsgi. Django is installed and mod_wsgi too. I can even see a 404 page when accessing the path and I can login to django admin. But if I want to install the tagging module I get the following error: Traceback (most recent call last): File "setup.py", line 49, in <module> version_tuple = __import__('tagging').VERSION File "/home/jim/django-tagging/tagging/__init__.py", line 3, in <module> from tagging.managers import ModelTaggedItemManager, TagDescriptor File "/home/jim/django-tagging/tagging/managers.py", line 5, in <module> from django.contrib.contenttypes.models import ContentType File "/usr/lib/python2.5/site-packages/django/contrib/contenttypes/models.py", line 1, in <module> from django.db import models File "/usr/lib/python2.5/site-packages/django/db/__init__.py", line 10, in <module> if not settings.DATABASE_ENGINE: File "/usr/lib/python2.5/site-packages/django/utils/functional.py", line 269, in __getattr__ self._setup() File "/usr/lib/python2.5/site-packages/django/conf/__init__.py", line 40, in _setup self._wrapped = Settings(settings_module) File "/usr/lib/python2.5/site-packages/django/conf/__init__.py", line 75, in __init__ raise ImportError, "Could not import settings '%s' (Is it on sys.path? Does it have syntax errors?): %s" % (self.SETTINGS_MODULE, e) ImportError: Could not import settings 'mysite.settings' (Is it on sys.path? Does it have syntax errors?): No module named mysite.settings My httpd.conf: Alias /media/ /home/jim/django/mysite/media/ <Directory /home/jim/django/mysite/media> Order deny,allow Allow from all </Directory> Alias /admin/media/ "/usr/lib/python2.5/site-packages/django/contrib/admin/media/" <Directory "/usr/lib/python2.5/site-packages/django/contrib/admin/media/"> Order allow,deny Allow from all </Directory> WSGIScriptAlias /dj /home/jim/django/mysite/apache/django.wsgi <Directory /home/jim/django/mysite/apache> Order deny,allow Allow from all </Directory> My django.wsgi: import sys, os sys.path.append('/home/jim/django') sys.path.append('/home/jim/django/mysite') os.chdir('/home/jim/django/mysite') os.environ['DJANGO_SETTINGS_MODULE'] = 'mysite.settings' import django.core.handlers.wsgi application = django.core.handlers.wsgi.WSGIHandler() I try to get this to work since a few days and have read several blogs and answers here on so but nothing worked.

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  • Django Error: NameError name 'current_datetime' is not defined

    - by Diego
    I'm working through the book "The Definitive Guide to Django" and am stuck on a piece of code. This is the code in my settings.py: ROOT_URLCONF = 'mysite.urls' I have the following code in my urls.py from django.conf.urls.defaults import * from mysite.views import hello, my_homepage_view urlpatterns = patterns('', ('^hello/$', hello), ) urlpatterns = patterns('', ('^time/$', current_datetime), ) And the following is the code in my views.py file: from django.http import HttpResponse import datetime def hello(request): return HttpResponse("Hello World") def current_datetime(request): now = datetime.datetime.now() html = "<html><body>It is now %s.</body></html>" % now return HttpResponse(html) Yet, I get the following error when I test the code in the development server. NameError at /time/ name 'current_datetime' is not defined Can someone help me out here? This really is just a copy-paste from the book. I don't see any mistyping.

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  • django views question

    - by Hulk
    In my django views i have the following def create(request): query=header.objects.filter(id=a)[0] a=query.criteria_set.all() logging.debug(a.details) I get an error saying 'QuerySet' object has no attribute 'details' in the debug statement .What is this error and what should be the correct statemnt to query this.And the model corresponding to this is as follows where as the models has the following: 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() Thanks..

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  • django: CheckboxMultiSelect problem with db queries

    - by xiackok
    firstly sorry for my bad english there is a simple model Person. That contains just languages: LANGUAGE_LIS = ( (1, 'English'), (2, 'Turkish'), (3, 'Spanish') ) class Person(models.Model): languages = models.CharField(max_length=100, choices=LANGUAGE_LIST) #languages is multi value (CheckBoxSelectMultiple) and here person_save_form: class person_save_form(forms.ModelForm): languages = forms.CharField(widget=forms.CheckBoxSelectMultiple(choices=LANGUAGE_LIST)) class Meta: model = Person it is ok. but how can i search persons for languages like "get persons who knows turkish and english" in the database (MySQL) record "languages" column seen like "[u'1', u'2']". but i want search persons like this: persons = Person.objects.filter(languages__in=request.POST.getlist('languages'))

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  • Prepopulating inlines based on the parent model in the Django Admin

    - by Alasdair
    I have two models, Event and Series, where each Event belongs to a Series. Most of the time, an Event's start_time is the same as its Series' default_time. Here's a stripped down version of the models. #models.py class Series(models.Model): name = models.CharField(max_length=50) default_time = models.TimeField() class Event(models.Model): name = models.CharField(max_length=50) date = models.DateField() start_time = models.TimeField() series = models.ForeignKey(Series) I use inlines in the admin application, so that I can edit all the Events for a Series at once. If a series has already been created, I want to prepopulate the start_time for each inline Event with the Series' default_time. So far, I have created a model admin form for Event, and used the initial option to prepopulate the time field with a fixed time. #admin.py ... import datetime class OEventInlineAdminForm(forms.ModelForm): start_time = forms.TimeField(initial=datetime.time(18,30,00)) class Meta: model = OEvent class EventInline(admin.TabularInline): form = EventInlineAdminForm model = Event class SeriesAdmin(admin.ModelAdmin): inlines = [EventInline,] I am not sure how to proceed from here. Is it possible to extend the code, so that the initial value for the start_time field is the Series' default_time?

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  • Django ModelForm for Many-to-Many fields

    - by theycallmemorty
    Consider the following models and form: class Pizza(models.Model): name = models.CharField(max_length=50) class Topping(models.Model): name = models.CharField(max_length=50) ison = models.ManyToManyField(Pizza, blank=True) class ToppingForm(forms.ModelForm): class Meta: model = Topping When you view the ToppingForm it lets you choose what pizzas the toppings go on and everything is just dandy. My questions is: How do I define a ModelForm for Pizza that lets me take advantage of the Many-to-Many relationship between Pizza and Topping and lets me choose what Toppings go on the Pizza?

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