Search Results

Search found 10967 results on 439 pages for 'django sites'.

Page 22/439 | < Previous Page | 18 19 20 21 22 23 24 25 26 27 28 29  | Next Page >

  • how to do server side form validation for dynamic inputs with Django

    - by Satoru.Logic
    Hi, all. I am using django.forms.Form to validate form data in a survey applications. In a survey-creating form, a user can submit multiple questions that belong to the survey being created. Names for the question inputs are in the form of 'question_seq' , where seq is maintained using Javascript. Back in the server side, my code doesn't know before hand how many such questions will be submitted. Is there any way to do this with Django form so that the form can automatically recognizes the questions and validate them?

    Read the article

  • Django - how to make ImageField/FileField optional?

    - by ilya
    class Product(models.Model): ... image = models.ImageField(upload_to = generate_filename, blank = True) When I use ImageField (blank=True) and do not select image into admin form, exception occures. In django code you can see this: class FieldFile(File): .... def _require_file(self): if not self: raise ValueError("The '%s' attribute has no file associated with it." % self.field.name) def _get_file(self): self._require_file() ... Django trac has ticket #13327 about this problem, but seems it can't be fixed soon. How to make these field optional?

    Read the article

  • Django: Data corrupted after loading? (possible programmer error)

    - by Rosarch
    I may be loading data the wrong way. excerpt of data.json: { "pk": "1", "model": "myapp.Course", "fields": { "name": "Introduction to Web Design", "requiredFor": [9], "offeringSchool": 1, "pre_reqs": [], "offeredIn": [1, 5, 9] } }, I run python manage.py loaddata -v2 data: Installed 36 object(s) from 1 fixture(s) Then, I go to check the above object using the Django shell: >>> info = Course.objects.filter(id=1) >>> info.get().pre_reqs.all() [<Course: Intermediate Web Programming>] # WRONG! There should be no pre-reqs >>> from django.core import serializers >>> serializers.serialize("json", info) '[{"pk": 1, "model": "Apollo.course", "fields": {"pre_reqs": [11], "offeredIn": [1, 5, 9], "offeringSchool": 1, "name": "Introduction to Web Design", "requiredFor": [9]}}]' The serialized output of the model is not the same as the input that was given to loaddata. The output has a non-empty pre_req list, whereas the input's pre_reqs field is empty. What am I doing wrong?

    Read the article

  • 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)

    Read the article

  • Django: Why Doesn't the Current URL Match any Patterns in urls.py

    - by austin_sherron
    I've found a few questions here related to my issue, but I haven't found anything that has helped me resolve my issue. I'm using Python 2.7.5 and Django 1.8.dev20140627143448. I have a view that's interacting with my database to delete objects, and it takes two arguments in addition to a request: def delete_data_item(request, dataclass_id, dataitem_id): form = AddDataItemForm(request.POST) data_set = get_object_or_404(DataClass, pk=dataclass_id) context = {'data_set': data_set, 'form': form} data_item = get_object_or_404(DataItem, pk=dataitem_id) data_item.delete() data_set.save() return HttpResponseRedirect(reverse('detail', args=(dataclass_id,))) The URL in myapp.urls.py looks something like this: url(r'^(?P<dataclass_id>[0-9]+)/(?P<dataitem_id>[0-9]+)/delete_data_item/$', views.delete_data_item, name='delete_data_item') and the portion of my template relevant to the view is: <a href="{% url 'delete_data_item' data_set.id data_item.id %}">DELETE</a> Whenever I click on the DELETE link, django tells me that the request URL: http://127.0.0.1:8000/myapp/5/%7B%%20url%20'delete_data_item'%20data_set.id%20data_item.id%20%%7D doesn't match any of my URL patterns. What am I missing? The URL on which the DELETE links exist is myapp/(<dataclass_id>[0-9]+)/

    Read the article

  • Active Directory Replication across Sites slow or not working

    - by neildeadman
    I've just inherited (isn't it always the way!) a Windows Domain. The domain is spread across 2 sites. Site01 has 3 DCs & Site02 has 2 DCs. If I create a user in either site, the other DCs in that site, immediately replicate and show the new user. The new user is not shown in the other site though. If I manually run the following command, everything syncs and the new user appears: repadmin /syncall issdc01 /APed In the Inter-Site Transports DEFAULTIPSITELINK the replicate every time value is set to 180 minutes. I thought this was the solution, but on another Windows Domain, this is the same, but replication takes place across sites immediately. What can I check to resolve this issue? We are running Windows Server 2008 Results of dcdiag /test:dns show a server that is no longer part of our domain: TEST: Delegations (Del) Error: DNS server: oldserver.win.domain.com IP: [Missing glue A record]

    Read the article

  • Getting a KeyError in DB backend of django-digest

    - by rtmie
    I have just started to integrate django_digest into my app. As a start I have added the @httpdigest decorator to one of my views. If I try to connect to it I get a KeyError exception thrown in django_digest/backend/db.py . Depending on which db I configure I get a different KeyError in a different location. I am using Django 1.2.1, with MySql (also tested with sqlite). I am using the default values for all the settings options. As far as I can see I have followed all instructions but am struggling all day with this. I am using the repository versions of django-digest and python-digest. Any steer would be greatly appreciated. Tracebacks for sqlite and mysql below: with sqlite: Traceback (most recent call last): File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/servers/basehttp.py", line 674, in __call__ return self.application(environ, start_response) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/handlers/wsgi.py", line 248, in __call__ signals.request_finished.send(sender=self.__class__) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/dispatch/dispatcher.py", line 162, in send response = receiver(signal=self, sender=sender, **named) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django_digest-1.8-py2.5.egg/django_digest/backend/db.py", line 16, in close_connection _connection.close() File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/db/backends/sqlite3/base.py", line 186, in close if self.settings_dict['NAME'] != ":memory:": KeyError: 'NAME' with mysql: Traceback (most recent call last): File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/servers/basehttp.py", line 674, in __call__ return self.application(environ, start_response) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/handlers/wsgi.py", line 241, in __call__ response = self.get_response(request) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/handlers/base.py", line 142, in get_response return self.handle_uncaught_exception(request, resolver, exc_info) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/handlers/base.py", line 166, in handle_uncaught_exception return debug.technical_500_response(request, *exc_info) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/handlers/base.py", line 80, in get_response response = middleware_method(request) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django_digest-1.8-py2.5.egg/django_digest/middleware.py", line 13, in process_request if (not self._authenticator.authenticate(request) and File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django_digest-1.8-py2.5.egg/django_digest/__init__.py", line 86, in authenticate partial_digest = self._account_storage.get_partial_digest(digest_response.username) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django_digest-1.8-py2.5.egg/django_digest/backend/db.py", line 97, in get_partial_digest cursor = get_connection().cursor() File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/db/backends/__init__.py", line 75, in cursor cursor = self._cursor() File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/db/backends/mysql/base.py", line 281, in _cursor if settings_dict['USER']: KeyError: 'USER'

    Read the article

  • Google App Engine with local Django 1.1 gets Intermittent Failures

    - by Jon Watte
    I'm using the Windows Launcher development environment for Google App Engine. I have downloaded Django 1.1.2 source, and un-tarrred the "django" subdirectory to live within my application directory (a peer of app.yaml) At the top of each .py source file, I do this: import settings import os os.environ["DJANGO_SETTINGS_MODULE"] = 'settings' In my file settings.py (which lives at the root of the app directory, as well), I do this: DEBUG = True TEMPLATE_DIRS = ('html') INSTALLED_APPS = ('filters') import os os.environ["DJANGO_SETTINGS_MODULE"] = 'settings' from google.appengine.dist import use_library use_library('django', '1.1') from django.template import loader Yes, this looks a bit like overkill, doesn't it? I only use django.template. I don't explicitly use any other part of django. However, intermittently I get one of two errors: 1) Django complains that DJANGO_SETTINGS_MODULE is not defined. 2) Django complains that common.html (a template I'm extending in other templates) doesn't exist. 95% of the time, these errors are not encountered, and they randomly just start happening. Once in that state, the local server seems "wedged" and re-booting it generally fixes it. What's causing this to happen, and what can I do about it? How can I even debug it? Here is the traceback from the error: Traceback (most recent call last): File "C:\code\kwbudget\edit_budget.py", line 34, in get self.response.out.write(t.render(template.Context(values))) File "C:\code\kwbudget\django\template\__init__.py", line 165, in render return self.nodelist.render(context) File "C:\code\kwbudget\django\template\__init__.py", line 784, in render bits.append(self.render_node(node, context)) File "C:\code\kwbudget\django\template\__init__.py", line 797, in render_node return node.render(context) File "C:\code\kwbudget\django\template\loader_tags.py", line 71, in render compiled_parent = self.get_parent(context) File "C:\code\kwbudget\django\template\loader_tags.py", line 66, in get_parent raise TemplateSyntaxError, "Template %r cannot be extended, because it doesn't exist" % parent TemplateSyntaxError: Template u'common.html' cannot be extended, because it doesn't exist And edit_budget.py starts with exactly the lines that I included up top. All templates live in a directory named "html" in my root directory, and "html/common.html" exists. I know the template engine finds them, because I start out with "html/edit_budget.html" which extends common.html. It looks as if the settings module somehow isn't applied (because that's what adds html to the search path for templates).

    Read the article

  • Contents farms, scrapers sites, aggregators real world examples? [closed]

    - by Marco Demaio
    Contents farm, scrappers, aggregators real world examples? Could you plz clarify me: efreedom.com is a scraper site, not a content farm? Because it simply copies and pastes contents from stackoverflow. ehow.com and squidoo.com are contents farm? They don't copy and paste contents they just generate fresh new user generated content, but too much and too quickly. expert-exchange.com is NOT a content farm or a scraper site, right?! It's simply that many people (an me too) hates it (they also wrote to Matt Cutts) because it shows up hight in Google providing a useless question with no answer. There are also many sites that act as 'contents aggregators in the form of specialized directories' (let's call them CASD), I don't know how to else define them. Do they have a specific definition? Anyway are these type of CASD contents farms or scrapers sites or what else? Basically these CASD search for all sites of the same type i.e. “restaurants websites”, they copy and paste the contents found in “Restaurant A” and create in their aggregator site a new page called “Restaurant A”, then they do the same for all websites of the same type, thus creating a sort of directory of restaurants. Later on these CASD also sends an email to the owner of “Restaurant A” (usually the email is on the website) with a user and password to let him modify/update its own page on the CASD site. Later on these CASD might ask for money to the owner of “Restaurant A” because they bring him traffic, otherwise they remove its page on the aggregator. Someone could call these simply directories, but I think a directory is different because is something you need to add your site into by filling a form and not something that steals contents from your existing site without a specific acceptance from the site's owner. I also really wonder how Google will sort out all these mess sites packed of contents that show up more and more and everywhere in search results.

    Read the article

  • django & postgres linux hosting (with SSH access) recommendations

    - by Justin Grant
    We're looking for a good place to host our custom Django app (a fork of OSQA) and its postgresql backend. Requirements include: Linux Python 2.6 or (ideally) Python 2.7 Django 1.2 Postgres 8.4 or later DB backup/restore handled by the hoster, not us OS & dev-platform-stack patching/maintenance handled by the hoster, not us SSH access (so we can pull source code from GitHub, so we can install python eggs, etc.) ability to set up cron jobs (e.g. to send out dail email updates) ability to send up to 10K emails/day good performance (not ganged up with a zillion other sites on one CPU, not starved for RAM) FTP or SCP access to web logs dedicated public IP SSL support Costs under $1000/month for a relatively small site (<5M pageviews/month) Good customer service We already have a prototype site running on EC2 on top of a Bitnami DjangoStack. The problem is that we have to patch the OS, patch postgres, etc. We'd really prefer a platform-as-a-service (PaaS) offering, like Heroku offers for Rails apps, where all we need to worry about is deploying our code instead of worrying about system software patching and maintenance. Google App Engine is closest to what we're looking for, but they don't offer relational DB access (not yet at least). Anyone have a recommendation?

    Read the article

  • Django running on Apache+WSGI and apache SSL proxying

    - by Lessfoe
    Hi all, I'm trying to rewrite all requests for my Django server running on apache+WSGI ( inside my local network) and configured as the WSGI's wiki how to, except that I set a virtualhost for it. The server which from I want to rewrite requests is another apache server listening on port 80. I can manage it to work well if I don't try to enable SSL connection as the required way to connect. But I need all requests to Django server encrypted with SSL so I generally used this directive to achieve this ( on my public webserver ): Alias /dirname "/var/www/dirname" SSLVerifyClient none SSLOptions +FakeBasicAuth SSLRequireSSL AuthName "stuff name" AuthType Basic AuthUserFile /etc/httpd/djangoserver.passwd require valid-user # redirect all request to django.test:80 RewriteEngine On RewriteRule (.*)$ http://django.test/$1 [P] This configuration works if I try to load a specific page trough the external server from my browser. It is not working clicking my django application urls ( even tough the url seems correct when I put my mouse over). The url my public server is trying to serve use http ( instead of https ) and the directory "dirname" I specified on my apache configuration disappear, so it says that the page was not found. I think it depends on Django and its WSGI handler . Does anybody went trough my same problem? PS: I have already tried to modify the WSGI script . I'm Using Django 1.0.3, Apache 2.2 on a Fedora10 (inside), Apache 2.2 on the public server. Thanks in advance for your help. Fab

    Read the article

  • 'NoneType' object has no attribute 'day'

    - by Asinox
    Hi guy, i dont know where is my error, but Django 1.2.1 is give this error: 'NoneType' object has no attribute 'day' when i try to save form from the Administrator Area models.py from django.db import models from django.contrib.auth.models import User class Editorial(models.Model): titulo = models.CharField(max_length=250,help_text='Titulo del editorial') editorial = models.TextField(help_text='Editorial') slug = models.SlugField(unique_for_date='pub_date') autor = models.ForeignKey(User) pub_date = models.DateTimeField(auto_now_add=True) activa = models.BooleanField(verbose_name="Activa") enable_comments = models.BooleanField(verbose_name="Aceptar Comentarios",default=False) editorial_html = models.TextField(editable=False,blank=True) def __unicode__(self): return unicode(self.titulo) def get_absolute_url(self): return "/editorial/%s/%s/" % (self.pub_date.strftime("%Y/%b/%d").lower(), self.slug) class Meta: ordering=['-pub_date'] verbose_name_plural ='Editoriales' def save(self,force_insert=False, force_update=False): from markdown import markdown if self.editorial: self.editorial_html = markdown(self.editorial) super(Editorial,self).save(force_insert,force_update) i dont know why this error, thanks guys sorry with my English

    Read the article

  • Returning user data for forms that have errors in when using ModelForms

    - by Sevenearths
    forms.py from django.forms import ModelForm from client.models import ClientDetails, ClientAddress, ClientPhone from snippets.UKPhoneNumberForm import UKPhoneNumberField class ClientDetailsForm(ModelForm): class Meta: model = ClientDetails class ClientAddressForm(ModelForm): class Meta: model = ClientAddress class ClientPhoneForm(ModelForm): number = UKPhoneNumberField() class Meta: model = ClientPhone views.py from django.shortcuts import render_to_response, redirect from django.template import RequestContext from client.forms import ClientDetailsForm, ClientAddressForm, ClientPhoneForm def new_client_view(request): formDetails = ClientDetailsForm(initial={'marital_status':'u'}) formAddress = ClientAddressForm() formHomePhone = ClientPhoneForm(initial={'phone_type':'home'}) formWorkPhone = ClientPhoneForm(initial={'phone_type':'work'}) formMobilePhone = ClientPhoneForm(initial={'phone_type':'mobi'}) return render_to_response('client/new_client.html', {'formDetails': formDetails, 'formAddress': formAddress, 'formHomePhone': formHomePhone, 'formWorkPhone': formWorkPhone, 'formMobilePhone': formMobilePhone}, context_instance=RequestContext(request)) (the new_client.html is nothing special) How should I write views.py so that if the user's data raises an error, instead of showing them the form again with the errors in but none of their original data, it shows them the form again with the errors AND their original data?

    Read the article

  • Google Sites page never shows up in Google Search organic results?

    - by gus
    I use Google Sites (i.e.: https://sites.google.com/site/EXAMPLE/ ) as a convenient way to maintain up-to-date info on several residential properties, info that's often requested by my property agents, its been around for about 1 year, but I still can never get it to appear in organic Google search results or Bing, even if I search the specific keywords such as the street names. I submitted the URL manually to search engines, knowing that my Sites page probably has very few incoming links. Is this expected behavior? The content of my page has simple formatted text, and outgoing links to Picasa/G+/imgur photo albums. Am I doing something wrong or do all GoogleSites pages have poor organic search rank? Thank you very much.

    Read the article

  • Foreign keys in django admin list display

    - by Olivier
    If a django model contains a foreign key field, and if that field is shown in list mode, then it shows up as text, instead of displaying a link to the foreign object. Is it possible to automatically display all foreign keys as links instead of flat text? (of course it is possible to do that on a field by field basis, but is there a general method?) Example: class Author(models.Model): ... class Post(models.Model): author = models.ForeignKey(Author) Now I choose a ModelAdmin such that the author shows up in list mode: class PostAdmin(admin.ModelAdmin): list_display = [..., 'author',...] Now in list mode, the author field will just use the __unicode__ method of the Author class to display the author. On the top of that I would like a link pointing to the url of the corresponding author in the admin site. Is that possible? Manual method: For the sake of completeness, I add the manual method. It would be to add a method author_link in the PostAdmin class: def author_link(self, item): return '<a href="../some/path/%d">%s</a>' % (item.id, unicode(item)) author_link.allow_tags = True That will work for that particular field but that is not what I want. I want a general method to achieve the same effect. (One of the problems is how to figure out automatically the path to an object in the django admin site.)

    Read the article

  • Re-ordering child nodes in django-MPTT

    - by Dominic Rodger
    I'm using Ben Firshman's fork of django-MPTT (hat tip to Daniel Roseman for the recommendation). I've got stuck trying to re-order nodes which share a common parent. I've got a list of primary keys, like this: ids = [5, 9, 7, 3] All of these nodes have a parent, say with primary key 1. At present, these nodes are ordered [5, 3, 9, 7], how can I re-order them to [5, 9, 7, 3]? I've tried something like this: last_m = MyModel.get(pk = ids.pop(0)) last_m.move_to(last_m.parent, position='first-child') for id in ids: m = MyModel.get(pk = id) m.move_to(last_m, position='right') Which I'd expect to do what I want, per the docs on move_to, but it doesn't seem to change anything. Sometimes it seems to move the first item in ids to be the first child of its parent, sometimes it doesn't. Am I right in my reading of the docs for move_to that calling move_to on a node n with position=right and a target which is a sibling of n will move n to immediately after the target? It's possible I've screwed up my models table in trying to figure this out, so maybe the code above is actually right. It's also possible there's a much more elegant way of doing this (perhaps one that doesn't involve O(n) selects and O(n) updates). Have I misunderstood something? Bonus question: is there a way of forcing django-MPTT to rebuild lft and rght values for all instances of a given model?

    Read the article

  • django-mptt fields showing up twice, breaking SQL

    - by Dominic Rodger
    I'm using django-mptt to manage a simple CMS, with a model called Page, which looks like this (most presumably irrelevant fields removed): class Page(mptt.Model, BaseModel): title = models.CharField(max_length = 20) slug = AutoSlugField(populate_from = 'title') contents = models.TextField() parent = models.ForeignKey('self', null=True, blank=True, related_name='children', help_text = u'The page this page lives under.') removed fields are called attachments, headline_image, nav_override, and published All works fine using SQLite, but when I use MySQL and try and add a Page using the admin (or using ModelForms and the save() method), I get this: ProgrammingError at /admin/mycms/page/add/ (1110, "Column 'level' specified twice") where the SQL generated is: 'INSERT INTO `kaleo_page` (`title`, `slug`, `contents`, `nav_override`, `parent_id`, `published`, `headline_image_id`, `lft`, `rght`, `tree_id`, `level`, `lft`, `rght`, `tree_id`, `level`) VALUES (%s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s)' for some reason I'm getting the django-mptt fields (lft, rght, tree_id and level) twice. It works in SQLite presumably because SQLite is more forgiving about what it accepts than MySQL. get_all_field_names() also shows them twice: >>> Page._meta.get_all_field_names() ['attachments', 'children', 'contents', 'headline_image', 'id', 'level', 'lft', 'nav_override', 'parent', 'published', 'rght', 'slug', 'title', 'tree_id'] Which is presumably why the SQL is bad. What could I have done that would result in those fields appearing twice in get_all_field_names()?

    Read the article

  • Django forms, inheritance and order of form fields

    - by Hannson
    I'm using Django forms in my website and would like to control the order of the fields. Here's how I define my forms: class edit_form(forms.Form): summary = forms.CharField() description = forms.CharField(widget=forms.TextArea) class create_form(edit_form): name = forms.CharField() The name is immutable and should only be listed when the entity is created. I use inheritance to add consistency and DRY principles. What happens which is not erroneous, in fact totally expected, is that the name field is listed last in the view/html but I'd like the name field to be on top of summary and description. I do realize that I could easily fix it by copying summary and description into create_form and loose the inheritance but I'd like to know if this is possible. Why? Imagine you've got 100 fields in edit_form and have to add 10 fields on the top in create_form - copying and maintaining the two forms wouldn't look so sexy then. (This is not my case, I'm just making up an example) So, how can I override this behavior? Edit: Apparently there's no proper way to do this without going through nasty hacks (fiddling with .field attribute). The .field attribute is a SortedDict (one of Django's internal datastructures) which doesn't provide any way to reorder key:value pairs. It does how-ever provide a way to insert items at a given index but that would move the items from the class members and into the constructor. This method would work, but make the code less readable. The only other way I see fit is to modify the framework itself which is less-than-optimal in most situations. In short the code would become something like this: class edit_form(forms.Form): summary = forms.CharField() description = forms.CharField(widget=forms.TextArea) class create_form(edit_form): def __init__(self,*args,**kwargs): forms.Form.__init__(self,*args,**kwargs) self.fields.insert(0,'name',forms.CharField()) That shut me up :)

    Read the article

  • Django's post_save signal behaves weirdly with models using multi-table inheritance

    - by hekevintran
    Django's post_save signal behaves weirdly with models using multi-table inheritance I am noticing an odd behavior in the way Django's post_save signal works when using a model that has multi-table inheritance. I have these two models: class Animal(models.Model): category = models.CharField(max_length=20) class Dog(Animal): color = models.CharField(max_length=10) I have a post save callback called echo_category: def echo_category(sender, **kwargs): print "category: '%s'" % kwargs['instance'].category post_save.connect(echo_category, sender=Dog) I have this fixture: [ { "pk": 1, "model": "animal.animal", "fields": { "category": "omnivore" } }, { "pk": 1, "model": "animal.dog", "fields": { "color": "brown" } } ] In every part of the program except for in the post_save callback the following is true: from animal.models import Dog Dog.objects.get(pk=1).category == u'omnivore' # True When I run syncdb and the fixture is installed, the echo_category function is run. The output from syncdb is: $ python manage.py syncdb --noinput Installing json fixture 'initial_data' from '~/my_proj/animal/fixtures'. category: '' Installed 2 object(s) from 1 fixture(s) The weird thing here is that the dog object's category attribute is an empty string. Why is it not 'omnivore' like it is everywhere else? As a temporary (hopefully) workaround I reload the object from the database in the post_save callback: def echo_category(sender, **kwargs): instance = kwargs['instance'] instance = sender.objects.get(pk=instance.pk) print "category: '%s'" % instance.category post_save.connect(echo_category, sender=Dog) This works but it is not something I like because I must remember to do it when the model inherits from another model and it must hit the database again. The other weird thing is that I must do instance.pk to get the primary key. The normal 'id' attribute does not work (I cannot use instance.id). I do not know why this is. Maybe this is related to the reason why the category attribute is not doing the right thing?

    Read the article

  • How can I filter these Django records?

    - by mipadi
    I have a set of Django models as shown in the following diagram (the names of the reverse relationships are shown in the yellow bubbles): In each relationship, a Person may have 0 or more of the items. Additionally, the slug field is (unfortunately) not unique; multiple Person records may have the same slug fields. Essentially these records are duplicates. I want to obtain a list of all records that meet the following criteria: All duplicate records (that is, having the same slug) with at least one Entry OR at least one Audio OR at least one Episode OR at least one Article. So far, I have the following query: Person.objects.values('slug').annotate(num_records=Count('slug')).filter(num_records__gt=1) This groups all records by slug, then adds a num_records attribute that says how many records have that slug, but the additional filtering is not performed (and I don't even know if this would work right anyway, since, given a set of duplicate records, one may have, e.g., and Entry and the other may have an Article). In a nutshell, I want to find all duplicate records and collapse them, along with their associated models, into one record. What's the best way to do this with Django?

    Read the article

  • Django sphinx works only after app restart.

    - by Lhiash
    Hi, I've set up django-sphinx in my project, which works perfectly only for some time. Later it always returns empty result set. Surprisingly restarting django app fixes it. And search works again but again only for short time (or very limiter number of queries). Heres my sphinx.conf: source src_questions { # data source type = mysql sql_host = xxxxxx sql_user = xxxxxx #replace with your db username sql_pass = xxxxxx #replace with your db password sql_db = xxxxxx #replace with your db name # these two are optional sql_port = xxxxxx #sql_sock = /var/lib/mysql/mysql.sock # pre-query, executed before the main fetch query sql_query_pre = SET NAMES utf8 # main document fetch query sql_query = SELECT q.id AS id, q.title AS title, q.tagnames AS tags, q.html AS text, q.level AS level \ FROM question AS q \ WHERE q.deleted=0 \ # optional - used by command-line search utility to display document information sql_query_info = SELECT title, id, level FROM question WHERE id=$id sql_attr_uint = level } index questions { # which document source to index source = src_questions # this is path and index file name without extension # you may need to change this path or create this folder path = /home/rafal/core_index/index_questions # docinfo (ie. per-document attribute values) storage strategy docinfo = extern # morphology morphology = stem_en # stopwords file #stopwords = /var/data/sphinx/stopwords.txt # minimum word length min_word_len = 3 # uncomment next 2 lines to allow wildcard (*) searches min_infix_len = 1 enable_star = 1 # charset encoding type charset_type = utf-8 } # indexer settings indexer { # memory limit (default is 32M) mem_limit = 64M } # searchd settings searchd { # IP address on which search daemon will bind and accept # optional, default is to listen on all addresses, # ie. address = 0.0.0.0 address = 127.0.0.1 # port on which search daemon will listen port = 3312 # searchd run info is logged here - create or change the folder log = ../log/sphinx.log # all the search queries are logged here query_log = ../log/query.log # client read timeout, seconds read_timeout = 5 # maximum amount of children to fork max_children = 30 # a file which will contain searchd process ID pid_file = searchd.pid # maximum amount of matches this daemon would ever retrieve # from each index and serve to client max_matches = 1000 } and heres my search part from views.py: content = Question.search.query(keywords) if level: content = content.filter(level=level)#level is array of integers There are no errors in any logs, it just isnt returning any results. All help would be most appreciated.

    Read the article

  • Order in many to many relation in Django model

    - by Pietro Speroni
    I am writing a small website to store the papers I have written. The relation papers<- author is important, but the order of the name of the authors (which one is First Author, which one is second order, and so on) is also important. I am just learning Django so I don't know much. In any case so far I have done: from django.db import models class author(models.Model): Name = models.CharField(max_length=60) URLField = models.URLField(verify_exists=True, null=True, blank=True) def __unicode__(self): return self.Name class topic(models.Model): TopicName = models.CharField(max_length=60) def __unicode__(self): return self.TopicName class publication(models.Model): Title = models.CharField(max_length=100) Authors = models.ManyToManyField(author, null=True, blank=True) Content = models.TextField() Notes = models.TextField(blank=True) Abstract = models.TextField(blank=True) pub_date = models.DateField('date published') TimeInsertion = models.DateTimeField(auto_now=True) URLField = models.URLField(verify_exists=True,null=True, blank=True) Topic = models.ManyToManyField(topic, null=True, blank=True) def __unicode__(self): return self.Title This work fine in the sense that I now can define who the authors are. But I cannot order them. How should I do that? Of course I could add a series of relations: first author, second author,... but it would be ugly, and would not be flexible. Any better idea? Thanks

    Read the article

  • Django filters - Using an AllValuesFilter (with a LinkWidget) on a ManyToManyField

    - by magnetix
    This is my first Stack Overflow question, so please let me know if I do anything wrong. I wish to create an AllValues filter on a ManyToMany field using the wonderful django-filters application. Basically, I want to create a filter that looks like it does in the Admin, so I also want to use the LinkWidget too. Unfortunately, I get an error (Invalid field name: 'operator') if I try this the standard way: # Models class Organisation(models.Model): name = models.CharField() ... class Sign(models.Model): name = models.CharField() operator = models.ManyToManyField('Organisation', blank=True) ... # Filter class SignFilter(LinkOrderFilterSet): operator = django_filters.AllValuesFilter(widget=django_filters.widgets.LinkWidget) class Meta: model = Sign fields = ['operator'] I got around this by creating my own filter with the many to many relationship hard coded: # Models class Organisation(models.Model): name = models.CharField() ... class Sign(models.Model): name = models.CharField() operator = models.ManyToManyField('Organisation', blank=True) ... # Filter class MyFilter(django_filters.ChoiceFilter): @property def field(self): cd = {} for row in self.model.objects.all(): orgs = row.operator.select_related().values() for org in orgs: cd[org['id']] = org['name'] choices = zip(cd.keys(), cd.values()) list.sort(choices, key=lambda x:(x[1], x[0])) self.extra['choices'] = choices return super(AllValuesFilter, self).field class SignFilter(LinkOrderFilterSet): operator = MyFilter(widget=django_filters.widgets.LinkWidget) I am new to Python and Django. Can someone think of a more generic/elegant way of doing this?

    Read the article

  • Django 1.1 template question

    - by Bovril
    Hi All, I'm a little stuck trying to get my head around a django template. I have 2 objects, a cluster and a node I would like a simple page that lists... [Cluster 1] [associated node 1] [associated node 2] [associated node 3] [Cluster 2] [associated node 4] [associated node 5] [associated node 6] I've been using Django for about 2 days so if i've missed the point, please be gentle :) Models - class Node(models.Model): name = models.CharField(max_length=30) description = models.TextField() cluster = models.ForeignKey(Cluster) def __unicode__(self): return self.name class Cluster(models.Model): name = models.CharField(max_length=30) description = models.TextField() def __unicode__(self): return self.name Views - def DSAList(request): clusterlist = Cluster.objects.all() nodelist = Node.objects.all() t = loader.get_template('dsalist.html') v = Context({ 'CLUSTERLIST' : clusterlist, 'NODELIST' : nodelist, }) return HttpResponse(t.render(v)) Template - <body> <TABLE> {% for cluster in CLUSTERLIST %} <tr> <TD>{{ cluster.name }}</TD> {% for node in NODELIST %} {% if node.cluster.id == cluster.id %} <tr> <TD>{{ node.name }}</TD> </tr> {% endif %} {% endfor %} </tr> {% endfor %} </TABLE> </body> Any ideas ?

    Read the article

  • Django sub-applications & module structure

    - by Rob Golding
    I am developing a Django application, which is a large system that requires multiple sub-applications to keep things neat. Therefore, I have a top level directory that is a Django app (as it has an empty models.py file), and multiple subdirectories, which are also applications in themselves. The reason I have laid my application out in this way is because the sub-applications are separated, but they would never be used on their own, outside the parent application. It therefore makes no sense to distribute them separately. When installing my application, the settings file has to include something like this: INSTALLED_APPS = ( ... 'myapp', 'myapp.subapp1', 'myapp.subapp2', ... ) ...which is obviously suboptimal. This also has the slightly nasty result of requiring that all the sub-applications are referred to by their "inner" name (i.e. subapp1, subapp2 etc.). For example, if I want to reset the database tables for subapp1, I have to type: python manage.py reset subapp1 This is annoying, especially because I have a sub-app called core - which is likely to conflict with another application's name when my application is installed in a user's project. Am I doing this completely wrongly, or is there away to force these "inner" apps to be referred to by their full name?

    Read the article

< Previous Page | 18 19 20 21 22 23 24 25 26 27 28 29  | Next Page >