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• Is this XML file correct for RSS feeding?

  - by Hermet

  Hi guys I am generating a XML-RSS type file from PHP. The output for example is like this <?xml version="1.0" encoding="iso-8859-1"?> <rss version="2.0"> <channel> <title>Mi web mola</title> <link>http://www.dominio.com/blog.php</link> <language>es-ES</language> <description>Mallas y eso</description> <generator>Autor</generator> <item> <title>Articulo de prueba</title> <link>http://www.midominio.com/2342</link> <pubDate>14/06/2010</pubDate> <description><![CDATA[Descripcion de prueba bla bla bla]]></description> <content:encoded><![CDATA[Contenido prueba]]></content:encoded> </item> </channel> </rss> ... and all I can see in the Firefox preview is the title and the description of the blog, not the items, but in the source it appears correctly, so I've thought it must be a parse error or something like that.. What could be wrong? Again, excuse me for my bad english, and thank you very much.

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• Exclude category from main RSS feed, but not all feeds

  - by jamEs

  I've got a blog that supplies content to multiple MailChimp newsletters via RSS. The first newsletter works fine, but the second I'm having issues with. The issue I have is that the second newsletter has "hidden" content. This content isn't meant for wide consumption, so it doesn't appear on the frontpage, but is accessible elsewhere on the site. The snafu with this is that not all of this content is hidden, just some of it, while other pieces of content for this newsletter could overlap with the first newsletter. This obviously makes excluding everything problematic, as they could be assigned multiple categories, some of which I wouldn't want hidden. The issue I'm running into is that I have a way to exclude this content from the frontpage, but not from the main RSS feed. I'm using WP Hide Post for this, which allows me to exclude from feed, which in turn removes it from all feeds, including the ones that feed the newsletter. I'm currently using /feed?cat=XXX to reference these feeds. Is there a way to make it so category feeds still work, just the main /feed RSS would exclude it?

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• Aggregating and displaying content from hundreds of RSS feeds

  - by Andrew LeClair

  I'd like to build a website that aggregates and displays content from hundreds of RSS feeds. The feeds will be from different sites: Twitter, Flickr, Tumblr, etc, so the content will be very heterogenous. In a perfect world — and this is more of a side issue — I would like to allow other people to help manage the list of feeds and assign tags to the content from each individual feed so that you can filter the items that are displayed. What I've tried so far: Google Feeds API – I thought this would be the answer, but unless I'm missing something, the FeedController will only output the collected feed content as separate lists. Is there any way to ask the Google Feeds API to aggregate and sort the content from many RSS feeds before displaying? Yahoo! Pipes – This also seemed like a good solution at first. I setup a Pipe that accesses a list of RSS feeds stored in a Google Doc spreadsheet and then aggregates the content. However, the output leaves a lot to be desired; Tumblr video posts, for example, only show a title and a permalink to the post, the embedded Youtube video is lost. PHP – I've seen this question, which looks like a good approach. I'm less proficient in PHP, so although I'm willing to learn, I'd ideally like to find a different approach. Any thoughts? Thanks.

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• 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'

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• 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).

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• 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?

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• 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

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• '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

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• 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?

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• 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.)

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• 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?

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• 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()?

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• 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 :)

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• 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?

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• 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?

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• 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.

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• 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

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• 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?

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• 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 ?

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• django dynamically deduce SITE_ID according to the domain

  - by dcrodjer

  I am trying to develop a site which will render multiple customized sites according to the domain name (subdomain to be more precise). My all the domain names are redirected to the So for each site there will be a corresponding model which defines how the site should look (SITE - SITE_SETTINGS) What will be the best way to utilize the django sites framework to get the SITE_ID of the current site from the domain name instead of hard-coding it in the settings files (django sites documentation) and run database queries, render the views accordingly? If using multiple settings file is my only option can this (wsgi script handle domain name) be done? Update So finally, following lukes answer, what I will do is define a custom middleware which makes the views available with the important vars required according to the domain. And as far as sitemaps and comments is concerned, I will have to customize sitemaps app and a custom sites model on which the other models of sites will be based. And since the comments system is based on the hard-coded sitemap ID I can use it just as is on the models (models will already be filtered according to the site based on my sites framework) though the permalink feature will have to be customized. So a lot of customization. Please suggest if I am going wrong anywhere in this because I have to ensure that the features of the project are optimized. Thanks!

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• 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?

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• Django Model Formset Pre-Filled Value Problem

  - by user552377

  Hi, i'm trying to use model formsets with Django. When i load forms template, i see that it's filled-up with previous values. Is there a caching mechanism that i should stop, or what? Thanks for your help, here is my code: models.py class FooModel( models.Model ): a_field = models.FloatField() b_field = models.FloatField() def __unicode__( self ): return self.a_field forms.py from django.forms.models import modelformset_factory FooFormSet = modelformset_factory(FooModel) views.py def foo_func(request): if request.method == 'POST': formset = FooFormSet(request.POST, request.FILES, prefix='foo_prefix' ) if formset.is_valid(): formset.save() return HttpResponseRedirect( '/true/' ) else: return HttpResponseRedirect( '/false/' ) else: formset = FooFormSet(prefix='foo_prefix') variables = RequestContext( request , { 'formset':formset , } ) return render_to_response('footemplate.html' , variables ) template: <form method="post" action="."> {% csrf_token %} <input type="submit" value="Submit" /> <table id="FormsetTable" border="0" cellpadding="0" cellspacing="0"> <tbody> {% for form in formset.forms %} <tr> <td>{{ form.a_field }}</td> <td>{{ form.b_field }}</td> </tr> {% endfor %} </tbody> </table> {{ formset.management_form }} </form>

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• Django: Create custom template tag -> ImportError

  - by Alexander Scholz

  I'm sorry to ask this again, but I tried several solutions from stack overflow and some tutorials and I couldn't create a custom template tag yet. All I get is ImportError: No module named test_tag when I try to start the server via python manage.py runserver. I created a very basic template tag (found here: django templatetag?) like so: My folder structure: demo manage.py test __init__.py settings.py urls.py ... templatetags __init__.py test_tag.py test_tag.py: from django import template register = template.Library() @register.simple_tag def test_tag(input): if "foo" == input: return "foo" if "bar" == input: return "bar" if "baz" == input: return "baz" return "" index.html: {% load test_tag %} <html> <head> ... </head> <body> {% cms_toolbar %} {% foobarbaz "bar" %} {% foobarbaz "elephant" %} {% foobarbaz "foo" %} </body> </html> and my settings.py: INSTALLED_APPS = ( ... 'test_tag', ... ) Please let me know if you need further information from my settings.py and what I did wrong so I can't even start my server. (If I delete 'test_tag' from installed apps I can start the server but I get the error that test_tag is not known, of course). Thanks

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• Performance Problems with Django's F() Object

  - by JayhawksFan93

  Has anyone else noticed performance issues using Django's F() object? I am running Windows XP SP3 and developing against the Django trunk. A snippet of the models I'm using and the query I'm building are below. When I have the F() object in place, each call to a QuerySet method (e.g. filter, exclude, order_by, distinct, etc.) takes approximately 2 seconds, but when I comment out the F() clause the calls are sub-second. I had a co-worker test it on his Ubuntu machine, and he is not experiencing the same performance issues I am with the F() clause. Anyone else seeing this behavior? class Move (models.Model): state_meaning = models.CharField( max_length=16, db_index=True, blank=True, default='' ) drop = models.ForeignKey( Org, db_index=True, null=False, default=1, related_name='as_move_drop' ) class Split(models.Model): state_meaning = models.CharField( max_length=16, db_index=True, blank=True, default='' ) move = models.ForeignKey( Move, related_name='splits' ) pickup = models.ForeignKey( Org, db_index=True, null=False, default=1, related_name='as_split_pickup' ) pickup_date = models.DateField( null=True, default=None ) drop = models.ForeignKey( Org, db_index=True, null=False, default=1, related_name='as_split_drop' ) drop_date = models.DateField( null=True, default=None, db_index=True ) def get_splits(begin_date, end_date): qs = Split.objects \ .filter(state_meaning__in=['INPROGRESS','FULFILLED'], drop=F('move__drop'), # <<< the line in question pickup_date__lte=end_date) elapsed = timer.clock() - start print 'qs1 took %.3f' % elapsed start = timer.clock() qs = qs.filter(Q(drop_date__gte=begin_date) | Q(drop_date__isnull=True)) elapsed = timer.clock() - start print 'qs2 took %.3f' % elapsed start = timer.clock() qs = qs.exclude(move__state_meaning='UNFULFILLED') elapsed = timer.clock() - start print 'qs3 took %.3f' % elapsed start = timer.clock() qs = qs.order_by('pickup_date', 'drop_date') elapsed = timer.clock() - start print 'qs7 took %.3f' % elapsed start = timer.clock() qs = qs.distinct() elapsed = timer.clock() - start print 'qs8 took %.3f' % elapsed

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• Django Save Incomplete Progress on Form

  - by jimbob

  I have a django webapp with multiple users logging in and fill in a form. Some users may start filling in a form and lack some required data (e.g., a grant #) needed to validate the form (and before we can start working on it). I want them to be able to fill out the form and have an option to save the partial info (so another day they can log back in and complete it) or submit the full info undergoing validation. Currently I'm using ModelForm for all the forms I use, and the Model has constraints to ensure valid data (e.g., the grant # has to be unique). However, I want them to be able to save this intermediary data without undergoing any validation. The solution I've thought of seems rather inelegant and un-django-ey: create a "Save Partial Form" button that saves the POST dictionary converts it to a shelf file and create a "SavedPartialForm" model connecting the user to partial forms saved in the shelf. Does this seem sensible? Is there a better way to save the POST dict directly into the db? Or is an add-on module that does this partial-save of a form (which seems to be a fairly common activity with webforms)? My biggest concern with my method is I want to eventually be able to do this form-autosave automatically (say every 10 minutes) in some ajax/jquery method without actually pressing a button and sending the POST request (e.g., so the user isn't redirected off the page when autosave is triggered). I'm not that familiar with jquery and am wondering if it would be possible to do this.

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