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  • Returning form errors for AJAX request in Django

    - by mridang
    Hi Guys, I've been finding my way around Django and jQuery. I've built a basic form in Django. On clicking submit, I'm using jQuey to make an AJAX request to the sever to post my data. This bit seems to work fine and I've managed to save the data. Django returns a ValidatioError when a form is invalid. Could anyone tell me how to return this set of error messages as a response to my AJAX request so i can easily iterate through it using JS and do whatever? I found this snippet. Looking at the JS bit (processJson) you'll see that he seems to get the error messages by extracting them from the response HTML. It seems kinda kludgy to me. Is this a best way to go about it? My apologies for any vagueness. Thanks in advance.

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  • django urlconf or .htaccess trouble

    - by Zayatzz
    Hello I am running my django project from subfolder of a website. Lets say the address where my project is meant to open from is. http://example.com/myproject/ the myproject folder is root folder for my user account. In that folder i have fcgi script that starts my project. The .htaccess file in the folder contains this: RewriteEngine On RewriteCond %{REQUEST_FILENAME} !-f RewriteRule ^(.*)$ mysite.fcgi/$1 [QSA,L] The trouble is, that at some cases, instead of redireting user to page like http://example.com/myproject/social/someurl/ it redirects to http://example.com/social/someurl/ which does not work. What i want to know is how to fix this problem. Is this django problem and i should change it with urconf and add myproject to all urls, or should i do this with .htaccess? I found similar question, which, sadly, remains unanswered: http://stackoverflow.com/questions/2321154/how-to-write-htaccess-if-django-project-is-in-subfolder-and-subdomain Alan.

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  • Django ORM and multiprocessing

    - by Ankur Gupta
    Hi, I am using Django ORM in my python script in a decoupled fashion i.e. it's not running in context of a normal Django Project. I am also using the multi processing module. And different process in turn are making queries. The process ran successfully for an hr and exited with this message "IOError: [Errno 32] Broken pipe" Upon futhur diagnosis and debugging this error pops up when I call save() on the model instance. I am wondering Is Django ORM Process save ? Why would this error arise else ? Cheers Ankur

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  • Using Django Admin for a custom database solution

    - by Prashanth Ellina
    A client wants to have a simple intranet application to manage his process. He runs a Quarry and wishes to track number of loads delivered per day and associated activities. Since I knew about Django's excellent Admin interface, I figured I could define the "Schema" in models.py and have Django Admin generate the forms. I did exactly that and the result is not bad at all. I've been able to customize the look and feel to suit the client's taste. Some questions: Is Django Admin the right choice for such a use-case? Will I run to problems in the future due to flexibility of the framework? Is there a better framework out there specifically designed for this use-case (general Database management for small businesses)? I prefer ones written in Python since I can hack it up to customize. Thanks!

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  • Django + Postgres: How to specify sequence for a field

    - by Giovanni Di Milia
    I have this model in django: class JournalsGeneral(models.Model): jid = models.AutoField(primary_key=True) code = models.CharField("Code", max_length=50) name = models.CharField("Name", max_length=2000) url = models.URLField("Journal Web Site", max_length=2000, blank=True) online = models.BooleanField("Online?") active = models.BooleanField("Active?") class Meta: db_table = u'journals_general' verbose_name = "Journal General" ordering = ['code'] def __unicode__(self): return self.name My problem is that in the DB (Postgres) the name of the sequence connected to jid is not journals_general_jid_seq as expected by Django but it has a different name. Is there a way to specify which sequence Django has to use for an AutoField? In the documentation I read I was not able to find an answer.

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  • Django Passing Custom Form Parameters to Formset

    - by Paolo Bergantino
    I have a Django Form that looks like this: class ServiceForm(forms.Form): option = forms.ModelChoiceField(queryset=ServiceOption.objects.none()) rate = forms.DecimalField(widget=custom_widgets.SmallField()) units = forms.IntegerField(min_value=1, widget=custom_widgets.SmallField()) def __init__(self, *args, **kwargs): affiliate = kwargs.pop('affiliate') super(ServiceForm, self).__init__(*args, **kwargs) self.fields["option"].queryset = ServiceOption.objects.filter(affiliate=affiliate) I call this form with something like this: form = ServiceForm(affiliate=request.affiliate) Where request.affiliate is the logged in user. This works as intended. My problem is that I now want to turn this single form into a formset. What I can't figure out is how I can pass the affiliate information to the individual forms when creating the formset. According to the docs to make a formset out of this I need to do something like this: ServiceFormSet = forms.formsets.formset_factory(ServiceForm, extra=3) And then I need to create it like this: formset = ServiceFormSet() Now how can I pass affiliate=request.affiliate to the individual forms this way?

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  • Cannot Extend Django 1.2.1 Admin Template

    - by jcady
    I am attempting to override/extend the header for the Django admin in version 1.2.1. However when I try to extend the admin template and simply change what I need documented here: http://docs.djangoproject.com/en/dev/ref/contrib/admin/#overriding-vs-replacing-an-admin-template), I run into a recursion problem. I have an index.html file in my project's templates/admin/ directory that starts with {% extends "admin/index.html" %} But it seems that this is referencing the local index file (a.k.a. itself) rather than the default Django copy. I want to extend the default Django template and simply change a few blocks. When I try this file, I get a recursion depth error. How can I extend parts of the admin? Thanks.

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  • Add a prefix do Django comment form

    - by Stefan Manastirliu
    I would like to add a prefix to each django comment form. I'm using multiply comment forms in the same page and depsite it's working well, i don't like having many input fields with the same id attribute like <input type="text" name="honeypot" id="id_honeypot" />. So, is there a way to tell django do add a prefix to each form instance? I know i can do it with other forms when i create a form instance in this waynewform = CustomForm(prefix="a") but using Django's comment system, this part is handled by a comment template tag {% get_comment_form for [object] as [varname] %}. Can I tell to the template tag to add a prefix?

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  • Prevent Django from redirecting to add trailing slash

    - by konrad
    UPDATED: Sorry, it looks like it's Apache that's rewriting it for some reason, not Django. I'll investigate further and post my findings. I need to add a /xmlrpc.php to my Byteflow installation to handle an application that is written for PHP blog engines and uses this hardcoded path. For some reason Byteflow appends a slash to this URL using a 301 Moved Permanently redirect, which breaks the application. It does not do so for the /robots.txt that is configured in a similar way. Relevant lines from the project urls.py: url(r'^xmlrpc.php$', 'django_xmlrpc.views.xmlrpc_handler'), url(r'^robots.txt$', include('robots.urls')), I read that the behavior was changed in the Django codebase in commit 6852 (in 2007) to prevent redirects being done for urls that have been explicitly configured not to contain any trailing slashes. I'm using Django 1.1. I assume that once I have fixed this problem, I should be able to use this application with Byteflow, because the application uses the MetaWeblog XML-RPC API. Any clue?

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  • Django: Geocoding an address on form submission?

    - by User
    Trying to wrap my head around django forms and the django way of doing things. I want to create a basic web form that allows a user to input an address and have that address geocoded and saved to a database. I created a Location model: class Location(models.Model): address = models.CharField(max_length=200) city = models.CharField(max_length=100) state = models.CharField(max_length=100, null=True) postal_code = models.CharField(max_length=100, null=True) country = models.CharField(max_length=100) latitude = models.DecimalField(max_digits=18, decimal_places=10, null=True) longitude = models.DecimalField(max_digits=18, decimal_places=10, null=True) And defined a form: class LocationForm(forms.ModelForm): class Meta: model = models.Location exclude = ('latitude','longitude') In my view I'm using form.save() to save the form. This works and saves an address to the database. I created a module to geocode an address. I'm not sure what the django way of doing things is, but I guess in my view, before I save the form, I need to geocode the address and set the lat and long. How do I set the latitude and longitude before saving?

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  • More than one profile in Django?

    - by JPC
    Is it possible to use Django's user authentication features with more than one profile? Currently I have a settings.py file that has this in it: AUTH_PROFILE_MODULE = 'auth.UserProfileA' and a models.py file that has this in it: from django.db import models from django.contrib.auth.models import User class UserProfileA(models.Model): company = models.CharField(max_length=30) user = models.ForeignKey(User, unique=True) that way, if a user logs in, I can easily get the profile because the User has a get_profile() method. However, I would like to add UserProfileB. From looking around a bit, it seems that the starting point is to create a superclass to use as the AUTH_PROFILE_MODULE and have both UserProfileA and UserProfileB inherit from that superclass. The problem is, I don't think the get_profile() method returns the correct profile. It would return an instance of the superclass. I come from a java background (polymorphism) so I'm not sure exactly what I should be doing. Thanks!

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  • Front-end structure of large scale Django project

    - by Saike
    Few days ago, I started to work in new company. Before me, all front-end and backend code was written by one man (oh my...). As you know, Django app contains two main directories for front-end: /static - for static(public) files and /templates - for django templates Now, we have large application with more than 10 different modules like: home, admin, spanel, mobile etc. This is current structure of files and directories: FIRST - /static directory. As u can see, it is mixed directories with some named like modules, some contains global libs. one more: SECOND - /templates directory. Some directories named like module with mixed templates, some depends on new version =), some used only in module, but placed globally. and more: I think, that this is ugly, non-maintable, put-in-stress structure! After some time spend, i suggest to use this scheme, that based on module-structure. At first, we have version directories, used for save full project backup, includes: /DEPRECATED directory - for old, unused files and /CURRENT (Active) directory, that contains production version of project. I think it's right, because we can access to older or newer version files fast and easy. Also, we are saved from broken or wrong dependencies between different versions. Second, in every version we have standalone modules and global module. Every module contains own /static and /templates directories. This structure used to avoid broken or wrong dependencies between different modules, because every module has own js app, css tables and local images. Global module contains all libraries, main stylesheets and images like logos or favicon. I think, this structure is much better to maintain, update, refactoring etc. My question is: How do you think, is this scheme better than current? Can this scheme live, or it is not possible to implement this in Django app?

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  • Django + gunicorn + virtualenv + Supervisord issue

    - by Florian Le Goff
    Dear all, I have a strange issue with my virtualenv + gunicorn setup, only when gunicorn is launched via supervisord. I do realize that it may very well be an issue with my supervisord and I would appreciate any feedback on a better place to ask for help... In a nutshell : when I run gunicorn from my user shell, inside my virtualenv, everything is working flawlessly. I'm able to access all the views of my Django project. When gunicorn is launched by supervisord at the system startup, everything is OK. But, if I have to kill the gunicorn_django processes, or if I perform a supervisord restart, once that gunicorn_django has relaunched, every request is answered with a weird Traceback : (...) File "/home/hc/prod/venv/lib/python2.6/site-packages/Django-1.2.5-py2.6.egg/django/db/__init__.py", line 77, in connection = connections[DEFAULT_DB_ALIAS] File "/home/hc/prod/venv/lib/python2.6/site-packages/Django-1.2.5-py2.6.egg/django/db/utils.py", line 92, in __getitem__ backend = load_backend(db['ENGINE']) File "/home/hc/prod/venv/lib/python2.6/site-packages/Django-1.2.5-py2.6.egg/django/db/utils.py", line 50, in load_backend raise ImproperlyConfigured(error_msg) TemplateSyntaxError: Caught ImproperlyConfigured while rendering: 'django.db.backends.postgresql_psycopg2' isn't an available database backend. Try using django.db.backends.XXX, where XXX is one of: 'dummy', 'mysql', 'oracle', 'postgresql', 'postgresql_psycopg2', 'sqlite3' Error was: cannot import name utils Full stack available here : http://pastebin.com/BJ5tNQ2N I'm running... Ubuntu/maverick (up-to-date) Python = 2.6.6 virtualenv = 1.5.1 gunicorn = 0.12.0 Django = 1.2.5 psycopg2 = '2.4-beta2 (dt dec pq3 ext)' gunicorn configuration : backlog = 2048 bind = "127.0.0.1:8000" pidfile = "/tmp/gunicorn-hc.pid" daemon = True debug = True workers = 3 logfile = "/home/hc/prod/log/gunicorn.log" loglevel = "info" supervisord configuration : [program:gunicorn] directory=/home/hc/prod/hc command=/home/hc/prod/venv/bin/gunicorn_django -c /home/hc/prod/hc/gunicorn.conf.py user=hc umask=022 autostart=True autorestart=True redirect_stderr=True Any advice ? I've been stuck on this one for quite a while. It seems like some weird memory limit, as I'm not enforcing anything special : $ ulimit -a core file size (blocks, -c) 0 data seg size (kbytes, -d) unlimited scheduling priority (-e) 20 file size (blocks, -f) unlimited pending signals (-i) 16382 max locked memory (kbytes, -l) 64 max memory size (kbytes, -m) unlimited open files (-n) 1024 pipe size (512 bytes, -p) 8 POSIX message queues (bytes, -q) 819200 real-time priority (-r) 0 stack size (kbytes, -s) 8192 cpu time (seconds, -t) unlimited max user processes (-u) unlimited virtual memory (kbytes, -v) unlimited file locks (-x) unlimited Thank you.

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  • Dynamically create and save image with Django and PIL/Django-Photologue

    - by Travis
    I want to generate a page of html with dynamic content and then save the result as an image as well as display the page to the user. So, user signs up to attend conference and gives their name and org. That data is combined with html/css elements to show what their id badge for the conference will look like (their name and org on top of our conference logo background) for preview. Upon approval, the page is saved on the server to an image format (PNG, PDF or JPG) to be printed onto a physical badge by an admin later. I am using Django and django-photologue powered by PIL. The view might look like this # app/views.py def badgepreview(request, attendee_id): u = User.objects.get(id=attendee_id) name = u.name org = u.org return render_to_response('app/badgepreview.html', {'name':name,'org':org,}, context_instance = RequestContext(request), ) The template could look like this {# templates/app/badgepreview.html #} {% extends "base.html" %} {% block page_body %} <div style='background:url(/site_media/img/logo_bg.png) no-repeat;'> <h4>{{ name }}</h4> <h4>{{ org }}</h4> </div> {% endblock %} simple, but how do I save the result? Or is there a better way to get this done?

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  • Django upload failing on request data read error

    - by Jake
    Hi All, I've got a Django app that accepts uploads from jQuery uploadify, a jQ plugin that uses flash to upload files and give a progress bar. Files under about 150k work, but bigger files always fail and almost always at around 192k (that's 3 chunks) completed, sometimes at around 160k. The Exception I get is below. exceptions.IOError request data read error File "/usr/lib/python2.4/site-packages/django/core/handlers/wsgi.py", line 171, in _get_post self._load_post_and_files() File "/usr/lib/python2.4/site-packages/django/core/handlers/wsgi.py", line 137, in _load_post_and_files self._post, self._files = self.parse_file_upload(self.META, self.environ[\'wsgi.input\']) File "/usr/lib/python2.4/site-packages/django/http/__init__.py", line 124, in parse_file_upload return parser.parse() File "/usr/lib/python2.4/site-packages/django/http/multipartparser.py", line 192, in parse for chunk in field_stream: File "/usr/lib/python2.4/site-packages/django/http/multipartparser.py", line 314, in next output = self._producer.next() File "/usr/lib/python2.4/site-packages/django/http/multipartparser.py", line 468, in next for bytes in stream: File "/usr/lib/python2.4/site-packages/django/http/multipartparser.py", line 314, in next output = self._producer.next() File "/usr/lib/python2.4/site-packages/django/http/multipartparser.py", line 375, in next data = self.flo.read(self.chunk_size) File "/usr/lib/python2.4/site-packages/django/http/multipartparser.py", line 405, in read return self._file.read(num_bytes) When running locally on the Django development server, big files work. I've tried setting my FILE_UPLOAD_HANDLERS = ("django.core.files.uploadhandler.TemporaryFileUploadHandler",) in case it was the memory upload handler, but it made no difference. Does anyone know how to fix this?

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  • Where does django look for sqlite3 installation/libraries?

    - by gath
    Am having a bit of a problem making my django application run in SUSE linux 9. I have Python2.5 installed well, Django 1.0 installed well. Am able to execute django command django-admin startproject fine But when i run the runserver command i get the error below. i have a folder with sqlite3, i can go in there and actually run the sqlite3* application, now am wondering where does Django look for the sqlite libraries? and how can i fix this? Validating models... Unhandled exception in thread started by <function inner_run at 0x2a96cb4f50> Traceback (most recent call last): File "/usr/local/lib/python2.5/site-packages/django/core/management/commands/runserver.py", line 48, in inner_run self.validate(display_num_errors=True) File "/usr/local/lib/python2.5/site-packages/django/core/management/base.py", line 122, in validate num_errors = get_validation_errors(s, app) File "/usr/local/lib/python2.5/site-packages/django/core/management/validation.py", line 22, in get_validation_errors from django.db import models, connection File "/usr/local/lib/python2.5/site-packages/django/db/__init__.py", line 16, in <module> backend = __import__('%s%s.base' % (_import_path, settings.DATABASE_ENGINE), {}, {}, ['']) File "/usr/local/lib/python2.5/site-packages/django/db/backends/sqlite3/base.py", line 27, in <module> raise ImproperlyConfigured, "Error loading %s module: %s" % (module, exc) django.core.exceptions.ImproperlyConfigured: Error loading sqlite3 module: No module named _sqlite3 Gath

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  • Django: DatabaseLockError exception with Djapian

    - by jul
    Hi, I've got the exception shown below when executing indexer.update(). I have no idea about what to do: it used to work and now index database seems "locked". Anybody can help? Thanks Environment: Request Method: POST Request URL: http://piem.org:8000/restaurant/add/ Django Version: 1.1.1 Python Version: 2.5.2 Installed Applications: ['django.contrib.auth', 'django.contrib.contenttypes', 'django.contrib.sessions', 'django.contrib.comments', 'django.contrib.sites', 'django.contrib.admin', 'registration', 'djapian', 'resto', 'multilingual'] Installed Middleware: ('django.middleware.common.CommonMiddleware', 'django.contrib.sessions.middleware.SessionMiddleware', 'django.contrib.auth.middleware.AuthenticationMiddleware', 'django.middleware.locale.LocaleMiddleware', 'multilingual.middleware.DefaultLanguageMiddleware') Traceback: File "/var/lib/python-support/python2.5/django/core/handlers/base.py" in get_response 92. response = callback(request, *callback_args, **callback_kwargs) File "/home/jul/atable/../atable/resto/views.py" in addRestaurant 639. Restaurant.indexer.update() File "/home/jul/python-modules/Djapian-2.3.1-py2.5.egg/djapian/indexer.py" in update 181. database = self._db.open(write=True) File "/home/jul/python-modules/Djapian-2.3.1-py2.5.egg/djapian/database.py" in open 20. xapian.DB_CREATE_OR_OPEN, File "/usr/lib/python2.5/site-packages/xapian.py" in __init__ 2804. _xapian.WritableDatabase_swiginit(self,_xapian.new_WritableDatabase(*args)) Exception Type: DatabaseLockError at /restaurant/add/ Exception Value: Unable to acquire database write lock on /home/jul/atable /djapian_spaces/resto/restaurant/resto.index.restaurantindexer: already locked

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  • Rebuilding old (2010) django project in 2012

    - by birgit
    I am trying to make an old Django project run again. After seemingly having solved issues with old sorl.thumbnail versions and deprecated expressions I now get this error when running python manage.py runserver I also tried to copy & paste my old files into a new Django project and get the exactly same error. Maybe someone here has a clue where the problem lies? Unhandled exception in thread started by <bound method Command.inner_run of <django.contrib.staticfiles.management.commands.runserver.Command object at 0x2a80510>> Traceback (most recent call last): File "/usr/lib/python2.7/dist-packages/django/core/management/commands/runserver.py", line 88, in inner_run self.validate(display_num_errors=True) File "/usr/lib/python2.7/dist-packages/django/core/management/base.py", line 249, in validate num_errors = get_validation_errors(s, app) File "/usr/lib/python2.7/dist-packages/django/core/management/validation.py", line 35, in get_validation_errors for (app_name, error) in get_app_errors().items(): File "/usr/lib/python2.7/dist-packages/django/db/models/loading.py", line 146, in get_app_errors self._populate() File "/usr/lib/python2.7/dist-packages/django/db/models/loading.py", line 61, in _populate self.load_app(app_name, True) File "/usr/lib/python2.7/dist-packages/django/db/models/loading.py", line 78, in load_app models = import_module('.models', app_name) File "/usr/lib/python2.7/dist-packages/django/utils/importlib.py", line 35, in import_module __import__(name) File "/home/me/Documents/wdws/wdws/../wdws/cityofwindows/models.py", line 73, in <module> class Image(models.Model): File "/home/me/Documents/wdws/wdws/../wdws/cityofwindows/models.py", line 83, in Image 'large': {'size': (640, 640)}, File "/usr/lib/python2.7/dist-packages/django/db/models/fields/files.py", line 233, in __init__ super(FileField, self).__init__(verbose_name, name, **kwargs) TypeError: __init__() got an unexpected keyword argument 'extra_thumbnails' I need to re-build the project just for visual documentation locally... so also any hints on how to quickly re-run outdated django-projects are very welcome!! Thanks a lot (using Ubuntu 12.04)

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  • In Django, what's the best way to handle optional url parameters from the template?

    - by Thierry Lam
    I have the following type of urls which are both valid: hello/ hello/1234/ My urls.py has the following: urlpatterns = patterns('hello.views', url(r'^$', 'index', name='index'), url(r'^(?P<user_id>\d+)/$', 'index', name='index'), ) In my views.py, when I pass user_id to the template, it defaults to 0 if not specified. My template looks like the following, I'm using namespace hello for my hello app: {% url hello:index user_id %} If user_id is not specified, the url defaults to hello/0/. The only way I can think of preventing the default 0 from showing in the url is by an if stmt: {% if user_id %} {% url hello:index user_id %} {% else %} {% url hello:index %} {% endif %} The above will give me hello/ if there are no user_id and hello/1234/ if it's present. Is the above solution the best way to solve this issue?

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  • Django's self.client.login(...) does not work in unit tests

    - by thebossman
    I have created users for my unit tests in two ways: 1) Create a fixture for "auth.user" that looks roughly like this: { "pk": 1, "model": "auth.user", "fields": { "username": "homer", "is_active": 1, "password": "sha1$72cd3$4935449e2cd7efb8b3723fb9958fe3bb100a30f2", ... } } I've left out the seemingly unimportant parts. 2) Use 'create_user' in the setUp function (although I'd rather keep everything in my fixtures class): def setUp(self): User.objects.create_user('homer', '[email protected]', 'simpson') Note that the password is simpson in both cases. I've verified that this info is correctly being loaded into the test database time and time again. I can grab the User object using User.objects.get. I can verify the password is correct using 'check_password.' The user is active. Yet, invariably, self.client.login(username='homer', password='simpson') FAILS. I'm baffled as to why. I think I've read every single Internet discussion pertaining to this. Can anybody help? The login code in my unit test looks like this: login = self.client.login(username='homer', password='simpson') self.assertTrue(login) Thanks.

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  • How can I display multiple django modelformset forms in a grouped fieldsets?

    - by JT
    I have a problem with needing to provide multiple model backed forms on the same page. I understand how to do this with single forms, i.e. just create both the forms call them something different then use the appropriate names in the template. Now how exactly do you expand that solution to work with modelformsets? The wrinkle, of course, is that each 'form' must be rendered together in the appropriate fieldset. For example I want my template to produce something like this: <fieldset> <label for="id_base-0-desc">Home Base Description:</label> <input id="id_base-0-desc" type="text" name="base-0-desc" maxlength="100" /> <label for="id_likes-0-icecream">Want ice cream?</label> <input type="checkbox" name="likes-0-icecream" id="id_likes-0-icecream" /> </fieldset> <fieldset> <label for="id_base-1-desc">Home Base Description:</label> <input id="id_base-1-desc" type="text" name="base-1-desc" maxlength="100" /> <label for="id_likes-1-icecream">Want ice cream?</label> <input type="checkbox" name="likes-1-icecream" id="id_likes-1-icecream" /> </fieldset> I am using a loop like this to process the results (after form validation) base_models = base_formset.save(commit=False) like_models = like_formset.save(commit=False) for base_model, likes_model in map(None, base_models, likes_models): which works as I'd expect (I'm using map because the # of forms can be different). The problem is that I can't figure out a way to do the same thing with the templating engine. The system does work if I layout all the base models together then all the likes models after wards, but it doesn't meet the layout requirements. EDIT: Updated the problem statement to be more clear about what exactly I'm processing (I'm processing models not forms in the for loop)

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  • How to put an InlineFormSet into a ModelFormSet in Django?

    - by Jannis
    Hi, I'd like to display a number of forms via a ModelFormSet where each one of the forms displays in turn InlineFormSets for all objects connected to the object. Now I'm not really sure how to provide the instances for each ModelFormSet. I thought about subclassing BaseModelFormSet but I have no clue on where to start and would like to know whether this is possible at all before I go through all the trouble. Thanks in advance!

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  • How can I display multiple django modelformset forms together?

    - by JT
    I have a problem with needing to provide multiple model backed forms on the same page. I understand how to do this with single forms, i.e. just create both the forms call them something different then use the appropriate names in the template. Now how exactly do you expand that solution to work with modelformsets? The wrinkle, of course, is that each 'form' must be rendered together in the appropriate fieldset. For example I want my template to produce something like this: <fieldset> <label for="id_base-0-desc">Home Base Description:</label> <input id="id_base-0-desc" type="text" name="base-0-desc" maxlength="100" /> <label for="id_likes-0-icecream">Want ice cream?</label> <input type="checkbox" name="likes-0-icecream" id="id_likes-0-icecream" /> </fieldset> <fieldset> <label for="id_base-1-desc">Home Base Description:</label> <input id="id_base-1-desc" type="text" name="base-1-desc" maxlength="100" /> <label for="id_likes-1-icecream">Want ice cream?</label> <input type="checkbox" name="likes-1-icecream" id="id_likes-1-icecream" /> </fieldset> I am using a loop like this to process the results for base_form, likes_form in map(None, base_forms, likes_forms): which works as I'd expect (I'm using map because the # of forms can be different). The problem is that I can't figure out a way to do the same thing with the templating engine. The system does work if I layout all the base models together then all the likes models after wards, but it doesn't meet the layout requirements.

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