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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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  • RSS feed created with PHP only shows the title in the feed reader

    - by James Simpson
    I am using the following PHP code to generate the XML for an RSS feed, but it doesn't seem to be working correctly. No short description is displayed in the feed reader, all I see is the title of the article. Also, all of the articles say they were published at the same time. This is the first time I have tried to setup an RSS feed, so I'm sure I've made several stupid mistakes. $result = mysql_query("SELECT * FROM blog ORDER BY id DESC LIMIT 10"); $date = date(DATE_RFC822); header('Content-type: text/xml'); echo ("<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n"); echo ("<rss version=\"2.0\">\n"); echo ("<channel>\n"); echo ("<lastBuildDate>$date</lastBuildDate>\n"); echo ("<pubDate>$date</pubDate>\n"); echo ("<title>my website name</title>\n"); echo ("<description><![CDATA[the description]]></description>\n"); echo ("<link>http://my-domain.com</link>\n"); echo ("<language>en</language>\n"); $ch=100; while ($a = mysql_fetch_array($result)) { $headline = htmlentities(stripslashes($a['subject'])); $posturl = $a[perm_link]; $content = $a['post']; $date = date(DATE_RFC822, $a['posted']); echo ("<item>\n"); echo ("<title>$headline</title>\n"); echo ("<link>$posturl</link>\n"); echo ("<description><![CDATA[$content]]></description>\n"); echo ("<guid isPermaLink=\"true\">$posturl</guid>\n"); echo ("<pubDate>$date2</pubDate>\n"); echo ("</item>\n"); } echo ("</channel>\n"); echo ("</rss>\n");

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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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  • looking to streamline my RSS feed mashup

    - by Mark Cejas
    Hello crafty developers, I have aggregated RSS feeds from various sources with RSSowl, fetching directly from the social mention API. The RSS feeds are categorized into the following major categories: blogs, news, twitter, Q&A and social networking sites. Each major category is nested with a common group of RSS feeds that represent a particular client/brand ontology. Merging these feeds into the RSSowl reader application, allows me to conduct and save refined search queries (from the aggregated data) into a single file - that I can then tag and further segment for analysis. This scheme is utilized for my own research needs and has helped me considerably. However, I find this RSS mashup scheme kinda clumsy, it requires quite a bit of time to initially organize all of the feeds and I would like to be able to do further natural language processing to the data as well as eventually be able to rank the collected list of URL's into some order of media prominence - right I don't want to pay the ridiculous radian6 web analytics fees, when my intuition is telling me that with a bit of 'elbow grease' I can maybe leverage some available resources online to develop a functional low scale web mining application and get some good intelligence from it. I am now starting to learn a little about computer science - my background is in physical science/statistics so is my thinking in the right track? So, I guess I am imagining an application that allows me to query in a refined manner. A manner that allows me to search for keyword combinations, applying AND/OR operators, selectively focus my queries into particular sources - like a collection of blogs or twitter, or social networking communities, then save the results of my queries into a structured format that can then be manipulated and explored. Am I dreaming? I just had to get all of this out. any bit of advice and insight would be hugely appreciated. my best, Mark

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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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  • Make Your Site's RSS Feed Shine

    RSS is a way of life these days, but there's no reason your site's RSS feed needs to look like it was written by a computer. Sure, in order for an RSS feed to be read by most RSS readers it needs to be valid XML, but who says you have to stop there?

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  • Subscribe to RSS Feeds in Chrome with a Single Click

    - by Asian Angel
    Do you have a Google Reader account and need a quick simple way to subscribe to new RSS feeds while you browse? Then you will definitely want to have a look at the Chrome Reader extension for Chrome. Before If you want to add a new feed to your Google Reader account in Chrome then you have to do it manually. A single feed now and then is not a problem but if you are wanting to build a serious set of RSS feeds quickly then not so good. Chrome Reader in Action Once the extension is installed you are ready to go. Any time that you visit a webpage with an RSS feed available you will see the familiar orange feed icon appear in your “Address Bar”. To add the feed to your Google Reader account just click on the orange feed icon. Note: You will need to be logged into your Google Reader account in your browser. When you click on the orange feed icon a small drop-down window will appear where you can modify the feed name and/or add it to a “custom folder” if desired. Notice that the orange feed icon has changed to the familiar Google Reader icon indicating that the feed has been added to the account. Now you are ready to continue browsing…no other actions are required. And now to subscribe to the Microsoft feed at Ars Technica. Once again a single click and all done. Refreshing our Google Reader page shows both of our new RSS feeds ready to enjoy. Conclusion The Chrome Reader extension makes it as simple as can be to add new RSS feeds to your Google Reader account while browsing with Chrome. Links Download the Chrome Reader extension (Google Chrome Extensions) Similar Articles Productive Geek Tips Access Your favorite RSS Feeds in Windows Media CenterChange Default Feed Reader in FirefoxUse Outlook 2007 as an RSS ReaderInstall Extensions in Google ChromeMake Outlook Stop Using Internet Explorer’s RSS Feeds TouchFreeze Alternative in AutoHotkey The Icy Undertow Desktop Windows Home Server – Backup to LAN The Clear & Clean Desktop Use This Bookmarklet to Easily Get Albums Use AutoHotkey to Assign a Hotkey to a Specific Window Latest Software Reviews Tinyhacker Random Tips DVDFab 6 Revo Uninstaller Pro Registry Mechanic 9 for Windows PC Tools Internet Security Suite 2010 Out of band Security Update for Internet Explorer 7 Cool Looking Screensavers for Windows SyncToy syncs Files and Folders across Computers on a Network (or partitions on the same drive) If it were only this easy Classic Cinema Online offers 100’s of OnDemand Movies OutSync will Sync Photos of your Friends on Facebook and Outlook

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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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  • Django: Is there any way to have "unique for date range"?

    - by tomwolber
    If my model for Items is: class Item(models.Model): name = models.CharField(max_length=500) startDate = models.DateField("Start Date", unique="true") endDate = models.DateField("End Date") Each Item needs to have a unique date range. for example, if i create an Item that has a date range of June 1st to June 8th, how can I keep and Item with a date range of June 3rd to June 5th from being created (or render an error with template logic)?

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  • Django - Can you use property as the field in an aggregation function?

    - by orokusaki
    I know the short answer because I tried it. Is there any way to accomplish this though (even if only on account of a hack)? class Ticket(models.Model): account = modelfields.AccountField() uuid = models.CharField(max_length=36, unique=True) created = models.DateTimeField(auto_now_add=True) class Meta: ordering = ['created'] @property def repair_cost(self): # cost is a @property of LineItem(models.Model) return self.lineitem_set.aggregate(models.Sum('cost'))

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  • Django: Can class-based views accept two forms at a time?

    - by Hooman
    If I have two forms: class ContactForm(forms.Form): name = forms.CharField() message = forms.CharField(widget=forms.Textarea) class SocialForm(forms.Form): name = forms.CharField() message = forms.CharField(widget=forms.Textarea) and wanted to use a class based view, and send both forms to the template, is that even possible? class TestView(FormView): template_name = 'contact.html' form_class = ContactForm It seems the FormView can only accept one form at a time. In function based view though I can easily send two forms to my template and retrieve the content of both within the request.POST back. variables = {'contact_form':contact_form, 'social_form':social_form } return render(request, 'discussion.html', variables) Is this a limitation of using class based view (generic views)? Many Thanks

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  • How to get a single widget to set 2 fields in Django?

    - by kender
    Hi, I got a model with 2 fields: latitude and longitude. Right now they're 2 CharFields, but I want to make a custom widget to set it in admin - was thinking about displaying Google Maps, then getting the coordinates of the marker. But can I have 1 widget (a single map) to set 2 different fields?

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  • How can I copy a queryset to a new model in django admin?

    - by user3806832
    I'm trying to write an action that allows the user to select the queryset and copy it to a new table. So: John, Mark, James, Tyler and Joe are in a table 1( called round 1) The user selects the action that say to "move to next round" and those same instances that were chosen are now also in the table for "round 2". I started trying with an action but don't really know where to go from here: def Round_2(modeladmin, request, queryset): For X in queryset: X.pk = None perform.short_description = "Move to Round 2" How can I copy them to the next table with all of their information (pk doesn't have to be the same)? Thanks

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