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  • Extending Django Flatpages to accept template tags

    - by Tristan
    I use django flatpages for a lot of content on our site, I'd like to extend it to accept django template tags in the content as well. I found this snippet but after much larking about I couldn't get it to work. Am I correct in assuming that you would need too "subclass" the django flatpages app to get this to work? Is this best way of doing it? I'm not quite sure how to structure it, as I don't really want to directly modify the django distribution.

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  • Django - I got TemplateSyntaxError when I try open the admin page. (http://DOMAIN_NAME/admin)

    - by user140827
    I use grappelly plugin. When I try open the admin page (/admin) I got TemplateSyntaxError. It says 'get_generic_relation_list' is invalid block tag. TemplateSyntaxError at /admin/ Invalid block tag: 'get_generic_relation_list', expected 'endblock' Request Method: GET Request URL: http://DOMAIN_NAME/admin/ Django Version: 1.4 Exception Type: TemplateSyntaxError Exception Value: Invalid block tag: 'get_generic_relation_list', expected 'endblock' Exception Location: /opt/python27/django/1.4/lib/python2.7/site-packages/django/template/base.py in invalid_block_tag, line 320 Python Executable: /opt/python27/django/1.4/bin/python Python Version: 2.7.0 Python Path: ['/home/vhosts/DOMAIN_NAME/httpdocs/media', '/home/vhosts/DOMAIN_NAME/private/new_malinnikov/lib', '/home/vhosts/DOMAIN_NAME/httpdocs/', '/home/vhosts/DOMAIN_NAME/private/new_malinnikov', '/home/vhosts/DOMAIN_NAME/private/new_malinnikov', '/home/vhosts/DOMAIN_NAME/private', '/opt/python27/django/1.4', '/home/vhosts/DOMAIN_NAME/httpdocs', '/opt/python27/django/1.4/lib/python2.7/site-packages/setuptools-0.6c12dev_r88846-py2.7.egg', '/opt/python27/django/1.4/lib/python2.7/site-packages/pip-0.8.1-py2.7.egg', '/opt/python27/django/1.4/lib/python27.zip', '/opt/python27/django/1.4/lib/python2.7', '/opt/python27/django/1.4/lib/python2.7/plat-linux2', '/opt/python27/django/1.4/lib/python2.7/lib-tk', '/opt/python27/django/1.4/lib/python2.7/lib-old', '/opt/python27/django/1.4/lib/python2.7/lib-dynload', '/opt/python27/lib/python2.7', '/opt/python27/lib/python2.7/plat-linux2', '/opt/python27/lib/python2.7/lib-tk', '/opt/python27/django/1.4/lib/python2.7/site-packages', '/opt/python27/lib/python2.7/site-packages/setuptools-0.6c11-py2.7.egg', '/opt/python27/lib/python2.7/site-packages/flup-1.0.3.dev_20100525-py2.7.egg', '/opt/python27/lib/python2.7/site-packages/virtualenv-1.5.1-py2.7.egg', '/opt/python27/lib/python2.7/site-packages/SQLAlchemy-0.6.4-py2.7.egg', '/opt/python27/lib/python2.7/site-packages/SQLObject-0.14.1-py2.7.egg', '/opt/python27/lib/python2.7/site-packages/FormEncode-1.2.3dev-py2.7.egg', '/opt/python27/lib/python2.7/site-packages/MySQL_python-1.2.3-py2.7-linux-x86_64.egg', '/opt/python27/lib/python2.7/site-packages/psycopg2-2.2.2-py2.7-linux-x86_64.egg', '/opt/python27/lib/python2.7/site-packages/pysqlite-2.6.0-py2.7-linux-x86_64.egg', '/opt/python27/lib/python2.7/site-packages', '/opt/python27/lib/python2.7/site-packages/PIL'] Server time: ???, 7 ??? 2012 04:19:42 +0700 Error during template rendering In template /home/vhosts/DOMAIN_NAME/httpdocs/templates/grappelli/admin/base.html, error at line 28 Invalid block tag: 'get_generic_relation_list', expected 'endblock' 18 <!--[if lt IE 8]> 19 <script src="http://ie7-js.googlecode.com/svn/version/2.0(beta3)/IE8.js" type="text/javascript"></script> 20 <![endif]--> 21 {% block javascripts %} 22 <script type="text/javascript" src="{% admin_media_prefix %}jquery/jquery-1.3.2.min.js"></script> 23 <script type="text/javascript" src="{% admin_media_prefix %}js/admin/Bookmarks.js"></script> 24 <script type="text/javascript"> 25 // Admin URL 26 var ADMIN_URL = "{% get_admin_url %}"; 27 // Generic Relations 28 {% get_generic_relation_list %} 29 // Get Bookmarks 30 $(document).ready(function(){ 31 $.ajax({ 32 type: "GET", 33 url: '{% url grp_bookmark_get %}', 34 data: "path=" + escape(window.location.pathname + window.location.search), 35 dataType: "html", 36 success: function(data){ 37 $('ul#bookmarks').replaceWith(data); 38 }

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  • Django ModelFormSet with Google app engine

    - by Eric
    I'm using Django with google app engine. I'm using the google furnished django app engine helper project. I'm attempting to create a Django modelformset like this: #MyModel inherits from BaseModel MyFormSet = modelformset_factory(models.MyModel) However, it's failing with this error: 'ModelOptions' object has no attribute 'fields' Apparently modelformset_factory() is expecting MyModel to implement a 'fields' accessor. Anybody successfully used a modelformset with GAE datastore? Or is this a fundamental incompatibility between Django and GAE?

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  • django-registration password reset custom template not loading

    - by ip.
    I'm using django-registration for registering users, however when I want to use my own template for password reset I get the admin template and not the template I created. My template is in myapp/templates/registration/password_reset_form.html and my template loaders are properly set: TEMPLATE_LOADERS = ( 'django.template.loaders.filesystem.Loader', 'django.template.loaders.app_directories.Loader', ) What could I be missing? I'm using Django 1.4

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  • Learning django by example source code (not examples)

    - by Bryce
    I'm seeking a nice complete open source django application to study and learn best practices from, or even use as a template. The tutorials only go so far, and django is super flexible which can lead one to paining themselves into a corner. Ideally such a template / example would: Ignore django admin, and implement full CRUD outside the admin. Be built like a large application in terms of best practices and patterns. Have a unit test Use at least one package (e.g. twitter integration or threaded comments) Implement some AJAX or Comet See also: Learning Django by example

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  • Visual web page designer for Django?

    - by Robert Oschler
    I'm just starting my Django learning so pardon me if any part of this question is off-base. I have done a lot of web searching for information on the equivalent of a visual web page designer for Django and I don't seem to be getting very far. I have experience with Delphi (Object Pascal), C, C++, Python, PHP, Java, and Javascript and have created and maintained several web sites that included MySQL database dependent content. For the longest time I've been using one of the standard WYSIWIG designers to design the actual web pages, with any needed back end programming done via Forms or AJAX calls that call server side PHP scripts. I have grown tired of the quirks, bugs, and other annoyances associated with the program. Also, I find myself hungry for the functionality and reliability a good MVC based framework would provide me so I could really express myself with custom code easily. So I am turning to Django/Python. However, I'm still a junkie for a good WYSIWIG designer for the layout of web pages. I know there are some out there that thrive on opening up a text editor, possibly with some code editor tools to assist, and pounding out pages. But I do adore a drag and drop editor for simple page layout, especially for things like embedded images, tables, buttons, etc. I found a few open source projects on GitHub but they seem to be focused on HTML web forms, not a generic web page editor. So can I get what I want here? The supreme goal would be to find not only a web page editor that creates Django compatible web pages, but if I dare say it, have a design editor that could add Python code stubs to various page elements in the style of the Delph/VCL or VB design editors. Note, I also have the Wing IDE Professional IDE, version 2.0. As a side note, if you know of any really cool, fun, or time-saving Python libraries that are designed for easy integration into Django please tell me about them. -- roschler

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  • Does this syntax for specifying Django conditional form display align with python/django convention?

    - by andy
    I asked a similar question on Stackoverflow and was told it was better asked here. So I'll ask it slightly rephrased. I am working on a Django project, part of which will become a distributable plugin that allows the python/django developer to specify conditional form field display logic in the form class or model class. I am trying to decide how the developer must specify that logic. Here's an example: class MyModel(models.Model): #these are some django model fields which will be used in a form yes_or_no = models.SomeField...choices are yes or no... why = models.SomeField...text, but only relevant if yes_or_no == yes... elaborate_even_more = models.SomeField...more text, just here so we can have multiple conditions #here i am inventing some syntax...i am looking for suggestions!! #this is one possibility why.show_if = ('yes_or_no','==','yes') elaborate_even_more.show_if = (('yes_or_no','==','yes'),('why','is not','None')) #help me choose a syntax that is *easy*...and Pythonic and...Djangonic...and that makes your fingers happy to type! #another alternative... conditions = {'why': ('yes_or_no','==','yes'), 'elaborate_even_more': (('yes_or_no','==','yes'),('why','is not','None')) } #or another alternative... """Showe the field whiche hath the name *why* only under that circumstance in whiche the field whiche hath the name *yes_or_no* hath the value *yes*, in strictest equality.""" etc... Those conditions will be eventually passed via django templates to some javascript that will show or hide form fields accordingly. Which of those options (or please propose a better option) aligns better with conventions such that it will be easiest for the python/django developer to use? Also are there other considerations that should impact what syntax I choose?

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  • Rebuilding website from Django 0.96 to Django 1.2

    - by Neytiri
    I've got a website done in Django 0.96 (done in 2007), and now we are thinking about rebuilding it (not just migrating) for Django 1.2 . Can anyone point me to the new (and worth the while) widgets, plugins and other stuff for Django 1.2 (released in april 2010). I've heard of "South" and of a widget for debugging (can't remember the name), but I'm a little lost here.

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  • Why use Django on Google App Engine?

    - by Travis Bradshaw
    When researching Google App Engine (GAE), it's clear that using Django is wildly popular for developing in Python on GAE. I've been scouring the web to find information on the costs and benefits of using Django, to find out why it's so popular. While I've been able to find a wide variety of sources on how to run Django on GAE and the various methods of doing so, I haven't found any comparative analysis on why Django is preferable to using the webapp framework provided by Google. To be clear, it's immediately apparent why using Django on GAE is useful for developers with an existing skillset in Django (a majority of Python web developers, no doubt) or existing code in Django (where using GAE is more of a porting exercise). My team, however, is evaluating GAE for use on an all-new project and our existing experience is with TurboGears, not Django. It's been quite difficult to determine why Django is beneficial to a development team when the BigTable libraries have replaced Django's ORM, sessions and authentication are necessarily changed, and Django's templating (if desirable) is available without using the entire Django stack. Finally, it's clear that using Django does have the advantage of providing an "exit strategy" if we later wanted to move away from GAE and need a platform to target for the exodus. I'd be extremely appreciative for help in pointing out why using Django is better than using webapp on GAE. I'm also completely inexperienced with Django, so elaboration on smaller features and/or conveniences that work on GAE are also valuable to me. Thanks in advance for your time!

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  • Django vs GAE + Django vs GAE + other framework

    - by Ilian Iliev
    I`m looking for opinion which one is better for building web applications(web sites). I have some experience with Django, and some with Google App Engine and App-Engine-Patch for Django. And it seems to me that only Django is working faster than the GAE implementation. Is there some other frameworks that simplify the developments process, providing forms creating, user management, url resolving etc. Thanks in advance, Ilian Iliev P.S. I am also interested in GAE and webapp framework case

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  • Google App Engine + Form Validation

    - by Iwona
    Hi, I would like to do google app engine form validation but I dont know how to do it? I tried like this: from google.appengine.ext.db import djangoforms from django import newforms as forms class SurveyForm(forms.Form): occupations_choices = ( ('1', ""), ('2', "Undergraduate student"), ('3', "Postgraduate student (MSc)"), ('4', "Postgraduate student (PhD)"), ('5', "Lab assistant"), ('6', "Technician"), ('7', "Lecturer"), ('8', "Other" ) ) howreach_choices = ( ('1', ""), ('2', "Typed the URL directly"), ('3', "Site is bookmarked"), ('4', "A search engine"), ('5', "A link from another site"), ('6', "From a book"), ('7', "Other") ) boxes_choices = ( ("des", "Website Design"), ("svr", "Web Server Administration"), ("com", "Electronic Commerce"), ("mkt", "Web Marketing/Advertising"), ("edu", "Web-Related Education") ) name = forms.CharField(label='Name', max_length=100, required=True) email = forms.EmailField(label='Your Email Address:') occupations = forms.ChoiceField(choices=occupations_choices, label='What is your occupation?') howreach = forms.ChoiceField(choices=howreach_choices, label='How did you reach this site?') # radio buttons 1-5 rating = forms.ChoiceField(choices=range(1,6), label='What is your occupation?', widget=forms.RadioSelect) boxes = forms.ChoiceField(choices=boxes_choices, label='Are you involved in any of the following? (check all that apply):', widget=forms.CheckboxInput) comment = forms.CharField(widget=forms.Textarea, required=False) And I wanted to display it like this: template_values = { 'url' : url, 'url_linktext' : url_linktext, 'userName' : userName, 'item1' : SurveyForm() } And I have this error message: Traceback (most recent call last): File "C:\Program Files\Google\google_appengine\google\appengine\ext\webapp_init_.py", line 515, in call handler.get(*groups) File "C:\Program Files\Google\google_appengine\demos\b00213576\main.py", line 144, in get self.response.out.write(template.render(path, template_values)) File "C:\Program Files\Google\google_appengine\google\appengine\ext\webapp\template.py", line 143, in render return t.render(Context(template_dict)) File "C:\Program Files\Google\google_appengine\google\appengine\ext\webapp\template.py", line 183, in wrap_render return orig_render(context) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 168, in render return self.nodelist.render(context) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 705, in render bits.append(self.render_node(node, context)) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 718, in render_node return(node.render(context)) File "C:\Program Files\Google\google_appengine\lib\django\django\template\defaulttags.py", line 209, in render return self.nodelist_true.render(context) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 705, in render bits.append(self.render_node(node, context)) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 718, in render_node return(node.render(context)) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 768, in render return self.encode_output(output) File "C:\Program Files\Google\google_appengine\lib\django\django\template_init_.py", line 757, in encode_output return str(output) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\util.py", line 26, in str return self.unicode().encode(settings.DEFAULT_CHARSET) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\forms.py", line 73, in unicode return self.as_table() File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\forms.py", line 144, in as_table return self._html_output(u'%(label)s%(errors)s%(field)s%(help_text)s', u'%s', '', u'%s', False) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\forms.py", line 129, in _html_output output.append(normal_row % {'errors': bf_errors, 'label': label, 'field': unicode(bf), 'help_text': help_text}) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\forms.py", line 232, in unicode value = value.str() File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\util.py", line 26, in str return self.unicode().encode(settings.DEFAULT_CHARSET) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\widgets.py", line 246, in unicode return u'\n%s\n' % u'\n'.join([u'%s' % w for w in self]) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\widgets.py", line 238, in iter yield RadioInput(self.name, self.value, self.attrs.copy(), choice, i) File "C:\Program Files\Google\google_appengine\lib\django\django\newforms\widgets.py", line 212, in init self.choice_value = smart_unicode(choice[0]) TypeError: 'int' object is unsubscriptable Do You have any idea how I can do this validation in different case? I have tried to do it using this kind of: class ItemUserAnswer(djangoforms.ModelForm): class Meta: model = UserAnswer But I dont know how to add extra labels to this form and it is displayed in one line. Do You have any suggestions? Thanks a lot as it making me crazy why it is still not working:/

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  • Django custom managers - how do I return only objects created by the logged-in user?

    - by Tom Tom
    I want to overwrite the custom objects model manager to only return objects a specific user created. Admin users should still return all objects using the objects model manager. Now I have found an approach that could work. They propose to create your own middleware looking like this: #### myproject/middleware/threadlocals.py try: from threading import local except ImportError: # Python 2.3 compatibility from django.utils._threading_local import local _thread_locals = local() def get_current_user(): return getattr(_thread_locals, 'user', None) class ThreadLocals(object): """Middleware that gets various objects from the request object and saves them in thread local storage.""" def process_request(self, request): _thread_locals.user = getattr(request, 'user', None) #### end And in the Custom manager you could call the get_current_user() method to return only objects a specific user created. class UserContactManager(models.Manager): def get_query_set(self): return super(UserContactManager, self).get_query_set().filter(creator=get_current_user()) Is this a good approach to this use-case? Will this work? Or is this like "using a sledgehammer to crack a nut" ? ;-) Just using: Contact.objects.filter(created_by= user) in each view doesn`t look very neat to me. EDIT Do not use this middleware approach !!! use the approach stated by Jack M. below After a while of testing this approach behaved pretty strange and with this approach you mix up a global-state with a current request. Use the approach presented below. It is really easy and no need to hack around with the middleware. create a custom manager in your model with a function that expects the current user or any other user as an input. #in your models.py class HourRecordManager(models.Manager): def for_user(self, user): return self.get_query_set().filter(created_by=user) class HourRecord(models.Model): #Managers objects = HourRecordManager() #in vour view you can call the manager like this and get returned only the objects from the currently logged-in user. hr_set = HourRecord.objects.for_user(request.user)

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  • Refactoring a custom User model to user UserProfile: Should I create a custom UserManager or add use

    - by BryanWheelock
    I have been refactoring an app that had customized the standard User model from django.contrib.auth.models by creating a UserProfile and defining it with AUTH_PROFILE_MODULE. The problem is the attributes in UserProfile are used throughout the project to determine the User sees. I had been creating tests and putting in this type of statement repeatedly: user = User.objects.get(pk=1) user_profile = user.get_profile() if user_profile.karma > 10: do_some_stuff() This is tedious and I'm now wondering if I'm violating the DRY principle. Would it make more sense to create a custom UserManager that automatically loads the UserProfile data when the user is requested. I could even iterate over the UserProfile attributes and append them to the User model. This would save me having to update all the references to the custom model attributes that litter the code. Of course, I'd have to reverse to process for to allow the User and UserProfile models to be updated correctly. Which approach is more Django-esque?

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  • Django simple syndication example gives: ImportError, cannot import name Feed

    - by AP257
    I'm trying to set up the simple syndication example from the Django docs, in a working project. But I'm getting an ImportError, even though I'm sure I've copied the example exactly. Here's what I have in feeds.py: from django.contrib.syndication.views import Feed class LatestEntriesFeed(Feed): # etc And here's what I have in urls.py: from election.feeds import LatestEntriesFeed #... further down, at the appropriate line... # RSS feed (r'^feed/$', LatestEntriesFeed()), But Django says it can't import the Feed class from django.contrib.syndication.views: ImportError at /feed/ cannot import name Feed ....feeds.py in <module> from django.contrib.syndication.views import Feed Any ideas? I'm baffled!

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  • url template tag in django template

    - by user192048
    guys: I was trying to use the url template tag in django, but no lucky, I defined my urls.py like this urlpatterns = patterns('', url(r'^analyse/$', views.home, name="home"), url(r'^analyse/index.html', views.index, name="index"), url(r'^analyse/setup.html', views.setup, name="setup"), url(r'^analyse/show.html', views.show, name="show"), url(r'^analyse/generate.html', views.generate, name="generate"), I defined the url pattern in my view like this {% url 'show'%} then I got this error message Caught an exception while rendering: Reverse for ''show'' with arguments '()' and keyword arguments '{}' not found. Original Traceback (most recent call last): File "/Library/Python/2.5/site-packages/django/template/debug.py", line 71, in render_node result = node.render(context) File "/Library/Python/2.5/site-packages/django/template/defaulttags.py", line 155, in render nodelist.append(node.render(context)) File "/Library/Python/2.5/site-packages/django/template/defaulttags.py", line 382, in render raise e NoReverseMatch: Reverse for ''show'' with arguments '()' and keyword arguments '{}' not found. I am wondering why django failed to render? what is the right way to define it in the tempalte?

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  • Django syncdb not making tables for my app

    - by Rosarch
    It used to work, and now it doesn't. python manage.py syncdb no longer makes tables for my app. From settings.py: INSTALLED_APPS = ( 'django.contrib.auth', 'django.contrib.contenttypes', 'django.contrib.sessions', 'django.contrib.sites', 'mysite.myapp', 'django.contrib.admin', ) What could I be doing wrong? The break appeared to coincide with editing this model in models.py, but that could be total coincidence. I commented out the lines I changed, and it still doesn't work. class MyUser(models.Model): user = models.ForeignKey(User, unique=True) takingReqSets = models.ManyToManyField(RequirementSet, blank=True) takingTerms = models.ManyToManyField(Term, blank=True) takingCourses = models.ManyToManyField(Course, through=TakingCourse, blank=True) school = models.ForeignKey(School) # minCreditsPerTerm = models.IntegerField(blank=True) # maxCreditsPerTerm = models.IntegerField(blank=True) # optimalCreditsPerTerm = models.IntegerField(blank=True) UPDATE: When I run python manage.py loadddata initial_data, it gives an error: DeserializationError: Invalid model identifier: myapp.SomeModel Loading this data had worked fine before. This error is thrown on the very first data object in the data file.

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  • Adding a generic image field onto a ModelForm in django

    - by Prairiedogg
    I have two models, Room and Image. Image is a generic model that can tack onto any other model. I want to give users a form to upload an image when they post information about a room. I've written code that works, but I'm afraid I've done it the hard way, and specifically in a way that violates DRY. Was hoping someone who's a little more familiar with django forms could point out where I've gone wrong. Update: I've tried to clarify why I chose this design in comments to the current answers. To summarize: I didn't simply put an ImageField on the Room model because I wanted more than one image associated with the Room model. I chose a generic Image model because I wanted to add images to several different models. The alternatives I considered were were multiple foreign keys on a single Image class, which seemed messy, or multiple Image classes, which I thought would clutter my schema. I didn't make this clear in my first post, so sorry about that. Seeing as none of the answers so far has addressed how to make this a little more DRY I did come up with my own solution which was to add the upload path as a class attribute on the image model and reference that every time it's needed. # Models class Image(models.Model): content_type = models.ForeignKey(ContentType) object_id = models.PositiveIntegerField() content_object = generic.GenericForeignKey('content_type', 'object_id') image = models.ImageField(_('Image'), height_field='', width_field='', upload_to='uploads/images', max_length=200) class Room(models.Model): name = models.CharField(max_length=50) image_set = generic.GenericRelation('Image') # The form class AddRoomForm(forms.ModelForm): image_1 = forms.ImageField() class Meta: model = Room # The view def handle_uploaded_file(f): # DRY violation, I've already specified the upload path in the image model upload_suffix = join('uploads/images', f.name) upload_path = join(settings.MEDIA_ROOT, upload_suffix) destination = open(upload_path, 'wb+') for chunk in f.chunks(): destination.write(chunk) destination.close() return upload_suffix def add_room(request, apartment_id, form_class=AddRoomForm, template='apartments/add_room.html'): apartment = Apartment.objects.get(id=apartment_id) if request.method == 'POST': form = form_class(request.POST, request.FILES) if form.is_valid(): room = form.save() image_1 = form.cleaned_data['image_1'] # Instead of writing a special function to handle the image, # shouldn't I just be able to pass it straight into Image.objects.create # ...but it doesn't seem to work for some reason, wrong syntax perhaps? upload_path = handle_uploaded_file(image_1) image = Image.objects.create(content_object=room, image=upload_path) return HttpResponseRedirect(room.get_absolute_url()) else: form = form_class() context = {'form': form, } return direct_to_template(request, template, extra_context=context)

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  • Django ORM and PostgreSQL connection limits

    - by bennylope
    I'm running a Django project on Postgresql 8.1.21 (using Django 1.1.1, Python2.5, psycopg2, Apache2 with mod_wsgi 3.2). We've recently encountered this lovely error: OperationalError: FATAL: connection limit exceeded for non-superusers I'm not the first person to run up against this. There's a lot of discussion about this error, specifically with psycopg, but much of it centers on older versions of Django and/or offer solutions involving edits to code in Django itself. I've yet to find a succinct explanation of how to solve the problem of the Django ORM (or psycopg, whichever is really responsible, in this case) leaving open Postgre connections. Will simply adding connection.close() at the end of every view solve this problem? Better yet, has anyone conclusively solved this problem and kicked this error's ass?

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  • How do I flag only one of the formsets in django admin?

    - by azuer88
    I have these (simplified) models: class Question(models.Model): question = models.CharField(max_length=60) class Choices(models.Model): question = models.ForeignKey(Question) text = models.CharField(max_length=60) is_correct = models.BooleanField(default=False) I've made Choices as an inline of Question (in admin). Is there a way to make sure that only one Choice will have is_correct = True? Ideally, is_correct will be displayed as a radio button when it is displayed in the admin formset (TabularInline). my admin.py has: from django.contrib import admin class OptionInline(admin.TabularInline): model = Option extra = 5 max_num = 5 class QuestionAdmin(admin.ModelAdmin): inlines = [OptionInline, ] admin.site.register(QType) admin.site.register(Question, QuestionAdmin)

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  • View centric design with Django

    - by wishi_
    Hi! I'm relatively new to Django and I'm designing a website that primarily needs usability experience, speaking of optimized CSS, HTML5 and UI stuff. It's very easy to use Django for data/Model centric design. Just designing a couple of Python classes and ./manage.py syncdb - there's your Model. But I'm dealing with a significant amount of View centric challenges. (Different user classes, different tasks, different design challenges.) The official Django tutorial cursorily goes through using a "Template". Is there any Design centric guide for Django, or a set of Templates that are ready and useable? I don't want to start from scratch using JS, HTML5, Ajax and everything. From the Model layer perspective Django is very rapid and delivering a working base system. I wonder whether there's something like that for the Views.

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  • How can I call model methods or properties from Django Admin?

    - by kg
    Is there a natural way to display model methods or properties in the Django admin site? In my case I have base statistics for a character that are part of the model, but other things such as status effects which affect the total calculation for that statistic: class Character(models.Model): base_dexterity = models.IntegerField(default=0) @property def dexterity(stat_name): total = self.base_dexterity total += sum(s.dexterity for s in self.status.all()]) return total It would be nice if I could display the total calculated statistic alongside the field to change the base statistic in the Change Character admin page, but it is not clear to me how to incorporate that information into the page.

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  • django newbie question : cant start a new project

    - by Moayyad Yaghi
    hello . I'm totally new to django . and I'm using its documentation to get help on how to use it but seems like something is missing. i installed django using setup.py install command and i added the ( django/bin ) to system path variable but. i still cant start a new project i use the following syntax to start a project : django-admin.py startproject myNewProject but it says Type 'django-admin.py help' for usage. 1 do i miss anything ? thank u

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  • Rails - Accessing model class methods from within ActiveRecord model

    - by aaronrussell
    I have a simple standalone model that doesn't inherit from ActiveRecord or anything else, called SmsSender. As the name suggests, it delivers text messages to an SMS gateway. I also have an ActiveRecord model called SmsMessage which has an instance method called deliver: def deliver SmsSender.deliver_message(self) self.update_attributes :status => "Sent" end The above is returning uninitialized constant SmsSender. I'm sure this is dead simple, but how can I access the SmsSender class from within my model?

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  • What's the straightforward way to implement one to many editing in list_editable in django admin?

    - by Nate Pinchot
    Given the following models: class Store(models.Model): name = models.CharField(max_length=150) class ItemGroup(models.Model): group = models.CharField(max_length=100) code = models.CharField(max_length=20) class ItemType(models.Model): store = models.ForeignKey(Store, on_delete=models.CASCADE, related_name="item_types") item_group = models.ForeignKey(ItemGroup) type = models.CharField(max_length=100) Inline's handle adding multiple item_types to a Store nicely when viewing a single Store. The content admin team would like to be able to edit stores and their types in bulk. Is there a simple way to implement Store.item_types in list_editable which also allows adding new records, similar to horizontal_filter? If not, is there a straightforward guide that shows how to implement a custom list_editable template? I've been Googling but haven't been able to come up with anything. Also, if there is a simpler or better way to set up these models that would make this easier to implement, feel free to comment.

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  • Django display manytomany field in form when definition is on other model

    - by John
    Hi I have the definition for my manytomany relationship on one model but want to display the field on the form of my other model. How do I do this? for example: # classes class modelA(models.Model): name = models.CharField(max_length=300) manytomany = models.ManyToManyField(modelA) class modelB(models.Model): name = models.CharField(max_length=300) # forms class modelBForm(forms.ModelForm): class Meta: model = modelB If I then used modelBForm it would show a select box with the models from modelA rather than just name. Thanks

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