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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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  • Jquery Widget Performance

    - by jamie-wilson
    I am working on an interface which involves ALOT of javascript. There is a calendar and blocks drawn on the calendar. The calendar is a jQuery widget, which works beautifully. The blocks drawn on top are also jQuery widgets. While it works - I am wondering, every time I create another block, is the widget fully duplicating, or is it referencing the widget? If I end up with 200 blocks on the screen, do I have 200 copies of the widget? Because if so i'm sure this will impact the performance quite heavily. Also it would determine whether I have functions inside the widget, or have them external to the widget looking in if that makes sense. Just putting some feelers out there for thoughts. I couldn't find anything by searching online.

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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 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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  • 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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  • 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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  • 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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  • Soft keyboard "del" key fails in EditText on Gallery widget

    - by droidful
    Hi, I am developing an application in Eclipse build ID 20090920-1017 using android SDK 2.2 and testing on a Google Nexus One. For the purposes of the tests below I am using the IME "Android keyboard" on a non-rooted phone. I have an EditText widget which exhibits some very strange behavior. I can type text, and then press the "del" key to delete that text; but after I enter a 'space' character, the "del" key will no longer remove characters before that space character. An example speaks a thousand words, so consider the following two incredibly simple applications... Example 1: An EditText in a LinearLayout widget: package com.example.linear.edit; import android.app.Activity; import android.os.Bundle; import android.view.ViewGroup.LayoutParams; import android.widget.EditText; import android.widget.Gallery; import android.widget.LinearLayout; public class LinearEdit extends Activity { @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); LinearLayout layout = new LinearLayout(getApplicationContext()); layout.setLayoutParams(new Gallery.LayoutParams(Gallery.LayoutParams.MATCH_PARENT, Gallery.LayoutParams.MATCH_PARENT)); EditText edit = new EditText(getApplicationContext()); layout.addView(edit, new LinearLayout.LayoutParams(LayoutParams.MATCH_PARENT, LayoutParams.WRAP_CONTENT)); setContentView(layout); } } Run the above application, enter text "edit example", then press the "del" key several times until the entire sentence is deleted. Everything Works fine. Now consider example 2: An EditText in a Gallery widget: package com.example.gallery.edit; import android.app.Activity; import android.os.Bundle; import android.view.View; import android.view.ViewGroup; import android.view.ViewGroup.LayoutParams; import android.widget.ArrayAdapter; import android.widget.EditText; import android.widget.Gallery; import android.widget.LinearLayout; public class GalleryEdit extends Activity { private final String[] galleryData = {"string1", "string2", "string3"}; @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); Gallery gallery = new Gallery(getApplicationContext()); gallery.setAdapter(new ArrayAdapter(getApplicationContext(), android.R.layout.simple_list_item_1, galleryData) { @Override public View getView(int position, View convertView, ViewGroup parent) { LinearLayout layout = new LinearLayout(getApplicationContext()); layout.setLayoutParams(new Gallery.LayoutParams(Gallery.LayoutParams.MATCH_PARENT, Gallery.LayoutParams.MATCH_PARENT)); EditText edit = new EditText(getApplicationContext()); layout.addView(edit, new LinearLayout.LayoutParams(LayoutParams.MATCH_PARENT, LayoutParams.WRAP_CONTENT)); return layout; } }); setContentView(gallery); } } Run the above application, enter text "edit example", then press the "del" key several times. If you are getting the same problem as me then you will find that you can't deleted past the 'space' character. All is not well. If anyone could shed some light on this issue I would be most appreciative. Regards

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  • android widget unresponsive

    - by John
    I have a widget that you press and it then it will update the text on the widget. I have set an on click listener to launch another activity to perform the text update, But for some reason it only works temporarily and then it will become unresponsive and not do anything when pressed. Does anyone know why it might be doing that? i have posted my widget code below in case it is helpful. @Override public void onUpdate(Context context, AppWidgetManager appWidgetManager,int[] appWidgetIds) { thisWidget = new ComponentName(context, MemWidget.class); Intent intent = new Intent(context, updatewidget.class); PendingIntent pendingIntent = PendingIntent.getActivity(context, 0, intent, 0); // Get the layout for the App Widget and attach an on-click listener to the button RemoteViews views = new RemoteViews(context.getPackageName(), R.layout.widget); views.setOnClickPendingIntent(R.id.ImageButton01, pendingIntent); // Tell the AppWidgetManager to perform an update on the current App Widget appWidgetManager.updateAppWidget(thisWidget, views); } @Override public void onReceive(Context context, Intent intent) { appWidgetManager = AppWidgetManager.getInstance(context); remoteViews = new RemoteViews(context.getPackageName(), R.layout.widget); thisWidget = new ComponentName(context, MemWidget.class); // v1.5 fix that doesn't call onDelete Action final String action = intent.getAction(); if (AppWidgetManager.ACTION_APPWIDGET_DELETED.equals(action)) { final int appWidgetId = intent.getExtras().getInt( AppWidgetManager.EXTRA_APPWIDGET_ID, AppWidgetManager.INVALID_APPWIDGET_ID); if (appWidgetId != AppWidgetManager.INVALID_APPWIDGET_ID) { this.onDeleted(context, new int[] { appWidgetId }); } } else { super.onReceive(context, intent); } }

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  • Django Form for date range

    - by gramware
    I am trying to come up with a form that lets the user select a date range to generate a web query in Django. I am having errors getting the date to filter with in my view, I am unable to strip the date. Here is my forms.py class ReportFiltersForm(forms.Form): start_date = forms.DateField(input_formats='%Y,%m,%d',widget=SelectDateWidget()) end_date = forms.DateField(input_formats='%Y,%m,%d',widget=SelectDateWidget()) And my view if request.method == 'POST': form = ReportFiltersForm(request.POST) sdy = request.POST['start_date_year'] sdm = request.POST['start_date_month'] sdd = request.POST['start_date_day'] edy = request.POST['end_date_year'] edm = request.POST['end_date_month'] edd = request.POST['end_date_day'] start_date= datetime.date(sdy, sdm, sdd) end_date= datetime.date(edy, edm,edd) Traceback Traceback (most recent call last): File "/usr/lib/python2.6/site-packages/django/core/servers/basehttp.py", line 651, in __call__ return self.application(environ, start_response) File "/usr/lib/python2.6/site-packages/django/core/handlers/wsgi.py", line 241, in __call__ response = self.get_response(request) File "/usr/lib/python2.6/site-packages/django/core/handlers/base.py", line 134, in get_response return self.handle_uncaught_exception(request, resolver, exc_info) File "/usr/lib/python2.6/site-packages/django/core/handlers/base.py", line 154, in handle_uncaught_exception return debug.technical_500_response(request, *exc_info) File "/usr/lib/python2.6/site-packages/django/core/handlers/base.py", line 92, in get_response response = callback(request, *callback_args, **callback_kwargs) File "/home/projects/acms/cms/views.py", line 470, in eventreports start_date= datetime.date(sdy, sdm, sdd) TypeError: an integer is required

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  • Detect the language & django locale-url

    - by mamcx
    I want to deploy a website in english & spanish and detect the user browser languaje & redirect to the correct locale site. My site is www.elmalabarista.com I install django-localeurl, but I discover that the languaje is not correctly detected. This are my middlewares: MIDDLEWARE_CLASSES = ( 'django.contrib.sessions.middleware.SessionMiddleware', 'django.middleware.locale.LocaleMiddleware', 'multilingual.middleware.DefaultLanguageMiddleware', 'middleware.feedburner.FeedburnerMiddleware', 'lib.threadlocals.ThreadLocalsMiddleware', 'middleware.url.UrlMiddleware', 'django.contrib.auth.middleware.AuthenticationMiddleware', 'maintenancemode.middleware.MaintenanceModeMiddleware', 'middleware.redirect.RedirectMiddleware', 'openidconsumer.middleware.OpenIDMiddleware', 'django.middleware.doc.XViewMiddleware', 'middleware.ajax_errors.AjaxMiddleware', 'pingback.middleware.PingbackMiddleware', 'localeurl.middleware.LocaleURLMiddleware', 'multilingual.flatpages.middleware.FlatpageFallbackMiddleware', 'django.middleware.common.CommonMiddleware', ) But ALWAYS the site get to US despite the fact my OS & Browser setup is spanish. LANGUAGES = ( ('en', ugettext('English')), ('es', ugettext('Spanish')), ) DEFAULT_LANGUAGE = 1 Then, I hack the middleware of locale-url and do this: def process_request(self, request): locale, path = self.split_locale_from_request(request) if request.META.has_key('HTTP_ACCEPT_LANGUAGE'): locale = utils.supported_language(request.META['HTTP_ACCEPT_LANGUAGE'].split(',')[0]) locale_path = utils.locale_path(path, locale) if locale_path != request.path_info: if request.META.get("QUERY_STRING", ""): locale_path = "%s?%s" % (locale_path, request.META['QUERY_STRING']) return HttpResponseRedirect(locale_path) request.path_info = path if not locale: locale = settings.LANGUAGE_CODE translation.activate(locale) request.LANGUAGE_CODE = translation.get_language() However, this detect fine the language but redirect the "en" urls to "es". So is impossible navigate in english. UPDATE: This is the final code (after the input from Carl Meyer) with a fix for the case of "/": def process_request(self, request): locale, path = self.split_locale_from_request(request) if (not locale) or (locale==''): if request.META.has_key('HTTP_ACCEPT_LANGUAGE'): locale = utils.supported_language(request.META['HTTP_ACCEPT_LANGUAGE'].split(',')[0]) else: locale = settings.LANGUAGE_CODE locale_path = utils.locale_path(path, locale) if locale_path != request.path_info: if request.META.get("QUERY_STRING", ""): locale_path = "%s?%s" % (locale_path, request.META['QUERY_STRING']) return HttpResponseRedirect(locale_path) request.path_info = path translation.activate(locale) request.LANGUAGE_CODE = translation.get_language()

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  • UnicodeDecodeError on attempt to save file through django default filebased backend

    - by Ivan Kuznetsov
    When i attempt to add a file with russian symbols in name to the model instance through default instance.file_field.save method, i get an UnicodeDecodeError (ascii decoding error, not in range (128) from the storage backend (stacktrace ended on os.exist). If i write this file through default python file open/write all goes right. All filenames in utf-8. I get this error only on testing Gentoo, on my Ubuntu workstation all works fine. class Article(models.Model): file = models.FileField(null=True, blank=True, max_length = 300, upload_to='articles_files/%Y/%m/%d/') Traceback: File "/usr/lib/python2.6/site-packages/django/core/handlers/base.py" in get_response 100. response = callback(request, *callback_args, **callback_kwargs) File "/usr/lib/python2.6/site-packages/django/contrib/auth/decorators.py" in _wrapped_view 24. return view_func(request, *args, **kwargs) File "/var/www/localhost/help/wiki/views.py" in edit_article 338. new_article.file.save(fp, fi, save=True) File "/usr/lib/python2.6/site-packages/django/db/models/fields/files.py" in save 92. self.name = self.storage.save(name, content) File "/usr/lib/python2.6/site-packages/django/core/files/storage.py" in save 47. name = self.get_available_name(name) File "/usr/lib/python2.6/site-packages/django/core/files/storage.py" in get_available_name 73. while self.exists(name): File "/usr/lib/python2.6/site-packages/django/core/files/storage.py" in exists 196. return os.path.exists(self.path(name)) File "/usr/lib/python2.6/genericpath.py" in exists 18. st = os.stat(path) Exception Type: UnicodeEncodeError at /edit/ Exception Value: ('ascii', u'/var/www/localhost/help/i/articles_files/2010/03/17/\u041f\u0440\u0438\u0432\u0435\u0442', 52, 58, 'ordinal not in range(128)')

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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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  • django form ModelChoiceField loads all state names while needed states which are mapped with current selected country

    - by Sonu
    I am using modelChoiceField to display country and state into address form class StateSelectionwidget(forms.Select): """ custom widget to state selection""" class Media: js = ('media/javascript/public/jquery-1.5.2.min.js', 'media/javascript/public/countrystateselection.js', ) class AddressForm(forms.Form): name = forms.CharField(max_length=30) country = forms.ModelChoiceField(queryset=[]) state = forms.ModelChoiceField(CountryState.objects, widget=StateSelectionwidget) def __init__(self, *args, **kwargs): super(AddressForm, self).__init__(*args, **kwargs) self.fields['country'].queryset = Country.objects.all() Country model is used to store country names. CountryState model is used to store all states which is foreign key to Country model At the time of form loading i am getting all state names in dropdown while i want field to be blank by default. If name field is empty at the time of form save i am getting error that name can not be empty but also getting all states into dropdown list while i want only the states which are mapped with current selected country.

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  • How to set HTMLField's widget's height in Admin?

    - by Georgie Porgie
    I have a HTMLField in a model as it's the laziest way to utilize tinymce widget in Admin. But the problem is that the textarea field doesn't have "rows" property set. So the textarea doesn't have enough height comfortable enough for editing in Admin. Is there any way to set the height of HTMLField without defining a ModelAdmin class? Update: I solved the problem by using the following code: def create_mce_formfield(db_field): return db_field.formfield(widget = TinyMCE( attrs = {'cols': 80, 'rows': 30}, mce_attrs = { 'external_link_list_url': reverse('tinymce.views.flatpages_link_list'), 'plugin_preview_pageurl': reverse('tinymce-preview', args= ('tinymce',)), 'plugins': "safari,pagebreak,style,layer,table,save,advhr,advimage,advlink,emotions,iespell,inlinepopups,insertdatetime,preview,media,searchreplace,print,contextmenu,paste,directionality,fullscreen,noneditable,visualchars,nonbreaking,xhtmlxtras,template", 'theme_advanced_buttons1': "save,newdocument,|,bold,italic,underline,strikethrough,|,justifyleft,justifycenter,justifyright,justifyfull,styleselect,formatselect,fontselect,fontsizeselect", 'theme_advanced_buttons2': "cut,copy,paste,pastetext,pasteword,|,search,replace,|,bullist,numlist,|,outdent,indent,blockquote,|,undo,redo,|,link,unlink,anchor,image,cleanup,help,code,|,insertdate,inserttime,preview,|,forecolor,backcolor", 'theme_advanced_buttons3': "tablecontrols,|,hr,removeformat,visualaid,|,sub,sup,|,charmap,emotions,iespell,media,advhr,|,print,|,ltr,rtl,|,fullscreen", 'theme_advanced_buttons4': "insertlayer,moveforward,movebackward,absolute,|,styleprops,|,cite,abbr,acronym,del,ins,attribs,|,visualchars,nonbreaking,template,pagebreak", 'theme_advanced_toolbar_location': "top", 'theme_advanced_toolbar_align': "left", 'theme_advanced_statusbar_location': "bottom", 'theme_advanced_resizing': True, 'extended_valid_elements': "iframe[src|title|width|height|allowfullscreen|frameborder|webkitAllowFullScreen|mozallowfullscreen|allowFullScreen]", }, )) class TinyMCEFlatPageAdmin(FlatPageAdmin): def formfield_for_dbfield(self, db_field, **kwargs): if db_field.name == 'content': return create_mce_formfield(db_field) return super(TinyMCEFlatPageAdmin, self).formfield_for_dbfield(db_field, **kwargs)

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  • Django facebook integration error

    - by Gaurav
    I'm trying to integrate facebook into my application so that users can use their FB login to login to my site. I've got everything up and running and there are no issues when I run my site using the command line python manage.py runserver But this same code refuses to run when I try and run it through Apache. I get the following error: Environment: Request Method: GET Request URL: http://helvetica/foodfolio/login Django Version: 1.1.1 Python Version: 2.6.4 Installed Applications: ['django.contrib.auth', 'django.contrib.contenttypes', 'django.contrib.sessions', 'django.contrib.sites', 'foodfolio.app', 'foodfolio.facebookconnect'] Installed Middleware: ('django.contrib.sessions.middleware.SessionMiddleware', 'facebook.djangofb.FacebookMiddleware', 'django.middleware.common.CommonMiddleware', 'django.contrib.auth.middleware.AuthenticationMiddleware', 'facebookconnect.middleware.FacebookConnectMiddleware') Template error: In template /home/swat/website-apps/foodfolio/facebookconnect/templates/facebook/js.html, error at line 2 Caught an exception while rendering: No module named app.models 1 : <script type="text/javascript"> 2 : FB_RequireFeatures(["XFBML"], function() {FB.Facebook.init("{{ facebook_api_key }}", " {% url facebook_xd_receiver %} ")}); 3 : 4 : function facebookConnect(loginForm) { 5 : FB.Connect.requireSession(); 6 : FB.Facebook.get_sessionState().waitUntilReady(function(){loginForm.submit();}); 7 : } 8 : function pushToFacebookFeed(data){ 9 : if(data['success']){ 10 : var template_data = data['template_data']; 11 : var template_bundle_id = data['template_bundle_id']; 12 : feedTheFacebook(template_data,template_bundle_id,function(){}); Traceback: File "/usr/lib/pymodules/python2.6/django/core/handlers/base.py" in get_response 92. response = callback(request, *callback_args, **callback_kwargs) File "/home/swat/website-apps/foodfolio/app/controller.py" in __showLogin__ 238. context_instance = RequestContext(request)) File "/usr/lib/pymodules/python2.6/django/shortcuts/__init__.py" in render_to_response 20. return HttpResponse(loader.render_to_string(*args, **kwargs), **httpresponse_kwargs) File "/usr/lib/pymodules/python2.6/django/template/loader.py" in render_to_string 108. return t.render(context_instance) File "/usr/lib/pymodules/python2.6/django/template/__init__.py" in render 178. return self.nodelist.render(context) File "/usr/lib/pymodules/python2.6/django/template/__init__.py" in render 779. bits.append(self.render_node(node, context)) File "/usr/lib/pymodules/python2.6/django/template/debug.py" in render_node 71. result = node.render(context) File "/usr/lib/pymodules/python2.6/django/template/__init__.py" in render 946. autoescape=context.autoescape)) File "/usr/lib/pymodules/python2.6/django/template/__init__.py" in render 779. bits.append(self.render_node(node, context)) File "/usr/lib/pymodules/python2.6/django/template/debug.py" in render_node 81. raise wrapped Exception Type: TemplateSyntaxError at /foodfolio/login Exception Value: Caught an exception while rendering: No module named app.models

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  • ClickableSpan on TextView in App Widget

    - by Dave Allison
    I have created an App Widget for Android 1.5. It uses a TextView to present a number of individual Text Links using ClickableSpans. However, the onClick event handler on the ClickableSpan is never called, it appears that you can not select individual components on the widget just the whole widget. This approach works fine for a normal app, so what I do I need to change to make this work for a widget?

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  • Determine if an element is a jQueryUI Widget

    - by Boycs
    I have written a jquery-ui widget using the Widget Factory... I need to be able to determine in code whether the element is already a widget or not... My investmentGrid widget is created on #container with $('#container').investmentGrid() I need to be able to determine elsewhere in the code if $('#container') is already an investmentGrid

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