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  • Access to field in extended flatpage in django

    - by Stanislav Feldman
    How to access field in extended flatpage in django? I wrote this: class ExtendedFlatPage(FlatPage): teaser = CharField(max_length=150) class ExtendedFlatPageForm(FlatpageForm): teaser = CharField(max_length=150) class Meta: model = ExtendedFlatPage class ExtendedFlatPageAdmin(FlatPageAdmin): form = ExtendedFlatPageForm fieldsets = ( (None, {'fields': ('url', 'title', 'teaser', 'content', 'sites',)}), ) admin.site.unregister(FlatPage) admin.site.register(ExtendedFlatPage, ExtendedFlatPageAdmin) And creation in admin is ok. But then in flatpages/default.html I tried this: <html> <body> <h1>{{ flatpage.title }}</h1> <strong>{{ flatpage.teaser }}</strong> <p>{{ flatpage.content }}</p> </body> </html> And there was no flatpage.teaser! What is wrong?

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  • (Django) Trim whitespaces from charField

    - by zardon
    How do I strip whitespaces (trim) from the end of a charField in Django? Here is my Model, as you can see I've tried putting in clean methods but these never get run. I've also tried doing name.strip(), models.charField().strip() but these do not work either. Is there a way to force the charField to trim automatically for me? Thanks. class Employee(models.Model): """(Workers, Staff, etc)""" name = models.CharField(blank=True, null=True, max_length=100) # This never gets run def clean_variable(self): data = self.cleaned_data['variable'].strip() return data def __unicode__(self): return self.name class Meta: verbose_name_plural = 'Employees' # This never gets run either class EmployeesForm(forms.ModelForm): class Meta: model = Employee def clean_description(self): #if not self.cleaned_data['description'].strip(): # raise forms.ValidationError('Your error message here') self.cleaned_data['name'].strip()

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  • in django am facing url problem.....

    - by dpaksp
    am using django.0.97 version i have model called profile in that i have few fields eg like name ,email...ects and it's backend also ready..i.e database . and all users profile is created...i have given all permission to all users. when i login ,click on profile..i able to see list of all user name thr when i click on it ,it's goin to model page where i can edit the user profile..instead of that i want to navigate to a template when i can display the user details ,i have set the URl for it so that when url of that type request comes it should call a view from view it will call my template to display user details,.....the problem is it's not calling my view.... i think my problem is brief....if any information still required ?? pls ask me....and help me

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  • Override Django inlineformset_factory has_changed() to always return True

    - by John
    Hi, I am using the django inlineformset_factory function. a = get_object_or_404(ModelA, pk=id) FormSet = inlineformset_factory(ModelA, ModelB) if request.method == 'POST': metaform = FormSet (instance=a, data=request.POST) if metaform.is_valid(): f = metaform.save(commit=False) for instance in f: instance.updated_by = request.user instance.save() else: metaform = FormSet(instance=a) return render_to_response('nodes/form.html', {'form':metaform}) What is happening is that if I change any of the data then everything works ok and all the data gets updated. However if I don't change any of the data then the data is not updated. i.e. only entries which are changed go through the for loop to be saved. I guess this makes sense as there is no point saving data if it has not changed. However I need to go through and save every object in the form regardless of whether it has any changes on not. So my question is how do I override this so that it goes through and saves every record whether it has any changes or not? Hope this makes sense Thanks

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  • Django: query spanning multiple many-to-many relationships

    - by Brant
    I've got some models set up like this: class AppGroup(models.Model): users = models.ManyToManyField(User) class Notification(models.Model): groups_to_notify = models.ManyToManyField(AppGroup) The User objects come from django's authentication system. Now, I am trying to get all the notifications pertaining to the groups that the current user is a part of. I have tried.. notifications = Notification.objects.filter(groups_to_notify=AppGroup.objects.filter(users=request.user)) But that gives an error: more than one row returned by a subquery used as an expression Which I suppose is because the groups_to_notify is checking against several groups. How can I grab all the notifications meant for the user based on the groups he is a part of?

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  • Django: where do I call settings.configure?

    - by RexE
    The Django docs say that I can call settings.configure instead of having a DJANGO_SETTINGS_MODULE. I would like my website's project to do this. In what file should I put the call to settings.configure so that my settings will get configured at the right time? Edit in response to Daniel Roseman's comment: The reason I want to do this is that settings.configure lets you pass in the settings variables as a kwargs dict, e.g. {'INSTALLED_APPS': ..., 'TEMPLATE_DIRS': ..., ...}. This would allow my app's users to specify their settings in a dict, then pass that dict to a function in my app that augments it with certain settings necessary to make my app work, e.g. adding entries to INSTALLED_APPS. What I envision looks like this. Let's call my app "rexe_app". In wsgi.py, my app's users would do: import rexe_app my_settings = {'INSTALLED_APPS': ('a','b'), ...} updated_settings = rexe_app.augment_settings(my_settings) # now updated_settings is {'INSTALLED_APPS': ('a','b','c'), 'SESSION_SAVE_EVERY_REQUEST': True, ...} settings.configure(**updated_settings)

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  • ModelName(django.contrib.auth.models.User) vs ModelName(models.Model)

    - by amr.negm
    I am developing a django project. I created some apps, some of those are related to User model, for instance, I have a feeds app that handles user feeds, and another app that deals with extra user data like age, contacts, and friends. for each of these, I created a table that should be connected to the User model, which I using for storing and authenticating users. I found two ways to deal with this issue. One, is through extending User model to be like this: ModelName(User): friends = models.ManyToMany('self') ..... Two, is through adding a foreign key to the new table like this: ModelName(models.Model): user = models.ForeignKey(User, unique=True) friends = friends = models.ManyToMany('self') ...... I can't decide which to use in which case. in other words, what are the core differences between both?

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  • Setting a preferred item of a many-to-one in Django

    - by Mike DeSimone
    I'm trying to create a Django model that handles the following: An Item can have several Names. One of the Names for an Item is its primary Name, i.e. the Name displayed given an Item. (The model names were changed to protect the innocent.) The models.py I've got looks like: class Item(models.Model): primaryName = models.OneToOneField("Name", verbose_name="Primary Name", related_name="_unused") def __unicode__(self): return self.primaryName.name class Name(models.Model): item = models.ForeignKey(Item) name = models.CharField(max_length=32, unique=True) def __unicode__(self): return self.name class Meta: ordering = [ 'name' ] The admin.py looks like: class NameInline(admin.TabularInline): model = Name class ItemAdmin(admin.ModelAdmin): inlines = [ NameInline ] admin.site.register(Item, ItemAdmin) It looks like the database schema is working fine, but I'm having trouble with the admin, so I'm not sure of anything at this point. My main questions are: How do I explain to the admin that primaryName needs to be one of the Names of the item being edited? Is there a way to automatically set primaryName to the first Name found, if primaryName is not set, since I'm using inline admin for the names?

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  • Django admin proper urls inside listview

    - by hinnye
    Hi, My current target is to give users the chance to download CSV files from the admin site of my application. I successfully managed to create an additional column in the model's list view this way: def doc_link(self): return '<a href="files/%s">%s</a>' % (self.output, self.output) doc_link.allow_tags = True This shows the file name and creates the link, but sadly - because it's inside my 'searches' view - it has an URL: my_site/my_app/searches/files/13.csv. This is my problem, I would like to have my files stored in the admin media directory, like this: http://my_site/media/files/13.csv Does somebody know how to give url which points "outer" from the model's directory? Maybe somehow tell Django to use the ADMIN_MEDIA_PREFIX in the link? I'd really appreciate any help, thanks!

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  • Saving data in a inherited django model

    - by aldeano
    I'm building an app to save data and some calculations made with those datas, the idea is keep the data in one model and the calculations in other. So, the models are like this: class FreshData(models.Model): name = models.CharField(max_length=20) one = models.IntegerField() two = models.IntegerField() def save(self, *args, **kwargs): Calculations() Calculations.three = self.one + self.two super(FreshData, self).save(*args, **kwargs) Calculations.save() class Calculations(FreshData): three = models.IntegerField() I've got a valueerror pointing out "self.one" and "self.two" as without value. I keep the idea in witch my design is wrong and django has a simpler way to store related data.

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  • Django: Site-Wide URL Prefix

    - by Tom
    I've built a Django site that will live at the root when it's live. Right now it's functioning perfectly at the IP address. For testing purposes, the client has pointed a proxy url at it, but the url has /folder/path in it, so none of the URL patterns match. I put (/folder/path)? into all the url patterns so they now respond, but all of the links are broken because I'm using the {% url %} tag and while the url patterns will match the optional path, they don't include it in that tag. Clearly I can just hard-code /folder/path into all of my urls (well, into all of the url includes) until testing is complete, but is there a better way to do this?

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  • Django model class and custom property

    - by dArignac
    Howdy - today a weird problem occured to me: I have a modle class in Django and added a custom property to it that shall not be saved into the database and therefore is not represent in the models structure: class Category(models.Model): groups = models.ManyToManyField(Group) title = defaultdict() Now, when I'm within the shell or writing a test and I do the following: c1 = Category.objects.create() c1.title['de'] = 'german title' print c1.title['de'] # prints "german title" c2 = Category.objects.create() print c2.title['de'] # prints "german title" <-- WTF? It seems that 'title' is kind of global. If I change title to a simple string it works as expected, so it has to do something with the dict? I also tried setting title as a property: title = property(_title) But that did not work, too. So, how can I solve this? Thank you in advance! enter code here

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  • Django 1.2: Dates in admin forms don't work with Locales (I10N=True)

    - by equalium
    I have an application in Django 1.2. Language is selectable (I18N and Locale = True) When I select the english lang. in the site, the admin works OK. But when I change to any other language this is what happens with date inputs (spanish example): Correctly, the input accepts the spanish format %d/%m/%Y (Even selecting from the calendar, the date inserts as expected). But when I save the form and load it again, the date shows in the english form: %Y-%m-%d The real problem is that when I load the form to change any other text field and try to save it I get an error telling me to enter a valid date, so I have to write all dates again or change the language in the site to use the admin. I haven't specified anything for DATE_INPUT_FORMATS in settings nor have I overridden forms or models. Surely I am missing something but I can't find it. Can anybody give me a hint?

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  • get foreign key objects in a single query - Django

    - by John
    Hi I have 2 models in my django code: class ModelA(models.Model): name = models.CharField(max_length=255) description = models.CharField(max_length=255) created_by = models.ForeignKey(User) class ModelB(models.Model): category = models.CharField(max_length=255) modela_link = models.ForeignKey(ModelA, 'modelb_link') functions = models.CharField(max_length=255) created_by = models.ForeignKey(User) Say ModelA has 100 records, all of which may or may not have links to ModelB Now say I want to get a list of every ModelA record along with the data from ModelB I would do: list_a = ModelA.objects.all() Then to get the data for ModelB I would have to do for i in list_a: i.additional_data = i.modelb_link.all() However this runs a query on every instance of i. Thus making 101 queries to run. Is there any way of running this all in just 1 query. Or at least less than the 101 queries. I've tried putting in ModelA.objects.select_related().all() but this didn't seem to have any effect. Thanks

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  • Django Template - Convert python list into a javascript object

    - by amcashcow
    I am working on a django / python website I have a page where I want to display a table of search results The list of results is passed in to the template as normal I also want to make this list of objects accessible to the javascript code My first solution was just create another view that returned json format. But each page load required calling the query twice. So then I tried only downloading the data using the json view and printing the table using javascript. but this is also not desirable as now the presentation layer is mixed into the javascript code. is there a way to create a javascript object from the python list as the page is rendered?

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  • Foreign Key Relationships

    - by Yehonathan Quartey
    I have two models class Subject(models.Model): name = models.CharField(max_length=100,choices=COURSE_CHOICES) created = models.DateTimeField('created', auto_now_add=True) modified = models.DateTimeField('modified', auto_now=True) syllabus = models.FileField(upload_to='syllabus') def __unicode__(self): return self.name and class Pastquestion(models.Model): subject=models.ForeignKey(Subject) year =models.PositiveIntegerField() questions = models.FileField(upload_to='pastquestions') def __unicode__(self): return str(self.year) Each Subject can have one or more past questions but a past question can have only one subject. I want to get a subject, and get its related past questions of a particular year. I was thinking of fetching a subject and getting its related past question. Currently am implementing my code such that I rather get the past question whose subject and year correspond to any specified subject like this_subject=Subject.objects.get(name=the_subject) thepastQ=Pastquestion.objects.get(year=2000,subject=this_subject) I was thinking there is a better way to do this. Or is this already a better way? Please Do tell ?

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  • Django: Converting an entire Model into a single dictionary

    - by LarrikJ
    Is there a good way in Django to convert an entire model to a dictionary? I mean, like this: class DictModel(models.Model): key = models.CharField(20) value = models.CharField(200) DictModel.objects.all().to_dict() ... with the result being a dictionary with the key/value pairs made up of records in the Model? Has anyone else seen this as being useful for them? Thanks. Update I just wanted to add is that my ultimate goal is to be able to do a simple variable lookup inside a Template. Something like: {{ DictModel.exampleKey }} With a result of DictModel.objects.get(key__exact=exampleKey).value Overall, though, you guys have really surprised me with how helpful allof your responses are, and how different the ways to approach it can be. Thanks a lot.

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  • Django, ModelForms, User and UserProfile - not hashing password

    - by IvanBernat
    I'm trying to setup a User - UserProfile relationship, display the form and save the data. When submitted, the data is saved, except the password field doesn't get hashed. Additionally, how can I remove the help_text from the username and password (inherited from the User model)? Full code is below, excuse me if it's too long. Models.py USER_IS_CHOICES = ( ('u', 'Choice A'), ('p', 'Choice B'), ('n', 'Ninja'), ) class UserProfile(models.Model): user = models.ForeignKey(User, unique=True) user_is = models.CharField(max_length=1, choices=USER_IS_CHOICES) Forms.py class UserForm(forms.ModelForm): class Meta: model = User fields = ["first_name", "last_name", "username", "email", "password"] def clean_username(self): username = self.cleaned_data['username'] if not re.search(r'^\w+$', username): raise forms.ValidationError('Username can contain only alphanumeric characters') try: User.objects.get(username=username) except ObjectDoesNotExist: return username raise forms.ValidationError('Username is already taken') class UserProfileForm(forms.ModelForm): class Meta: model = UserProfile fields = ['user_is'] Views.py if request.method == 'POST': uform = UserForm(request.POST) pform = UserProfileForm(request.POST) if uform.is_valid() and pform.is_valid(): user = uform.save() profile = pform.save(commit = False) profile.user = user profile.save() return HttpResponseRedirect('/') else: uform = UserForm() pform = UserProfileForm() variables = RequestContext(request, { 'uform':uform, 'pform':pform }) return render_to_response('registration/register.html', variables)

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  • Implement a generic C++/CLI interface in a C# generic interface

    - by Florent
    I have a C++/CLI generic interface like this : using namespace System::Collections::Generic; namespace Test { generic <class T> public interface class IElementList { property List<T>^ Elements; }; } and I want to implement it in a C# generic interface like this : using Test; namespace U { public interface IElementLightList<T> : IElementList<T> where T : IElementLight { bool isFrozen(); bool isPastable(); } } This don't work, Visual Studio is not able to see C++/CLI IElementList interface ! I tested with a not generic C++/CLI interface and this work. What I missed ?

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  • Django - transactions in the model?

    - by orokusaki
    Models (disregard typos / minor syntax issues. It's just pseudo-code): class SecretModel(models.Model): some_unique_field = models.CharField(max_length=25, unique=True) # Notice this is unique. class MyModel(models.Model): secret_model = models.OneToOneField(SecretModel, editable=False) # Not in the form spam = models.CharField(max_length=15) foo = models.IntegerField() def clean(self): SecretModel.objects.create(some_unique_field=self.spam) Now if I go do this: MyModel.objects.create(spam='john', foo='OOPS') # Obviously foo won't take "OOPS" as it's an IntegerField. #.... ERROR HERE MyModel.objects.create(spam='john', foo=5) # So I try again here. #... IntegrityError because SecretModel with some_unique_field = 'john' already exists. I understand that I could put this into a view with a request transaction around it, but I want this to work in the Admin, and via an API, etc. Not just with forms, or views. How is it possible?

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  • Django Initial for a ManyToMany Field

    - by gramware
    I have a form that edits an instance of my model. I would like to use the form to pass all the values as hidden with an inital values of username defaulting to the logged in user so that it becomes a subscribe form. The problem is that the normal initial={'field':value} doesn't seem to work for manytomany fields. how do i go about it? my views.py @login_required def event_view(request,eventID): user = UserProfile.objects.get(pk=request.session['_auth_user_id']) event = events.objects.get(eventID = eventID) if request.method == 'POST': form = eventsSusbcribeForm( request.POST,instance=event) if form.is_valid(): form.save() return HttpResponseRedirect('/events/') else: form = eventsSusbcribeForm(instance=event) return render_to_response('event_view.html', {'user':user,'event':event, 'form':form},context_instance = RequestContext( request )) my forms.py class eventsSusbcribeForm(forms.ModelForm): eventposter = forms.ModelChoiceField(queryset=UserProfile.objects.all(), widget=forms.HiddenInput()) details = forms.CharField(widget=forms.Textarea(attrs={'cols':'50', 'rows':'5'}),label='Enter Event Description here') date = forms.DateField(widget=SelectDateWidget()) class Meta: model = events exclude = ('deleted') def __init__(self, *args, **kwargs): super(eventsSusbcribeForm, self).__init__(*args, **kwargs) self.fields['username'].initial = (user.id for user in UserProfile.objects.filter())

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  • Django: How/Where to store a value for a session without unnecessary DB hits

    - by GerardJP
    Hi all, I have an extended userprofile with AUTH_PROFILE_MODULE (ref: http://tinyurl.com/yhracqq) I would like to set a user.is_guru() method similar to user.is_active(). This would results for al views (or rather templates) to e.g. disable/enable certain user messages, displaying of widgets, etc. The boolean is stored in the extended user profile model, but I want to avoid hitting the DB for every view. So the questions is .. Do I use a context_processor, a template tag, session_dict or what have you to, possible cached, store this info for the duration of the users visit. Note: I dont have performance issues, so it's definitely filed under premature optimization. I just want to avoid generating extra work in the future :). Any pointers are very welcome. Thanx and greetz! Gerard.

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  • I want to select the distict value from models field and then update them (django)

    - by qulzam
    I have models... class Item(models.Model): name = models.CharField('Item Name', max_length = 30) item_code = models.CharField(max_length = 10) color = models.CharField(max_length = 150, null = True, blank = True) size = models.CharField(max_length = 30, null = True, blank = True) fabric_code = models.CharField(max_length = 30, null = True, blank = True) I have values in Item. in Item model name field has the similar values..(but the other values of record are change). I want to select the name field values distinctly(ie similar values select only ones). in one box(like combo box). What kind of form or views i use??

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  • Deploying a Django application in a virtual Ubuntu Server

    - by mfsaint
    I have a virtualbox machine running Ubuntu Server 10.04LTS. My intention is to this machine to work like a VPS, this way I can learn and prepare for when I get a VPS service. Apache+mod_wsgi for deploying the Django app seems the right choice to me. I have the domain (marianofalcon.com.ar) but nothing else, no DNS. The problem is that I'm pretty lost with all the deployment stuff. I know how to configure mod_wsgi(with the django.wsgi file) and apache(creating a VirtualHost). Something is missing and I don't know what it is. I think that I lack networking skills ant that's the big problem. Trying to host the app on a virtualbox adds some difficulty because I don't know well what IP to use. This is what I've got: file placed at: /etc/apache2/sites-available: NameVirtualHost *:80 <VirtualHost *:80> ServerAdmin [email protected] ServerName www.my-domain.com ServerAlias my-domain.com Alias /media /path/to/my/project/media DocumentRoot /path/to/my/project WSGIScriptAlias / /path/to/your/project/apache/django.wsgi ErrorLog /var/log/apache2/error.log LogLevel warn CustomLog /var/log/apache2/access.log combined </VirtualHost> django.wsgi file: import os, sys wsgi_dir = os.path.abspath(os.path.dirname(__file__)) project_dir = os.path.dirname(wsgi_dir) sys.path.append(project_dir) project_settings = os.path.join(project_dir,'settings') os.environ['DJANGO_SETTINGS_MODULE'] = 'myproject.settings' import django.core.handlers.wsgi application = django.core.handlers.wsgi.WSGIHandler()

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  • Komodo Edit 5.2 Django Template Syntax Error - Info: <head> previously mentioned

    - by Lance McNearney
    I am using Komodo Edit 5.2 for editing html and Django template files. It always shows a single syntax error inside the first {% block %} area on the first tag of my template. For example: {% extends "base.html" %} {% load i18n %} {% block title %}Hello{% endblock %} {% block content %} <p>Hello</p> <-- Syntax error on this single line <p>Other lines have no errors</p> {% endblock %} {% block footer %} <p>No errors here</p> {% endblock %} The syntax error given is: Info: <head> previously mentioned I know for a fact that the error has nothing to do with my <head> tag since it occurs in the base template and in child templates (and the IDE isn't smart enough to process the base templates when in a child, etc.) All of my html tags are closed properly and everything validates for XHTML strict. This forum post mentions a similar problem but offers no solution (and may be specific to Smarty syntax highlighting). Any ideas on how to resolve this error (or disable it from being shown)?

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