You can delete any QuerySet you&#39;d like. For example, to delete all blog posts with some Post model

    Post.objects.all().delete()

and to delete any Post with a future publication date

    Post.objects.filter(pub_date__gt=datetime.now()).delete()

You do, however, need to come up with a way to narrow down your QuerySet. If you just want a view to delete a particular object, look into the [delete generic view][1].

EDIT:

Sorry for the misunderstanding. I think the answer is somewhere between. To implement your own, combine `ModelForm`s and [generic views][2]. Otherwise, look into 3rd party apps that provide similar functionality. In a [related question][3],  the recommendation was [django-filter][4].


  [1]: https://docs.djangoproject.com/en/dev/ref/class-based-views/#deleteview
  [2]: https://docs.djangoproject.com/en/dev/ref/class-based-views/
  [3]: https://stackoverflow.com/questions/2315500/reusing-django-changelist-outside-of-admin-site
  [4]: http://github.com/alex/django-filter

I&#39;d like to have all of the menu items that don&#39;t fit into the ActionBar go into the overflow menu (the one that is reached from the Action Bar not the menu button) *even on devices that **do** have a Menu button*.  This seems much more intuitive for users than throwing them into a separate menu list that requires the user to jump from a touch(screen) interaction to a button based interaction simply because the layout of the ActionBar can&#39;t fit them on the bar. 

On the emulator I can set the &quot;Hardware Back/Home Keys&quot; value to &quot;no&quot; and get this effect. 
I&#39;ve searched for a way to do this in code for an actual device that has a menu button but can&#39;t fine one.  Can anyone help me?

How to force use of overflow menu on devices with menu button

I was just studying OCPJP questions and I found this strange code:

    public static void main(String a[]) {
        System.out.println(Double.NaN==Double.NaN);
    	System.out.println(Double.NaN!=Double.NaN);
    }

When I ran the code, I got:

    false
    true

How is the output `false` when we&#39;re comparing two things that look the same as each other? What does `NaN` mean?

Why does Double.NaN==Double.NaN return false?

I need to select several objects to be deleted from my database in django using a webpage. There is no category to select from so I can&#39;t delete from all of them like that. Do I have to implement my own delete form and process it in django or does django have a way to already do this? As its implemented in the admin interface.

Delete multiple objects in django

<p>I need to select several objects to be deleted from my database in django using a webpage. There is no category to select from so I can't delete from all of them like that. Do I have to implement my own delete form and process it in django or does django have a way to already do this? As its implemented in the admin interface.</p>


I&#39;ve tried prefetch_related() in django 1.4 from trunk and can&#39;t make it to prefetch reverse lookup.

My simplified models (each book has many prices):

    class Book(models.Model):
        # some fields
    
    class Price(models.Model):
        book = models.ForeignKey(Book)


My view&#39;s query:
    
    books = Book.objects.prefetch_related(&#39;price&#39;)
    
Then, I got the AttributeError message:
    
    AttributeError: Cannot find &#39;price&#39; on Book object, &#39;price&#39; is an invalid parameter to prefetch_related()


How to make it work?
Thanks.


Django: Does prefetch_related() follow reverse relationship lookup?

This is related to this question: https://stackoverflow.com/questions/9260249/django-return-json-and-html-depending-on-client-python

----------


I have a command line Python API for a Django app. When I access the app through the API it should return JSON and with a browser it should return HTML. I can use different URLs to access the different versions but how do I render the HTML template and JSON in the **views.py** with just one template?

To render the HTML I would use:

    return render_to_response(&#39;sample/sample.html....&#39;)

But how would I do the same for JSON without putting a JSON template? (the `content-type` should be `application/json` instead of `text/html`)

What would determine the JSON and HTML outputs?

So in my **views.py**:

    if something:
        return render_to_response(&#39;html_template&#39;,.....)
    else:
        return HttpReponse(jsondata,mimetype=&#39;application/json&#39;)


How do I return JSON without using a template in Django?

I&#39;m sure this is a trivial operation, but I can&#39;t figure out how it&#39;s done.

There&#39;s got to be something smarter than this:

    ids = [1, 3, 6, 7, 9]

    for id in ids:
        MyModel.objects.filter(pk=id)

I&#39;m looking to get them all in one query with something like:

    MyModel.objects.filter(pk=[1, 3, 6, 7, 9])

How can I filter a Django query with a list of values?

How can I filter a Django query with a list of values?

What is the best way to set a default value for a foreign key field in a model? Suppose I have two models, `Student` and `Exam` with student having `exam_taken` as foreign key. How would I ideally set a default value for it? Here&#39;s a log of my effort

    class Student(models.Model):
       ....
       .....
       exam_taken = models.ForeignKey(&quot;Exam&quot;, default=1)

Works, but have a hunch there&#39;s a better way.


    def get_exam():
        return Exam.objects.get(id=1)
    
    class Student(models.Model):
        ....
        .....
        exam_taken = models.ForeignKey(&quot;Exam&quot;, default=get_exam)

But this fails with tables does not exist error while syncing.

Any help would be appreciated.

Setting default value for Foreign Key attribute

Hi I am using the django model class with some field and a password field. Instead of displaying regular plain text I want to display password input. I created a model class like this:

    class UserForm(ModelForm):
        class Meta:
            password = forms.CharField(widget=forms.PasswordInput)
            model = User
            widgets = {
                &#39;password&#39;: forms.PasswordInput(),
            }
But i am getting the following error: NameError: name &#39;forms&#39; is not defined. 

I am using django version 1.4.0. I followed this link : https://stackoverflow.com/questions/3532354/django-password-problems

Still getting the same error. What should i do. Where am i getting wrong.Please help

How to create password input field in django

Say, I have the following mixins that overlaps with each other by touching `dispatch()`:

    class FooMixin(object):
        def dispatch(self, *args, **kwargs):
            # perform check A
            ...
            return super(FooMixin, self).dispatch(*args, **kwargs)

    class BarMixin(object):
        def dispatch(self, *args, **kwargs):
            # perform check B
            ...
            return super(FooMixin, self).dispatch(*args, **kwargs)

If I want my view to go through the order, check A -&gt; check B, should my code be `MyView(FooMixin, BarMixin, View)` or `MyView(BarMixin, FooMixin, View)`?

And why do we always put `View` or its subclasses after mixins? (I have noticed this from reading the source code of the django generic views, but I don&#39;t know the rationale behind it, if any) 

How does the order of mixins affect the derived class?

I ran into an issue where I assigned `request.user` to a variable called `prior_user`, then essentially authenticated the user, then checked to see if `request.user != prior_user`. I expected them not to be the same and that `prior_user` should contain `AnonymousUser. To my surprise, they were the same. 

Sample code:

    prior_user = request.user   # request object, obtained froma  view
    authenticate_user(request)   # some function that authenticates
    print prior_user.username != request.user.username   # returns False i.e.they are the same!

I then discovered prior_user actually contains an instance of django.utils.functional.SimpleLazyObject so I assume it is some sort of lazy lookup type thing i.e. prior_user&#39;s value isn&#39;t looked up until actually used. Looking at the source code, I cannot confirm this.

Anyone with django experience can tell me what is going on and why it is needed? 

This leaves me a little shaken, because the usual assignment statement doesn&#39;t work the way I expect and what else within Django acts like this? Nor did I see this described in the [docs][1].

So anyone with super human knowledge of django can provide some clarity?

  [1]: https://docs.djangoproject.com/en/dev/ref/request-response/#httprequest-objects

django: Purpose of django.utils.functional.SimpleLazyObject?

I have a custom class-based view

    # myapp/views.py
    from django.views.generic import *

    class MyView(DetailView):
        template_name = &#39;detail.html&#39;
        model = MyModel

        def get_object(self, queryset=None):
            return queryset.get(slug=self.slug)

I want to pass in the slug parameter (or other parameters to the view) like this

    MyView.as_view(slug=&#39;hello_world&#39;)

Do I need to override any methods to be able to do this?

Content Type	Original Author	Original Content on Stackoverflow
Question	Dean	View Question on Stackoverflow
Solution 1 - Django	Matt Luongo	View Answer on Stackoverflow

Delete multiple objects in django

Django Problem Overview

Django Solutions

Solution 1 - Django

Why does Double.NaN==Double.NaN return false?

How to force use of overflow menu on devices with menu button

Attributions