Define a method in your ModelAdmin-class and set its `allow_tags` attribute to `True`. This will allow the method to return unescaped HTML for display in the column.

Then list it as an entry in the ModelAdmin.list_display attribute.

**Example:**

    class YourModelAdmin(admin.ModelAdmin):
        list_display = (&#39;my_url_field&#39;,)

        def my_url_field(self, obj):
            return &#39;&lt;a href=&quot;%s%s&quot;&gt;%s&lt;/a&gt;&#39; % (&#39;http://url-to-prepend.com/&#39;, obj.url_field, obj.url_field)
        my_url_field.allow_tags = True
        my_url_field.short_description = &#39;Column description&#39;

See the documentation for [ModelAdmin.list_display](http://docs.djangoproject.com/en/dev/ref/contrib/admin/#django.contrib.admin.ModelAdmin.list_display) for more details.

Use the `format_html` utility. This will escape any html from parameters and mark the string as safe to use in templates. The `allow_tags` method attribute has been deprecated in Django 1.9.

    from django.utils.html import format_html
    from django.contrib import admin
    from django.utils.translation import gettext_lazy as _

    class MyModelAdmin(admin.ModelAdmin):
        list_display = [&#39;show_url&#39;, ...]
        # ...

        @admin.display(description=_(&quot;Column title&quot;))
        def show_url(self, obj):
            return format_html(&quot;&lt;a href=&#39;http://pre.com{0}&#39;&gt;{0}&lt;/a&gt;&quot;, obj.url)


Now your admin users are safe even in the case of:

    url == &#39;&lt;script&gt;eval(...);&lt;/script&gt;&#39;

See [the documentation][1] for more info.


  [1]: https://docs.djangoproject.com/en/stable/ref/contrib/admin/#django.contrib.admin.ModelAdmin.list%5Fdisplay

I am using &#39;instance&#39; instead of &#39;obj&#39;. Seppo Ervi&#228;l&#228;&#39;s answer helped me the most as I&#39;m using Django 3.0.1.

    def get_facebook(self, instance):
        return format_html(&quot;&lt;a target=&#39;_blank&#39; href=&#39;{0}&#39;&gt;{0}&lt;/a&gt;&quot;, instance.profile.facebook)
    get_facebook.short_description = &#39;Facebook&#39;

&gt; **Possible Duplicates:**  
&gt; [is &amp;ldquo;else if&amp;rdquo; faster than &amp;ldquo;switch() case&amp;rdquo; ?](https://stackoverflow.com/questions/767821/is-else-if-faster-than-switch-case)  
&gt; [What is the relative performance of if/else vs. switch in Java?](https://stackoverflow.com/questions/2086529/what-is-the-relative-performance-of-if-else-vs-switch-in-java)  

&lt;!-- End of automatically inserted text --&gt;

Ive been coding-in-the-run again....when the debugger steps through a case statement it jumps to the item that matches the conditions immediately, however when the same logic is specified using if/else it steps through every if statement until it finds the winner. Is the case statement more efficient, or is my debugger just optimizing the step through? (don&#39;t worry about the syntax/errors, i typed this in SO, don&#39;t know if it will compile, its the principle i&#39;m after, I didn&#39;t want to do them as ints cause i vaguely remember something about case using an offset with ints) I use C#, but im interested in a general answer across programming languages.

    switch(myObject.GetType()){
        
        case typeof(Car):
            //do something
            break;
        
        case typeof(Bike):
            //do something
            break;
        
        case typeof(Unicycle):
            //do something
            break;
        
        case default:
            break;
    }

**VS**

       Type myType = myObject.GetType();
            
       if (myType == typeof(Car)){
                //do something
       }
              
       else if (myType == typeof(Bike)){
                //do something
       }
              
       else if (myType == typeof(Unicycle)){
                //do something
       }
       else{
    
       }

   


Case vs If Else If: Which is more efficient?

I like constructor-based injection as it allows me to make injected fields `final`. I also like annotation driven injection as it simplifies my `context.xml`. I can mark my constructor with `@Autowired` and everything works fine, as long as I don&#39;t have two parameters of the same type. For example, I have a class:

&lt;!-- language: lang-java --&gt;

    @Component
    public class SomeClass {
        @Autowired(required=true)
        public SomeClass(OtherClass bean1, OtherClass bean2) {
            …
        }
    }

and an application context with:

&lt;!-- language: lang-xml --&gt;

    &lt;bean id=&quot;bean1&quot; class=&quot;OtherClass&quot; /&gt;
    &lt;bean id=&quot;bean2&quot; class=&quot;OtherClass&quot; /&gt;

There should be a way to specify the bean ID on the constructor of the class `SomeClass`, but I can&#39;t find it in the documentation.  Is it possible, or am I dreaming of a solution that does not exist yet?



  [1]: https://stackoverflow.com/a/2153680/1281433

Spring -- inject 2 beans of same type

I have a django admin interface and in the model listing I want a custom column that will be a hyperlink using one of the fields values.  Basically one of the models&#39; fields is a url and i&#39;d like the column to have that URL in a clickable hyperlink.  This link will need to have additional URL prepended to it as its a relative path in the model field.

How do I add a custom column with a hyperlink in the django admin interface?

<p>I have a django admin interface and in the model listing I want a custom column that will be a hyperlink using one of the fields values.  Basically one of the models' fields is a url and i'd like the column to have that URL in a clickable hyperlink.  This link will need to have additional URL prepended to it as its a relative path in the model field.</p>


My code in myapp_extras.py:

    from django import template
    
    register = template.Library()
    
    @register.inclusion_tag(&#39;new/userinfo.html&#39;)
    def address():
    	address = request.session[&#39;address&#39;]
    	return {&#39;address&#39;:address}

in &#39;settings.py&#39;:

    TEMPLATE_CONTEXT_PROCESSORS =(
    	&quot;django.core.context_processors.auth&quot;,
    	&quot;django.core.context_processors.debug&quot;,
    	&quot;django.core.context_processors.i18n&quot;,
    	&quot;django.core.context_processors.media&quot;,
    	&#39;django.core.context_processors.request&#39;
    )


but I got an error:

    TemplateSyntaxError at /items/
    
    Caught an exception while rendering: global name &#39;request&#39; is not defined
    
    Original Traceback (most recent call last):
      File &quot;C:\Python25\lib\site-packages\django\template\debug.py&quot;, line 71, in render_node
        result = node.render(context)
      File &quot;C:\Python25\lib\site-packages\django\template\__init__.py&quot;, line 915, in render
        dict = func(*args)
      File &quot;C:\p4\projects\myproject\..\myproject\invoice\templatetags\myapp_extras.py&quot;, line 9, in address
        address = request.session[&#39;address&#39;]
    NameError: global name &#39;request&#39; is not defined

I referenced this one
https://stackoverflow.com/questions/335231/in-django-is-it-possible-to-access-the-current-user-session-from-within-a-custom.


Access request in django custom template tags

I&#39;m trying to find a way to implement both a custom `QuerySet` and a custom `Manager` without breaking DRY.  This is what I have so far:

    class MyInquiryManager(models.Manager):
        def for_user(self, user):
            return self.get_query_set().filter(
                        Q(assigned_to_user=user) |
                        Q(assigned_to_group__in=user.groups.all())
                    )
    
    class Inquiry(models.Model):   
        ts = models.DateTimeField(auto_now_add=True)
        status = models.ForeignKey(InquiryStatus)
        assigned_to_user = models.ForeignKey(User, blank=True, null=True)
        assigned_to_group = models.ForeignKey(Group, blank=True, null=True)
        objects = MyInquiryManager()

This works fine, until I do something like this:

    inquiries = Inquiry.objects.filter(status=some_status)
    my_inquiry_count = inquiries.for_user(request.user).count()

This promptly breaks everything because the `QuerySet` doesn&#39;t have the same methods as the `Manager`.  I&#39;ve tried creating a custom `QuerySet` class, and implementing it in `MyInquiryManager`, but I end up replicating all of my method definitions.

I also found [this snippet][1] which works, but I need to pass in the extra argument to `for_user` so it breaks down because it relies heavily on redefining `get_query_set`.

Is there a way to do this without redefining all of my methods in both the `QuerySet` and the `Manager` subclasses?


  [1]: http://adam.gomaa.us/blog/2009/feb/16/subclassing-django-querysets/

Custom QuerySet and Manager without breaking DRY?

I&#39;m working on a large Django app, the vast majority of which requires a login to access.  This means that all throughout our app we&#39;ve sprinkled:

    @login_required
    def view(...):

That&#39;s fine, and it works great *as long as we remember to add it everywhere*!  Sadly sometimes we forget, and the failure often isn&#39;t terribly evident.  If the only link to a view is on a @login_required page then you&#39;re not likely to notice that you can actually reach that view without logging in.  But the bad guys might notice, which is a problem.

My idea was to reverse the system.  Instead of having to type @login_required everywhere, instead I&#39;d have something like:

    @public
    def public_view(...):

Just for the public stuff.  I tried to implement this with some middleware and I couldn&#39;t seem to get it to work.  Everything I tried interacted badly with other middleware we&#39;re using, I think.  Next up I tried writing something to traverse the URL patterns to check that everything that&#39;s not @public was marked @login_required - at least then we&#39;d get a quick error if we forgot something.  But then I couldn&#39;t figure out how to tell if @login_required had been applied to a view...

So, what&#39;s the right way to do this?  Thanks for the help!

Best way to make Django&#39;s login_required the default

When writing django queries one can use both id/pk as query parameters.

    Object.objects.get(id=1)
    Object.objects.get(pk=1)


I know that pk stands for Primary Key and is just a shortcut, according to django&#39;s documentation. However it is not clear when one should be using id or pk.

Django queries - id vs pk

    # admin.py
    class CustomerAdmin(admin.ModelAdmin):  
        list_display = (&#39;foo&#39;, &#39;number_of_orders&#39;)
    

    # models.py
    class Order(models.Model):
        bar = models.CharField[...]
        customer = models.ForeignKey(Customer)

    class Customer(models.Model):
        foo = models.CharField[...]

        def number_of_orders(self):
            return u&#39;%s&#39; % Order.objects.filter(customer=self).count()  

How could I sort Customers, depending on `number_of_orders` they have?  

`admin_order_field` property can&#39;t be used here, as it requires a database field to sort on. Is it possible at all, as Django relies on the underlying DB to perform sorting? Creating an aggregate field to contain the number of orders seems like an overkill here.

The fun thing: if you change url by hand in the browser to sort on this column - it works as expected!

Django admin: how to sort by one of the custom list_display fields that has no database field

How can I create more than one ModelAdmin for the same model, each customised differently and linked to different URLs?

Let&#39;s say I have a Django model called Posts. By default, the admin view of this model will list all Post objects.

I know I can customise the list of objects displayed on the page in various ways by setting variables like list_display or overriding the `queryset` method in my ModelAdmin like so:


    class MyPostAdmin(admin.ModelAdmin):
        list_display = (&#39;title&#39;, &#39;pub_date&#39;)

        def queryset(self, request):
            request_user = request.user
            return Post.objects.filter(author=request_user)

    admin.site.register(MyPostAdmin, Post)

By default, this would be accessible at the URL `/admin/myapp/post`. However I would like to have multiple views/ModelAdmins of the same model. e.g `/admin/myapp/post` would list all post objects, and `/admin/myapp/myposts` would list all posts belonging to the user, and `/admin/myapp/draftpost` might list all posts that have not yet been published. (these are just examples, my actual use-case is more complex)

You cannot register more than one ModelAdmin for the same model (this results in an `AlreadyRegistered` exception). Ideally I&#39;d like to achieve this **without** putting everything into a single ModelAdmin class and writing my own &#39;urls&#39; function to return a different queryset depending on the URL.

I&#39;ve had a look at the Django source and I see functions like `ModelAdmin.changelist_view` that could be somehow included in my urls.py, but I&#39;m not sure exactly how that would work.

**Update**: I&#39;ve found one way of doing what I want (see below), but I&#39;d still like to hear other ways of doing this.

Multiple ModelAdmins/views for same model in Django admin

I tried following Code with the help of the **django.contrib.auth.User** and **django.contrib.auth.Group**

    for g in request.user.groups:
        l.append(g.name)

But that failed and I received following **Error**:

    TypeError at /
    &#39;ManyRelatedManager&#39; object is not iterable
    Request Method:	GET
    Request URL:	http://localhost:8000/
    Exception Type:	TypeError
    Exception Value:	
    &#39;ManyRelatedManager&#39; object is not iterable
    Exception Location:	C:\p4\projects\...\users.py in permission, line 55


Thanks for any help!

is there a simple way to get group names of a user in django

I just want to add the subscription date in the User list in the Django CRUD&#160;Administration site.
How can I do that ?

Thank you for your help

How to customize the auth.User Admin page in Django CRUD?

By default, in Django-admin there is Users, Groups, and Sites apps. How can I remove Groups and Sites?

I tried to remove `admin.autodiscover()` from root urls. Then, when I added something like `admin.site.register(User, UserAdmin)` somewhere in my app models I got an `AlreadyRegistered` exception (this is fairly right - models users already registered in django.contrib.auth).


Content Type	Original Author	Original Content on Stackoverflow
Question	mikec	View Question on Stackoverflow
Solution 1 - Django	user257858	View Answer on Stackoverflow
Solution 2 - Django	Seppo Erviälä	View Answer on Stackoverflow
Solution 3 - Django	Alex Winkler	View Answer on Stackoverflow

How do I add a custom column with a hyperlink in the django admin interface?

Django Problem Overview

Django Solutions

Solution 1 - Django

Solution 2 - Django

Solution 3 - Django

Spring -- inject 2 beans of same type

Case vs If Else If: Which is more efficient?

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