The Python Oracle

Display tick and cross icons for a property in the Django administration console

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Chapters
00:00 Question
00:48 Accepted answer (Score 34)
02:01 Answer 2 (Score 2)
02:22 Thank you

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Full question
https://stackoverflow.com/questions/7474...

Accepted answer links:
[list_display]: https://docs.djangoproject.com/en/1.0/re...

Answer 2 links:
[@admin.display]: https://docs.djangoproject.com/en/3.2/re...

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https://meta.stackexchange.com/help/lice...

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Tags
#python #django #djangoadmin

#avk47



ACCEPTED ANSWER

Score 38


You don't want to use list_filter. The property you're looking for is list_display. The documentation offers an example of how you can create a column that behaves like a boolean in the display. In short, you do something like this:

  1. Create a method in the class:

    def is_activated(self)
        if self.bar == 'something':
            return True
        return False
    
  2. add the .boolean method attribute directly below the is_activated method:

    is_activated.boolean = True
    
  3. Add the method as a field in list_display:

    class MyAdmin(ModelAdmin): list_display = ['name', 'is_activated']

  4. You'll notice the column name is probably now "Is Activated" or something like that. If you want the column heading to change, you use the short_description method attribute:

    is_activated.short_description = "Activated"
    



ANSWER 2

Score 9


The correct way to do this now in Django 3.0+ is with @admin.display.

    @admin.display(
      boolean=True,
      ordering='-publish_date',
      description='Is Published?', 
    ) 
    def is_published(self, obj):
        return obj.publish_date is not None