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# Dica 29 - Django Admin: Criando actions no Admin

[![](https://980509322-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2F-Mfu9nhT31FhtY3TMBTi%2Fuploads%2Fgit-blob-b3d2f4124239955eec7072e99d8d0f8ab8cac2b9%2Fyoutube.png?alt=media)](https://youtu.be/WofnItMvqKU)

<https://docs.djangoproject.com/en/3.1/ref/contrib/admin/actions/>

Em `models.py` considere

```python
# models.py
STATUS_CHOICES = (
    ('d', 'Rascunho'),
    ('p', 'Publicado'),
    ('w', 'Retirado'),
)


class Article(models.Model):
    ...
    status = models.CharField(max_length=1, choices=STATUS_CHOICES)
```

E em `admin.py`

```python
# admin.py
@admin.register(Article)
class ArticleAdmin(admin.ModelAdmin):
    ...
    actions = ('make_published',)

    def make_published(self, request, queryset):
        count = queryset.update(status='p')

        if count == 1:
            msg = '{} artigo foi publicado.'
        else:
            msg = '{} artigos foram publicados.'

        self.message_user(request, msg.format(count))

    make_published.short_description = "Publicar artigos"
```