flask-admin how to access model in viewmodel

Question

I want to access the model in a descendant ModelView like so:

class MyAppLibraryView(MyModelView):
    if model != Salary:
        column_searchable_list = ['name', ]
    form_excluded_columns = ['date_created', 'date_modified', ]
    column_display_pk = True

This class is used as follows:

admin.add_view(MyAppLibraryView(Section, db.session))
admin.add_view(MyAppLibraryView(Office, db.session))
admin.add_view(MyAppLibraryView(Salary_reference, db.session))
admin.add_view(MyAppLibraryView(Salary, db.session))

In MyAppLibraryView, model is unknown to the class. How do I access the models passed to the class MyAppLibraryView?

Answer 1

As you can see in flask-admin code, you can access model with self.model.

You have some ways to override column_searchable_list based on the model.

1. Subclass (the way I recommend)

class MyAppLibraryView(MyModelView):
    form_excluded_columns = ['date_created', 'date_modified', ]
    column_display_pk = True


class MyAppLibrarySalaryView(MyAppLibraryView):
    column_searchable_list = ['name', ]


admin.add_view(MyAppLibrarySalaryView(Salary, db.session))

2. you can set it in __init__.

class MyAppLibraryView(MyModelView):
    def __init__(self, model, *args, **kwargs):
        if model != Salary:
            self.column_searchable_list = ['name',]
]
        super(MyAppLibraryView, self).__init__(model, *args, **kwargs)