.. versionadded:: 3.2

.. _wizard_how_to:

#########################################
How to implement content creation wizards
#########################################

django CMS offers a framework for creating 'wizards' - helpers - for content editors.

They provide a simplified workflow for common tasks.

A django CMS Page wizard already exists, but you can create your own for other content types very easily.


********************************
Create a content-creation wizard
********************************

Creating a CMS content creation wizard for your own module is fairly easy.

To begin, create a file in the root level of your module called ``forms.py``
to create your form(s)::

    # my_apps/forms.py

    from django import forms

    class MyAppWizardForm(forms.ModelForm):
        class Meta:
            model = MyApp
            exclude = []

Now create another file in the root level called ``cms_wizards.py``.
In this file, import ``Wizard`` as follows::

    from cms.wizards.wizard_base import Wizard
    from cms.wizards.wizard_pool import wizard_pool

Then, simply subclass ``Wizard``, instantiate it, then register it. If you were to
do this for ``MyApp``, it might look like this::


    # my_apps/cms_wizards.py

    from cms.wizards.wizard_base import Wizard
    from cms.wizards.wizard_pool import wizard_pool

    from .forms import MyAppWizardForm

    class MyAppWizard(Wizard):
        pass

    my_app_wizard = MyAppWizard(
        title="New MyApp",
        weight=200,
        form=MyAppWizardForm,
        description="Create a new MyApp instance",
    )

    wizard_pool.register(my_app_wizard)

.. note::

    If your model doesn't define a ``get_absolute_url`` function then your wizard
    will require a :ref:`get_success_url` method.

    ..  code-block:: python

        class MyAppWizard(Wizard):

            def get_success_url(self, obj, **kwargs):
                """
                This should return the URL of the created object, «obj».
                """
                if 'language' in kwargs:
                    with force_language(kwargs['language']):
                        url = obj.get_absolute_url()
                else:
                    url = obj.get_absolute_url()

                return url

That's it!

.. note::

    The module name ``cms_wizards`` is special, in that any such-named modules in
    your project's Python path will automatically be loaded, triggering the
    registration of any wizards found in them. Wizards may be declared and
    registered in other modules, but they might not be automatically loaded.

The above example is using a ``ModelForm``, but you can also use ``forms.Form``.
In this case, you **must** provide the model class as another keyword argument
when you instantiate the Wizard object.

For example::

    # my_apps/forms.py

    from django import forms

    class MyAppWizardForm(forms.Form):
        name = forms.CharField()


    # my_apps/cms_wizards.py

    from cms.wizards.wizard_base import Wizard
    from cms.wizards.wizard_pool import wizard_pool

    from .forms import MyAppWizardForm
    from .models import MyApp

    class MyAppWizard(Wizard):
        pass

    my_app_wizard = MyAppWizard(
        title="New MyApp",
        weight=200,
        form=MyAppWizardForm,
        model=MyApp,
        description="Create a new MyApp instance",
    )

    wizard_pool.register(my_app_wizard)

You must subclass ``cms.wizards.wizard_base.Wizard`` to use it. This is because
each wizard's uniqueness is determined by its class and module name.

See the :ref:`Reference section on wizards <wizard_reference>` for technical details of the wizards
API.