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git-svn-id: http://code.djangoproject.com/svn/django/branches/newforms-admin@6656 bcc190cf-cafb-0310-a4f2-bffc1f526a37
98 lines
3.8 KiB
Plaintext
98 lines
3.8 KiB
Plaintext
============
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Form preview
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============
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Django comes with an optional "form preview" application that helps automate
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the following workflow:
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"Display an HTML form, force a preview, then do something with the submission."
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To force a preview of a form submission, all you have to do is write a short
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Python class.
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Overview
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=========
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Given a ``django.newforms.Form`` subclass that you define, this application
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takes care of the following workflow:
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1. Displays the form as HTML on a Web page.
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2. Validates the form data when it's submitted via POST.
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a. If it's valid, displays a preview page.
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b. If it's not valid, redisplays the form with error messages.
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3. When the "confirmation" form is submitted from the preview page, calls
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a hook that you define -- a ``done()`` method that gets passed the valid
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data.
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The framework enforces the required preview by passing a shared-secret hash to
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the preview page via hidden form fields. If somebody tweaks the form parameters
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on the preview page, the form submission will fail the hash-comparison test.
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How to use ``FormPreview``
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==========================
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1. Point Django at the default FormPreview templates. There are two ways to
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do this:
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* Add ``'django.contrib.formtools'`` to your ``INSTALLED_APPS``
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setting. This will work if your ``TEMPLATE_LOADERS`` setting includes
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the ``app_directories`` template loader (which is the case by
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default). See the `template loader docs`_ for more.
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* Otherwise, determine the full filesystem path to the
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``django/contrib/formtools/templates`` directory, and add that
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directory to your ``TEMPLATE_DIRS`` setting.
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2. Create a ``FormPreview`` subclass that overrides the ``done()`` method::
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from django.contrib.formtools.preview import FormPreview
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from myapp.models import SomeModel
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class SomeModelFormPreview(FormPreview):
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def done(self, request, cleaned_data):
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# Do something with the cleaned_data, then redirect
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# to a "success" page.
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return HttpResponseRedirect('/form/success')
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This method takes an ``HttpRequest`` object and a dictionary of the form
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data after it has been validated and cleaned. It should return an
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``HttpResponseRedirect`` that is the end result of the form being
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submitted.
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3. Change your URLconf to point to an instance of your ``FormPreview``
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subclass::
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from myapp.preview import SomeModelFormPreview
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from myapp.models import SomeModel
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from django import newforms as forms
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...and add the following line to the appropriate model in your URLconf::
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(r'^post/$', SomeModelFormPreview(forms.models.form_for_model(SomeModel))),
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Or, if you already have a Form class defined for the model::
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(r'^post/$', SomeModelFormPreview(SomeModelForm)),
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4. Run the Django server and visit ``/post/`` in your browser.
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.. _template loader docs: ../templates_python/#loader-types
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``FormPreview`` classes
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=======================
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A ``FormPreview`` class is a simple Python class that represents the preview
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workflow. ``FormPreview`` classes must subclass
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``django.contrib.formtools.preview.FormPreview`` and override the ``done()``
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method. They can live anywhere in your codebase.
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``FormPreview`` templates
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=========================
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By default, the form is rendered via the template ``formtools/form.html``, and
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the preview page is rendered via the template ``formtools.preview.html``.
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These values can be overridden for a particular form preview by setting
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``preview_template`` and ``form_template`` attributes on the FormPreview
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subclass. See ``django/contrib/formtools/templates`` for the default templates.
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