Flask-Admin-Subview
===================
Embed Flask-Admin list views to an arbitrary page:
.. image:: https://raw.githubusercontent.com/artemShelest/flask-admin-subview/master/res/screen1.png
.. image:: https://raw.githubusercontent.com/artemShelest/flask-admin-subview/master/res/screen2.png
Limitations
===========
- Inline edits are not supported
- Bootstrap3 templates only
Installation
============
.. code-block:: console
pip install flask-admin-subview
Integration
===========
The easiest way to integrate is to use helpers for the details view of the model. The following example demonstrates
integration of subview to show relations of SQLAlchemy model in the details page.
**DB Schema**
.. code-block:: python
class ContentModel(db.Model):
__table__ = "content"
id = db.Column(db.Integer, primary_key=True)
container_id = db.Column(db.Integer, db.ForeignKey("container.id"), nullable=False)
class ContainerModel(db.Model):
__table__ = "container"
id = db.Column(db.Integer, primary_key=True)
content = db.relationship(ContentModel)
**Prepare your subview**
It is a good idea to subclass existing view of your model:
.. code-block:: python
import flask_admin_subview
class ContentModelSubview(flask_admin_subview.View, ContentModelView):
pass
Or you can create a brand-new view for the subview:
.. code-block:: python
from flask_admin.contrib.sqla import ModelView
import flask_admin_subview
class ContentModelSubview(flask_admin_subview.View, ModelView):
pass
Add query filter to show content for certain container only, container id will be passed as a URL parameter:
.. code-block:: python
class ContentModelSubview(...):
def get_query(self):
return self._extend_query(super(ContentModelSubview, self).get_query())
def get_count_query(self):
return self._extend_query(super(ContentModelSubview, self).get_count_query())
def _extend_query(self, query):
container_id = request.args.get('id')
if container_id is None:
abort(400, "Container id required")
return query.filter(ContentModel.container_id == container_id)
**Initialize an extension**
.. code-block:: python
from flask_admin_subview import Subview
app = Flask(__name__)
admin = Admin(app, template_mode="bootstrap3")
# only supports bootstrap3 mode
Subview(app, template_mode="bootstrap3")
**Add your subview as a blueprint**
.. code-block:: python
app = Flask(__name__)
# ...
app.register_blueprint(
ContentModelSubview(Content, db.session, "Content", endpoint="content_subview").
create_blueprint(admin))
**Prepare container view**
Use helper to display subview in the model's details:
.. code-block:: python
from flask_admin_subview import SubviewContainerMixin, SubviewEntry
class ContainerView(SubviewContainerMixin, ModelView):
can_view_details = True
subviews = (
# specify that we need to pass id from the location URL to the subview
SubviewEntry("/admin/content_subview/", "Content Subview", "id"),
)
TODO
====
- Add tests
- Add example app code comments
- Add Bootstrap2 templates
- Possibly, support inline edits
- Describe advanced usage
