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Embedded Flask-Admin list views.

Project description


Embed Flask-Admin list views to an arbitrary page:


  • Inline edits are not supported

  • Bootstrap3 templates only


pip install flask-admin-subview


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

class ContentModel(db.Model):
    __table__ = "content"
    id = db.Column(db.Integer, primary_key=True)
    container_id = db.Column(db.Integer, db.ForeignKey(""), 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:

import flask_admin_subview

class ContentModelSubview(flask_admin_subview.View, ContentModelView):

Or you can create a brand-new view for the subview:

from flask_admin.contrib.sqla import ModelView
import flask_admin_subview

class ContentModelSubview(flask_admin_subview.View, ModelView):

Add query filter to show content for certain container only, container id will be passed as a URL parameter:

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

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

app = Flask(__name__)
# ...
    ContentModelSubview(Content, db.session, "Content", endpoint="content_subview").

Prepare container view

Use helper to display subview in the model’s details:

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"),


  • Add tests

  • Add example app code comments

  • Add Bootstrap2 templates

  • Possibly, support inline edits

  • Describe advanced usage

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