Serve static files from blueprints without url pre

Question

For development, I'd like to serve static files from the blueprint's static folder. However, I don't want to load all files via the url prefix of the blueprint. Is there a possibility to configure flask, so it looks in all blueprint's static folders and returns the first file it finds? The reason for this is that when I deploy the application, the build process will pull all the files our of the blueprints and put them into a separate folder (similar to Django's collectstatics).

In other words, I have 2 blueprints: foo and bar:

app
  +--- foo
  |    +--- assets
  |    |     +--- css
  |    |     |
  |    |     +--- js
  |    +
  |    |
  |    +--- __init__.py:  foo_blueprint = Blueprint('foo', __name__, static_folder='assets')
  |
  +--- bar
       +
       |
       +--- __init__.py: bar_blueprint = Blueprint('bar', __name__, static_folder='assets')

I want all js files in the particular js subfolders to be available from the url /static/js/file.js. Is this possible?

Answer 1

You can't with default flask mechanism.

In your your example you will create 3 rules for static forlder:

Rule "/assets/<path:filename>" for "static" endpoint
Rule "/assets/<path:filename>" for "foo.static" endpoint
Rule "/assets/<path:filename>" for "bar.static" endpoint

When flask will match your URL to rule it take first match and return it. In this case it return static endpoint rule. After will dispatched function for this endpoint, e.g. one folder.

PS. I don't sure that exactly static endpoint, better check Map.update method.

But you always can write own request descriptor, which will look at blue prints folders:

class MyApp(Flask):
    def static_dispatchers(self):
        yield super(MyApp, self).send_static_file
        for blueprint in self.blueprints.values():
            yield blueprint.send_static_file

    def send_static_file(self, filename):
        last_exception = None
        for static_dispatcher in self.static_dispatchers():
            try:
                return static_dispatcher(filename)
            except NotFound as e:
                last_exception = e
        raise last_exception

PS. This example doesn't include the blueprint registration sequence, because it stored in dict.

But if you have two files with same name, then first file will processed. For example if you have next structure:

/static
/static/app.js "console.log('app');"
/foo/static/app.js "console.log('foo app');"
/foo/static/blue.js "console.log('foo blue');"
/foo/static/foo.js "console.log('foo self');"
/bar/static/app.js "console.log('bar app');"
/bar/static/blue.js "console.log('foo blue');"
/bar/static/bar.js "console.log('bar self');"

And include scripts to page:

<script src="{{ url_for('static', filename='app.js') }}"></script>
<script src="{{ url_for('foo.static', filename='app.js') }}"></script>
<script src="{{ url_for('bar.static', filename='app.js') }}"></script>
<script src="{{ url_for('foo.static', filename='blue.js') }}"></script>
<script src="{{ url_for('bar.static', filename='blue.js') }}"></script>
<script src="{{ url_for('foo.static', filename='foo.js') }}"></script>
<script src="{{ url_for('bar.static', filename='bar.js') }}"></script>

You will have next result:

app
app
app
foo blue (or bar blue)
foo blue (or bar blue)
foo self
bar self

For js and css can concatenate files with same path, but can't do this for images.

However I prefer use unique URL prefix for each blueprint, because it simple with url_for.