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-1

Background

I'm trying to set up a blueprint whose name matches the filename it resides in, so that when I reference it in my app.py I know where the blueprint comes from. This should be possible because the example on exploreflask uses the same pattern. Still, I cannot figure out how to make this work with my structure.

File structure

├── app.py
├── frontend
    ├── __init__.py
    └── views
    ├── home.py
    └── __init__.py

Example

frontend/views/home.py

from flask import Blueprint, render_template

home = Blueprint('home', __name__)
home1 = Blueprint('home1', __name__)

frontend/views/__init__.py

from .home import home
from .home import home1

app.py

from flask import Flask

from frontend.views import home
from frontend.views import home1

print (type(home)) --> <class 'function'> 
print (type(home1)) --> <class 'flask.blueprints.Blueprint'>

As home1 registers correctly as a Blueprint but home does not I suspect that
there is a name collision but I don't know how to resolve it despite looking into
this excellent article on importing conventions.

As a result, when I try to register my blueprints with the app
this will work:

app.register_blueprint(home1, url_prefix='/home1') --> Fine

but this won't:

app.register_blueprint(home, url_prefix='/home')
--> AttributeError: 'function' object has no attribute 'name'

Why not just go along with using home1?

1. I want to understand how the collision can be resolved


2. I want to be able to use route names that are the same as the filename they are in like so:

frontend/views/home.py

from flask import Blueprint, render_template

home = Blueprint('home', __name__)

@home.route('/')
def home():
 pass

python flask python-import

share|improve this question

edited Mar 24 at 16:28

pfabri

asked Mar 24 at 15:34

pfabripfabri

136112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Are you trying to use the Divisional or Functional structure described in exporeflask?
  
  – PGHE
  Mar 25 at 0:12
  

  
  
  
  

add a comment | 

-1

Background

I'm trying to set up a blueprint whose name matches the filename it resides in, so that when I reference it in my app.py I know where the blueprint comes from. This should be possible because the example on exploreflask uses the same pattern. Still, I cannot figure out how to make this work with my structure.

File structure

├── app.py
├── frontend
    ├── __init__.py
    └── views
    ├── home.py
    └── __init__.py

Example

frontend/views/home.py

from flask import Blueprint, render_template

home = Blueprint('home', __name__)
home1 = Blueprint('home1', __name__)

frontend/views/__init__.py

from .home import home
from .home import home1

app.py

from flask import Flask

from frontend.views import home
from frontend.views import home1

print (type(home)) --> <class 'function'> 
print (type(home1)) --> <class 'flask.blueprints.Blueprint'>

As home1 registers correctly as a Blueprint but home does not I suspect that
there is a name collision but I don't know how to resolve it despite looking into
this excellent article on importing conventions.

As a result, when I try to register my blueprints with the app
this will work:

app.register_blueprint(home1, url_prefix='/home1') --> Fine

but this won't:

app.register_blueprint(home, url_prefix='/home')
--> AttributeError: 'function' object has no attribute 'name'

Why not just go along with using home1?

1. I want to understand how the collision can be resolved


2. I want to be able to use route names that are the same as the filename they are in like so:

frontend/views/home.py

from flask import Blueprint, render_template

home = Blueprint('home', __name__)

@home.route('/')
def home():
 pass

python flask python-import

share|improve this question

edited Mar 24 at 16:28

pfabri

asked Mar 24 at 15:34

pfabripfabri

136112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Are you trying to use the Divisional or Functional structure described in exporeflask?
  
  – PGHE
  Mar 25 at 0:12
  

  
  
  
  

add a comment | 

Background

I'm trying to set up a blueprint whose name matches the filename it resides in, so that when I reference it in my app.py I know where the blueprint comes from. This should be possible because the example on exploreflask uses the same pattern. Still, I cannot figure out how to make this work with my structure.

File structure

├── app.py
├── frontend
    ├── __init__.py
    └── views
    ├── home.py
    └── __init__.py

Example

frontend/views/home.py

from flask import Blueprint, render_template

home = Blueprint('home', __name__)
home1 = Blueprint('home1', __name__)

frontend/views/__init__.py

from .home import home
from .home import home1

app.py

from flask import Flask

from frontend.views import home
from frontend.views import home1

print (type(home)) --> <class 'function'> 
print (type(home1)) --> <class 'flask.blueprints.Blueprint'>

As home1 registers correctly as a Blueprint but home does not I suspect that
there is a name collision but I don't know how to resolve it despite looking into
this excellent article on importing conventions.

As a result, when I try to register my blueprints with the app
this will work:

app.register_blueprint(home1, url_prefix='/home1') --> Fine

but this won't:

app.register_blueprint(home, url_prefix='/home')
--> AttributeError: 'function' object has no attribute 'name'

Why not just go along with using home1?

1. I want to understand how the collision can be resolved


2. I want to be able to use route names that are the same as the filename they are in like so:

frontend/views/home.py

from flask import Blueprint, render_template

home = Blueprint('home', __name__)

@home.route('/')
def home():
 pass

python flask python-import

share|improve this question

edited Mar 24 at 16:28

pfabri

asked Mar 24 at 15:34

pfabripfabri

136112

├── app.py
├── frontend
    ├── __init__.py
    └── views
    ├── home.py
    └── __init__.py

from flask import Blueprint, render_template

home = Blueprint('home', __name__)
home1 = Blueprint('home1', __name__)

from .home import home
from .home import home1

from flask import Flask

from frontend.views import home
from frontend.views import home1

print (type(home)) --> <class 'function'> 
print (type(home1)) --> <class 'flask.blueprints.Blueprint'>

app.register_blueprint(home1, url_prefix='/home1') --> Fine

app.register_blueprint(home, url_prefix='/home')
--> AttributeError: 'function' object has no attribute 'name'

from flask import Blueprint, render_template

home = Blueprint('home', __name__)

@home.route('/')
def home():
 pass

share|improve this question

edited Mar 24 at 16:28

pfabri

asked Mar 24 at 15:34

pfabripfabri

136112

share|improve this question

edited Mar 24 at 16:28

pfabri

asked Mar 24 at 15:34

pfabripfabri

136112

asked Mar 24 at 15:34

pfabripfabri

136112

asked Mar 24 at 15:34

pfabripfabri

136112

1 Answer
1

active

oldest

votes

-1

Try to use Capital letters in the Blueprint Module.

also you can use the url_prefix in the module.

Home = Blueprint("Home", __name__, url_prefix="/home")

@Home.route("/")
def home():
 pass

share|improve this answer

edited Mar 24 at 16:16

answered Mar 24 at 15:59

JoySalomonJoySalomon

15

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Using capital letters could be a hacky workaround, but that's exactly what I'd like to avoid. Besides it violates PEP8: capitals are for classes. The url_prefix is unrelated to the problem (and I'm using it already, anyway).
  
  – pfabri
  Mar 24 at 16:11
  
  

  
  
  
  

add a comment | 

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-1

Try to use Capital letters in the Blueprint Module.

also you can use the url_prefix in the module.

Home = Blueprint("Home", __name__, url_prefix="/home")

@Home.route("/")
def home():
 pass

share|improve this answer

edited Mar 24 at 16:16

answered Mar 24 at 15:59

JoySalomonJoySalomon

15

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Using capital letters could be a hacky workaround, but that's exactly what I'd like to avoid. Besides it violates PEP8: capitals are for classes. The url_prefix is unrelated to the problem (and I'm using it already, anyway).
  
  – pfabri
  Mar 24 at 16:11
  
  

  
  
  
  

add a comment | 

-1

Try to use Capital letters in the Blueprint Module.

also you can use the url_prefix in the module.

Home = Blueprint("Home", __name__, url_prefix="/home")

@Home.route("/")
def home():
 pass

share|improve this answer

edited Mar 24 at 16:16

answered Mar 24 at 15:59

JoySalomonJoySalomon

15

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Using capital letters could be a hacky workaround, but that's exactly what I'd like to avoid. Besides it violates PEP8: capitals are for classes. The url_prefix is unrelated to the problem (and I'm using it already, anyway).
  
  – pfabri
  Mar 24 at 16:11
  
  

  
  
  
  

add a comment | 

Try to use Capital letters in the Blueprint Module.

also you can use the url_prefix in the module.

Home = Blueprint("Home", __name__, url_prefix="/home")

@Home.route("/")
def home():
 pass

share|improve this answer

edited Mar 24 at 16:16

answered Mar 24 at 15:59

JoySalomonJoySalomon

15

Home = Blueprint("Home", __name__, url_prefix="/home")

@Home.route("/")
def home():
 pass

share|improve this answer

edited Mar 24 at 16:16

answered Mar 24 at 15:59

JoySalomonJoySalomon

15

answered Mar 24 at 15:59

JoySalomonJoySalomon

15

answered Mar 24 at 15:59

JoySalomonJoySalomon

15