python - Flask 路由到不同模块中具有相同名称的函数

Question

如果我有 2 个文件，例如:

moduleA.py

from MyPackage import app

@app.route('/moduleA/get')
def get():
    return "a"

moduleB.py

from MyPackage import app

@app.route('/moduleB/get')
def get():
    return "b"

并在 __init__.py

from flask import Flask
import IPython
import logging

app = Flask(__name__,
        static_url_path='', 
        static_folder='static',
        template_folder='templates')
from MyPackage import moduleA, moduleB

然后flask会抛出错误AssertionError: View function mapping is overwriting an existing endpoint function: get

我认为 python 本身在这里没有看到冲突，因为函数在 2 个独立的模块中，但是 flask 有。有没有更好的标准可以在这里使用，还是我必须将函数名称命名为 def getModuleA？

Answer 1

可以考虑使用Blueprint

Factor an application into a set of blueprints. This is ideal for larger applications; a project could instantiate an application object, initialize several extensions, and register a collection of blueprints.

例子:

# module_a.py
from flask import Blueprint

blueprint = Blueprint('module_a', __name__)

@blueprint.route('/get')
def get():
    return 'a'

# module_b.py
from flask import Blueprint

blueprint = Blueprint('module_b', __name__)

@blueprint.route('/get')
def get():
    return 'b'

# app.py
from flask import Flask
from module_a import blueprint as blueprint_a
from module_b import blueprint as blueprint_b


app = Flask(__name__)
app.register_blueprint(blueprint_a, url_prefix='/a')
app.register_blueprint(blueprint_b, url_prefix='/b')

# serves:
#  - /a/get
#  - /b/get