为所有Flask路由添加前缀
为所有Flask路由添加前缀
提问于 2013-09-24 03:39:27
我有一个前缀，我想添加到每个路由。现在，我在每个定义中将一个常量添加到路由。有没有办法自动做到这一点？
PREFIX = "/abc/123"
@app.route(PREFIX + "/")
def index_page():
return "This is a website about burritos"
@app.route(PREFIX + "/about")
def about_page():
return "This is a website about burritos"回答 7
Stack Overflow用户
发布于 2013-09-24 05:24:13
您可以将您的路线放在蓝图中：
bp = Blueprint('burritos', __name__,
template_folder='templates')
@bp.route("/")
def index_page():
return "This is a website about burritos"
@bp.route("/about")
def about_page():
return "This is a website about burritos"然后使用前缀将蓝图注册到应用程序中：
app = Flask(__name__)
app.register_blueprint(bp, url_prefix='/abc/123')Stack Overflow用户
发布于 2016-05-19 03:52:08
因此，我认为一个有效的答案是:前缀应该在开发完成后使用的实际服务器应用程序中进行配置。Apache、nginx等。
但是，如果您希望在调试中运行Flask应用程序时在开发过程中使用此方法，请查看this gist。
Flask的DispatcherMiddleware来拯救你了！
我将把代码复制到这里以供后人使用：
"Serve a Flask app on a sub-url during localhost development."
from flask import Flask
APPLICATION_ROOT = '/spam'
app = Flask(__name__)
app.config.from_object(__name__) # I think this adds APPLICATION_ROOT
# to the config - I'm not exactly sure how!
# alternatively:
# app.config['APPLICATION_ROOT'] = APPLICATION_ROOT
@app.route('/')
def index():
return 'Hello, world!'
if __name__ == '__main__':
# Relevant documents:
# http://werkzeug.pocoo.org/docs/middlewares/
# http://flask.pocoo.org/docs/patterns/appdispatch/
from werkzeug.serving import run_simple
from werkzeug.wsgi import DispatcherMiddleware
app.config['DEBUG'] = True
# Load a dummy app at the root URL to give 404 errors.
# Serve app at APPLICATION_ROOT for localhost development.
application = DispatcherMiddleware(Flask('dummy_app'), {
app.config['APPLICATION_ROOT']: app,
})
run_simple('localhost', 5000, application, use_reloader=True)现在，当将上面的代码作为独立的Flask应用程序运行时，http://localhost:5000/spam/将显示Hello, world!。
在对另一个答案的评论中，我表达了我希望做这样的事情：
from flask import Flask, Blueprint
# Let's pretend module_blueprint defines a route, '/record/<id>/'
from some_submodule.flask import module_blueprint
app = Flask(__name__)
app.config['APPLICATION_ROOT'] = '/api'
app.register_blueprint(module_blueprint, url_prefix='/some_submodule')
app.run()
# I now would like to be able to get to my route via this url:
# http://host:8080/api/some_submodule/record/1/将DispatcherMiddleware应用到我设计的示例中：
from flask import Flask, Blueprint
from flask.serving import run_simple
from flask.wsgi import DispatcherMiddleware
# Let's pretend module_blueprint defines a route, '/record/<id>/'
from some_submodule.flask import module_blueprint
app = Flask(__name__)
app.config['APPLICATION_ROOT'] = '/api'
app.register_blueprint(module_blueprint, url_prefix='/some_submodule')
application = DispatcherMiddleware(Flask('dummy_app'), {
app.config['APPLICATION_ROOT']: app
})
run_simple('localhost', 5000, application, use_reloader=True)
# Now, this url works!
# http://host:8080/api/some_submodule/record/1/Stack Overflow用户
发布于 2019-01-10 19:30:28
from flask import Flask
app = Flask(__name__)
app.register_blueprint(bp, url_prefix='/abc/123')
if __name__ == "__main__":
app.run(debug='True', port=4444)
bp = Blueprint('burritos', __name__,
template_folder='templates')
@bp.route('/')
def test():
return "success"页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/18967441
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