Python 如何用WSGI实现Flask应用程序的路径调度?
我希望使用单个域作为多个flask应用程序的暂存环境,这些应用程序最终将在它们自己的域上运行 比如:Python 如何用WSGI实现Flask应用程序的路径调度?,python,path,flask,wsgi,dispatch,Python,Path,Flask,Wsgi,Dispatch,我希望使用单个域作为多个flask应用程序的暂存环境,这些应用程序最终将在它们自己的域上运行 比如: 示例_staging.com/app1 示例_staging.com/app2 示例_staging.com/app3 其中: 示例_staging.com/app1的作用与app1.example_staging.com相同 示例_staging.com/app2的作用与app2.example_staging.com相同 示例_staging.com/app3的作用与app3.exam
- 示例_staging.com/app1
- 示例_staging.com/app2
- 示例_staging.com/app3
- 示例_staging.com/app1的作用与app1.example_staging.com相同
- 示例_staging.com/app2的作用与app2.example_staging.com相同
- 示例_staging.com/app3的作用与app3.example_staging.com相同
- 示例_staging.com/app1的作用与app1.com相同
- 示例_staging.com/app2的作用与app2.com相同
- 示例_staging.com/app3的作用与app3.com相同
from flask import Flask
app = Flask(__name__)
@app.route('/')
def hello_world():
return 'Hello from Flask!'
WSGI启动程序配置文件:
import sys
project_home = u'/home/path/sample1'
if project_home not in sys.path:
sys.path = [project_home] + sys.path
from app import app as application
提及:
我不知道在哪里添加文档中给出的代码作为示例,以及create_app、default_app、get_user_for_prefix应该是什么样子
注意:使用pythonawhere
解决方案
Glenns输入后的WSGI配置文件:
import sys
# add your project directory to the sys.path
project_home = u'/home/path/app1'
if project_home not in sys.path:
sys.path = [project_home] + sys.path
from werkzeug.wsgi import DispatcherMiddleware
from app import app as app1
from app2.app import app as app2
from app3.app import app as app3
application = DispatcherMiddleware(app1, {
'/app2': app2,
'/app3': app3
})
文件夹结构:
app1 folder
app2 folder
app3 folder
这里需要注意的关键是,你实际上有4个应用程序(3个单独的应用程序和一个组合应用程序)。这忽略了staging/live的区别,因为staging和live只是不同目录中彼此的副本 创建每个应用程序,并让它们在各自的域上做出响应。然后创建一个新的应用程序,从各个应用程序导入
应用程序
变量,并使用DispatcherMiddle软件
将它们组合起来,就像您链接到的文档页面上标题“组合应用程序”下的示例一样。这对我来说很有效:
文件夹结构
DISPATCHER (folder)
dispatcher.py
app1 (folder)
__init__.py
app2 (folder)
__init__.py
app3 (folder)
__init__.py
dispatcher.py
from flask import Flask
from werkzeug.wsgi import DispatcherMiddleware #use the commented version below for Werkzeug >= v.1.0
#from werkzeug.middleware.dispatcher import DispatcherMiddleware
from werkzeug.exceptions import NotFound
from app1 import app as app1
from app2 import app as app2
from app3 import app as app3
app = Flask(__name__)
app.wsgi_app = DispatcherMiddleware(NotFound(), {
"/app1": app1,
'/app2': app2,
'/app3': app3
})
if __name__ == "__main__":
app.run()
from flask import Flask
app = Flask(__name__)
@app.route("/")
def index_one():
return "Hi im 1 or 2 or 3"
if __name__ == "__main__":
app.run()
app1到app3init.py
from flask import Flask
from werkzeug.wsgi import DispatcherMiddleware #use the commented version below for Werkzeug >= v.1.0
#from werkzeug.middleware.dispatcher import DispatcherMiddleware
from werkzeug.exceptions import NotFound
from app1 import app as app1
from app2 import app as app2
from app3 import app as app3
app = Flask(__name__)
app.wsgi_app = DispatcherMiddleware(NotFound(), {
"/app1": app1,
'/app2': app2,
'/app3': app3
})
if __name__ == "__main__":
app.run()
from flask import Flask
app = Flask(__name__)
@app.route("/")
def index_one():
return "Hi im 1 or 2 or 3"
if __name__ == "__main__":
app.run()
工作
python dispatcher.py
localhost:5000/app1 "Hi im one"
localhost:5000/app2 "Hi im two"
localhost:5000/app3 "Hi im three"
另一种配置
您可以导入另一个应用程序,如app0,并向应用程序添加菜单,使用NotFound()
这有帮助
有没有可能与大家分享一个更加通俗易懂的例子?这看起来非常好,除非你说
python app.py
。您没有名为app.py的文件