有两种不同的方法可以指定一个应用程序modjy ：
$ b $ ol

对于第一种方法，简单地创建一个导入Flask应用程序对象的文件。

从my_flask_app导入应用程序作为应用程序

然后在您的web.xml设置正确的init-param：

 < init-param> 
< param-name> app_import_name< / param-name> 
< param-value> wsgi.application< / param-value> 
< / init-param> 
 

对于第二种方法，您可以使用在servlet上下文根中定义application.py的modjy约定调用Flask WSGI应用程序的单个处理程序方法：
$ b $ pre $ $ $ c $ def处理程序（environ，start_response）：
返回应用程序。 wsgi_app（environ，start_response）

I successfully deployed the demo web app that comes with Jython. It uses modjy which is a Jython WSGI gateway. I'm now trying to hook modjy to my Flask app. I get a handler not defined error.

The full traceback is here: http://pastie.org/2810227

There are two different ways you can specify an application to modjy:

1. Using the app_import_name mechanism
2. Using a combination of app_directory/app_filename/app_callable_name

For the first method simply create a file that imports your Flask app object.

from my_flask_app import app as application

Then in your web.xml set the proper init-param:

<init-param>
  <param-name>app_import_name</param-name>
  <param-value>wsgi.application</param-value>
</init-param>

For the second method you can use the modjy convention of defining application.py in the servlet context root with a single handler method that invokes the Flask WSGI app:

def handler(environ, start_response):
    return application.wsgi_app(environ, start_response)