Question

我只是试图通过服务器运行Hello World并遇到wsgi错误。

我的应用程序位于/ var / www / testapp中我在此路径下的文件是testapp.py 代码是...

from flask import Flask
app = Flask(__name__)

@app.route('/')
def hello_world():
        return 'Hello, World'

if __name__ == "__main_":
    app.run()

足够容易。如果我从CLI（127.0.0.1:5000）启动它，它将运行正常

问题是当我尝试通过apache运行它时。我在同一文件夹中有一个wsgi文件，名为testapp.wsgi

代码在这里。...

import sys
sys.path.insert(0, "/var/www/")
from testapp import app as application

Apache conf文件是这个。.

<VirtualHost *>
        ServerName example.com
        WSGIDaemonProcess testapp
        WSGIScriptAlias /  /var/www/testapp/testapp.wsgi
        <Directory /var/www/testapp>
                WSGIProcessGroup testapp
                WSGIApplicationGroup %{GLOBAL}
                Order deny,allow
                Allow from all
        </Directory>
        LogLevel warn
</VirtualHost>

我得到的错误是在Apache日志中...

[wsgi:error] [pid 28774:tid 139666678216448] [remote xxxxxx:56870] mod_wsgi (pid=28774): Target WSGI script '/var/www/testapp/testapp.wsgi' cannot be loaded as Python module.
[Fri Feb 15 19:53:01.769229 2019] 139666678216448] [remote xxxxxx:56870] mod_wsgi (pid=28774): Exception occurred processing WSGI script '/var/www/testapp/testapp.wsgi'.
[Fri Feb 15 19:53:01.774590 2019] [wsgi:error] [pid 28774:tid 139666678216448] [remote xxxxxx:56870] Traceback (most recent call last):
[Fri Feb 15 19:53:01.774716 2019] [wsgi:error] [pid 28774:tid 139666678216448] [remote xxxxxx:56870]   File "/var/www/testapp/testapp.wsgi", line 3, in <module>
[Fri Feb 15 19:53:01.774785 2019] [wsgi:error] [pid 28774:tid 139666678216448] [remote xxxxxx:56870]     from testapp import app as application
[Fri Feb 15 19:53:01.774875 2019] [wsgi:error] [pid 28774:tid 139666678216448] [remote xxxxxx:56870] ImportError: cannot import name 'app'

我忍不住简单地认为这很荒谬。有人可以帮我指出正确的方向吗？ JW

如何通过wsgi运行Flask？

0 个答案: