Question

我正在尝试在 IIS 上托管简单的 FastAPI 应用程序。

app.py

from fastapi.middleware.wsgi import WSGIMiddleware
from flask import Flask, escape, request

flask_app = Flask(__name__)


@flask_app.route("/")
def flask_main():
    name = request.args.get("name", "World")
    return f"Hello, {escape(name)} from Flask!"


app = FastAPI()


@app.get("/v2")
def read_main():
    return {"message": "Hello World"}


app.mount("/v1", WSGIMiddleware(flask_app))

web.config

<configuration>  
  <system.webServer>
    <handlers>
        <add name="Python FastCGI" path="*" verb="*" modules="FastCgiModule" 
        scriptProcessor="C:\Users\oniel\AppData\Local\Programs\Python\Python39\python.exe|C:\Users\oniel\AppData\Local\Programs\Python\Python39\Lib\site-packages\wfastcgi.py" 
        resourceType="Unspecified" requireAccess="Script" />
    </handlers>
  </system.webServer>
  <appSettings>
    <add key="WSGI_HANDLER" value="my_app.app" />
    <add key="PYTHONPATH" value="C:\inetpub\wwwroot\Flask_Demo" />
    <!-- <add key="WSGI_LOG" value="C:\inetpub\wwwroot\Flask_Demo\app.log" />-->
  </appSettings>
</configuration>

将 WSGI_HANDLER 更改为 my_app.flask_app，然后就可以了。

为 fastAPI 给出的错误：

Error occurred:

Traceback (most recent call last):
  File "c:\users\oniel\appdata\local\programs\python\python39\lib\site-packages\wfastcgi.py", line 847, in main
    result = handler(record.params, response.start)
TypeError: __call__() missing 1 required positional argument: 'send'


StdOut: 

StdErr:

我正在关注此链接：https://fastapi.tiangolo.com/advanced/wsgi/

使用 WSGI 在 IIS 上托管 FastAPI 时出现问题

0 个答案: