Python + Flask 网络应用报告 "[Errno 9] Bad file descriptor"与 pexpect 模块

Question

我正在使用 Python + Flask 开发网络应用程序。从最简单的意义上讲，客户端对特定 URL 的请求将触发应用程序登录到服务器端的远程机器，执行一系列 shell 命令，并将输出(JSON 格式)解析并发送给客户端作为响应的一部分。这些命令相当简单。要登录到远程机器，我唯一可用的方法是 rlogin，所以我使用了 pexpect 模块，因为我 couldn't find rlogin 的任何标准 Python 模块。

现在的问题是，虽然我能够从 pexpect/rlogin 获得正确的输出，但将该输出(字符串)作为响应发送会导致错误:

----------------------------------------
Exception happened during processing of request from ('171.71.55.54', 62736)
Traceback (most recent call last):
  File "/isan/python/python2.7/SocketServer.py", line 295, in _handle_request_noblock
    self.process_request(request, client_address)
  File "/isan/python/python2.7/SocketServer.py", line 321, in process_request
    self.finish_request(request, client_address)
  File "/isan/python/python2.7/SocketServer.py", line 334, in finish_request
    self.RequestHandlerClass(request, client_address, self)
  File "/isan/python/python2.7/SocketServer.py", line 651, in __init__
    self.finish()
  File "/isan/python/python2.7/SocketServer.py", line 710, in finish
    self.wfile.close()
  File "/isan/python/python2.7/socket.py", line 279, in close
    self.flush()
  File "/isan/python/python2.7/socket.py", line 303, in flush
    self._sock.sendall(view[write_offset:write_offset+buffer_size])
error: [Errno 9] Bad file descriptor
----------------------------------------

我已将代码精简到重现错误所需的最低限度。对此应用程序的 HTTP 请求导致“文件描述符错误”错误。抱歉缩进不同，我使用了两个不同的编辑器来修改代码!

import os
import subprocess
import pexpect

from flask import Flask

app = Flask(__name__)

class rlogin:
        def __init__(self, host, prompt):
                self.child = pexpect.spawn("rlogin " + host)
                self.prompt = prompt
                self.child.expect_exact(self.prompt)

        def command(self, command):
                self.child.sendline(command)
                self.child.expect_exact(self.prompt)
                response = self.child.before
                return response

        def close(self):
                self.child.close()

@app.route('/')
def index():
        rl = rlogin("myserver", "root@myserver:~# ")
        output = rl.command("pwd")
        rl.close()
        # The output of the next line is just as I expect:
        print output
        # This is probably where it fails:
        return output

if __name__ == '__main__':
    app.run(host='0.0.0.0', debug=True)

删除对 rlogin()、rl.command() 和 rl.close() 的调用，并返回一个简单的字符串，例如“A”解决了错误。我已经在这个问题上停留了一段时间，非常感谢任何帮助。谢谢!

Answer 1

您必须确保返回类型是 str、unicode、响应类或 WSGI 函数。看起来您的输出不属于任何可接受的 Flask 路由返回类型。

@app.route('/')
def index():
        rl = rlogin("myserver", "...")
        output = rl.command("pwd")
        rl.close()
        print type(output)
        # Convert output into something that flask can understand
        value = str(output)
        resp = make_response(value)
        resp.headers['Content-Type'] = 'text/plain'
        return resp

您可以在 http://flask.pocoo.org/docs/0.10/api/#flask.Flask.make_response 阅读更多相关信息