为什么Python的HTTP服务器引起的Andr​​oid VolleyError [英] Why Python http server cause android VolleyError

问题描述

我有以下简单的Python脚本3.4：

I have following simple Python 3.4 script:

LISTEN_PORT = 7000

class MyHandler(http.server.SimpleHTTPRequestHandler):

    def do_GET(self):
        print("DO GET")
        self.send_response(200, "OK")



def run():

    handler = MyHandler
    print("Server Started")
    httpd = socketserver.TCPServer(("0.0.0.0", LISTEN_PORT), handler)

    try:
        print("serving at port", LISTEN_PORT)
        httpd.serve_forever()
    except KeyboardInterrupt:
        httpd.socket.close()


run()

当我在Android拨打：

When I call from android:

final StringRequest request = new StringRequest(Request.Method.GET, context.getString(R.string.domain)+"/", new Response.Listener<String>() {
        @Override
        public void onResponse(String s) {

        }
    }, new Response.ErrorListener() {
        @Override
        public void onErrorResponse(VolleyError volleyError) {

            Log.e(TAG, "Can not send token" + volleyError);
            Toast.makeText(context, "Can not send", Toast.LENGTH_LONG).show();
        }
    });

    Volley.newRequestQueue(context).add(request);

Android的：
我看吐司无法发送和logcat的控制台上我有以下信息：

Android: I see Toast "Can not send" and on logcat console I have following information:

06-27 18:05:16.332  28946-28946/com.example.code E/NETWORK_CONNECTOR﹕ Can not send token Code:com.android.volley.NoConnectionError: java.io.EOFException

的Python：

192.168.12.246 - - [27/Jun/2014 18:15:55] "GET / HTTP/1.1" 200 -
DO GET

什么是错的Python脚本？

What is wrong with Python script?

推荐答案

什么是错的python脚本？

What is wrong with python script ?

它会以某种方式与python3.0-3.2工作，但不会与python3.3工作+

it would somehow work with python3.0-3.2 but won't work with python3.3+

从<一个href=\"https://docs.python.org/3/library/http.server.html#http.server.BaseHTT$p$pquestHandler.send_response\"相对=nofollow> send_response文档

在变更3.3版：头被存储到内部缓冲区和end_headers（）需要显式调用

Changed in version 3.3: Headers are stored to an internal buffer and end_headers() needs to be called explicitly.

在当前设置HTTP应答的头部部分是根本无法发送。

in the current setup the header part of http answer is simply not sent.

这样做的python3.3 +正确的方法是

so the right way to do it in python3.3+ is

def do_GET(self):

    print("DO GET")
    self.send_response(200, "OK")
    self.end_headers()

    self.wfile.write(b"Response body\n") #optional

但它可以在你的脚本来改善一些东西。

but there is several thing which could be improved in your script

• 使用http.server.BaseHTT prequestHandler代替http.server.SimpleHTT prequestHandler因为SimpleHTT prequestHandler添加一些魔法BaseHTT prequestHandler处理GET和 HEAD 请求。如你覆盖do_GET，更换SimpleHTT prequestHandler的行为，但由于您没有为do_HEAD做到这一点，你的服务器会回答HEAD请求，因为这将是SimpleHTT prequestHandler这是不是一个好主意


• httpd.server_close（），而不是httpd.socket.close（），即使不改变的东西太多，因为它基本上是在当前的Python一个空操作

• use http.server.BaseHTTPRequestHandler instead of http.server.SimpleHTTPRequestHandler because SimpleHTTPRequestHandler add some magic to BaseHTTPRequestHandler to process GET and HEAD request. as you override do_GET, you replace the behavior of SimpleHTTPRequestHandler but as you don't do it for do_HEAD, your server would answer to HEAD requests as it would be SimpleHTTPRequestHandler which is not a good idea
• httpd.server_close() instead httpd.socket.close() even if that don't change thing too much as it is basically a noop in the current python

在脚本的顶部添加进口瞧一个完整的工作脚本

adding an import at top of the script and voilà a full working script

import http.server
LISTEN_PORT = 7000

class MyHandler(http.server.BaseHTTPRequestHandler):

    def do_GET(self):

        print("DO GET")
        self.send_response(200, "OK")
        self.end_headers()

        self.wfile.write(b"Response body\n") #optional


def run():

    handler = MyHandler
    print("Server Started")
    httpd = http.server.HTTPServer(("0.0.0.0", LISTEN_PORT), handler)

    try:
        print("serving at port", LISTEN_PORT)
        httpd.serve_forever()
    except KeyboardInterrupt:
        httpd.server_close()


if __name__ == "__main__":
    run()

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