Python:运行SimpleHTTPServer并在脚本中向它发出请求

Question

我想编写Python脚本,它将:

1. 在某个端口上启动SimpleHTTPServer
2. 向服务器发出请求并通过POST发送一些数据
3. 在交互模式下通过shell运行脚本并与之交互

现在代码看起来像这样:

import SimpleHTTPServer
import SocketServer
import urllib
import urllib2

# Variables
URL = 'localhost:8000'
PORT = 8000

# Setup simple sever
Handler = SimpleHTTPServer.SimpleHTTPRequestHandler
httpd = SocketServer.TCPServer(("", PORT), Handler)
print "serving at port", PORT
httpd.serve_forever()

# Getting HTML from the target page
values = {
    'name': 'Thomas Anderson',
    'location': 'unknown'
}
data = urlilib.urlencode(values)
req = urllib2.Request(url, data)
response = urllib2.urlopen(req)
html = response.read()

问题是,一旦我运行我的脚本

python -i foo.py

它打印serving at port 8000然后冻结.我敢打赌这对于这里的Python专家来说是微不足道的,但是我们将不胜感激.

Answer 1

将服务器作为不同的进程运行，这将允许您运行脚本的其余部分。

我也宁愿使用requests而不是 urllib。

import SocketServer
import SimpleHTTPServer

import requests
import multiprocessing

# Variables
PORT = 8000
URL = 'localhost:{port}'.format(port=PORT)

# Setup simple sever
Handler = SimpleHTTPServer.SimpleHTTPRequestHandler
httpd = SocketServer.TCPServer(("", PORT), Handler)
print "Serving at port", PORT

# start the server as a separate process
server_process = multiprocessing.Process(target=httpd.serve_forever)
server_process.daemon = True
server_process.start()

# Getting HTML from the target page
values = {
    'name': 'Thomas Anderson',
    'location': 'unknown'
}

r = requests.post(URL, data=values)
r.text

# stop the server
server_process.terminate()