Python: run SimpleHTTPServer and make request to it in a script

Question

I would like to write Python script that would:

Start SimpleHTTPServer on some port
Make a request to the server and send some data via POST
Run script via shell in interactive mode and interact with it

Right now the code looks like this:

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()

The problem is that once I run my script

python -i foo.py

It prints serving at port 8000 and then freezes. I bet this is something trivial for Python gurus here, but help would be appreciated.

Utah_Dave · Accepted Answer · 2017-03-02 19:35:55Z

4

Run the server as a different process, that will allow you to run the rest of your script.

I would also rather use requests than 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()

edited Mar 2, 2017 at 19:35

Utah_Dave

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answered Jan 21, 2016 at 19:22

Dušan Maďar

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4 Comments

0leg Over a year ago

Thank you, this is what I was looking for, but couldn't get due to python-experience limit.

0leg Over a year ago

One thing I've noticed, is an error while trying to execute this: socket.error: [Errno 10048] Only one usage of each socket address (protocol/network address/port) is normally permit ted

NarūnasK Over a year ago

This is most likely due to the fact, that you are not shutting down your http server, at least in your example. Please see my answer for more detail.

Dušan Maďar Over a year ago

Do you get this error when you try to restart the script? So yes, as @NarūnasK suggests, it may be the case that the server is still running and therefore the port 8000 is taken. The answer was edited.

NarūnasK · Accepted Answer · 2016-01-21 19:32:37Z

Alternatively run it in the separate thread:

import SimpleHTTPServer
import SocketServer
import urllib2
from threading import Thread
from datetime import time

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

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

simple_sever_T = Thread(target=simple_sever, name='simple_sever')
simple_sever_T.daemon = True
simple_sever_T.start()

while not simple_sever_T.is_alive():
    time.sleep(1)

# Getting test file
req = urllib2.Request('http://localhost:%s/test_file' % PORT)
response = urllib2.urlopen(req)
print response.read()
httpd.shutdown()

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