http - Is it possible to run python SimpleHTTPServer on localhost only?

94
votes

I have a vpn connection and when I'm running python -m SimpleHTTPServer, it serves on 0.0.0.0:8000, which means it can be accessed via localhost and via my real ip. I don't want robots to scan me and interested that the server will be accessed only via localhost.

Is it possible? 

python -m SimpleHTTPServer 127.0.0.1:8000  # doesn't work.

Any other simple http server which can be executed instantly using the command line is also welcome.

3
pythonhttpcommand-linepython-2.xsimplehttpserver
You could simply block outside connections on that port from your firewall/router. – Burhan Khalid
While a good question for python2, it may be noted here that in python3 the replacement http.server allows binding right away, e.g python3 -m http.server --bind 127.0.0.1 8000 would suffice – humanityANDpeace
Sidenote: SimpleHTTPServer is single-threading and blocking, meaning you won't be able to do another request until the previous request is over. And it has no range support e.g. for streaming/seeking a media file from a specific position. A better alternative is twisted (pip install twisted) which you can run with twistd -n web --path /. It can also do anonymous FTP with twistd -n ftp -p 2121 -r /. More http server one-liners: gist.github.com/willurd/5720255. – ccpizza

3 Answers

56
votes

If you read the source you will see that only the port can be overridden on the command line. If you want to change the host it is served on, you will need to implement the test() method of the SimpleHTTPServer and BaseHTTPServer yourself. But that should be really easy.

Here is how you can do it, pretty easily:

import sys
from SimpleHTTPServer import SimpleHTTPRequestHandler
import BaseHTTPServer


def test(HandlerClass=SimpleHTTPRequestHandler,
         ServerClass=BaseHTTPServer.HTTPServer):

    protocol = "HTTP/1.0"
    host = ''
    port = 8000
    if len(sys.argv) > 1:
        arg = sys.argv[1]
        if ':' in arg:
            host, port = arg.split(':')
            port = int(port)
        else:
            try:
                port = int(sys.argv[1])
            except:
                host = sys.argv[1]

    server_address = (host, port)

    HandlerClass.protocol_version = protocol
    httpd = ServerClass(server_address, HandlerClass)

    sa = httpd.socket.getsockname()
    print "Serving HTTP on", sa[0], "port", sa[1], "..."
    httpd.serve_forever()


if __name__ == "__main__":
    test()

And to use it: 

> python server.py 127.0.0.1     
Serving HTTP on 127.0.0.1 port 8000 ...

> python server.py 127.0.0.1:9000
Serving HTTP on 127.0.0.1 port 9000 ...

> python server.py 8080          
Serving HTTP on 0.0.0.0 port 8080 ...
112
votes

In Python versions 3.4 and higher, the http.server module accepts a bind parameter.

According to the docs:

python -m http.server 8000

By default, server binds itself to all interfaces. The option -b/--bind specifies a specific address to which it should bind. For example, the following command causes the server to bind to localhost only:

python -m http.server 8000 --bind 127.0.0.1

New in version 3.4: --bind argument was introduced.

77
votes

As @sberry explained, simply doing it by using the nice python -m ... method won't be possible, because the IP address is hardcoded in the implementation of the BaseHttpServer.test function.

A way of doing it from the command line without writing code to a file first would be

python -c 'import BaseHTTPServer as bhs, SimpleHTTPServer as shs; bhs.HTTPServer(("127.0.0.1", 8888), shs.SimpleHTTPRequestHandler).serve_forever()'

If that still counts as a one liner depends on your terminal width ;-) It's certainly not very easy to remember.