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.
In Python versions 3.4 and higher, the http.server module accepts a bind parameter.
According to the docs:
python -m http.server 8000By 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 127.0.0.1New in version 3.4: --bind argument was introduced.
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.
H. Thanks! -- gist.github.com/cmawhorter/f2a09bcf63c68b0cff10
python -c "import http.server as hs; hs.HTTPServer(('127.0.0.1', 8888), hs.SimpleHTTPRequestHandler).serve_forever()" Note the change in quotation and the fact that Base and Simple HTTP Server are now in http.server.
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 ...
http.serverallows binding right away, e.gpython3 -m http.server --bind 127.0.0.1 8000would sufficeSimpleHTTPServeris 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 istwisted(pip install twisted) which you can run withtwistd -n web --path /. It can also do anonymous FTP withtwistd -n ftp -p 2121 -r /. More http server one-liners: gist.github.com/willurd/5720255.