Reputation: 33069
Consider the following example:
import string,cgi,time
from os import curdir, sep
from BaseHTTPServer import BaseHTTPRequestHandler, HTTPServer
class MyHandler(BaseHTTPRequestHandler):
def do_GET(self):
try:
if self.path.endswith(".html"):
f = open(curdir + sep + self.path) #self.path has /test.html
#note that this potentially makes every file on your computer readable by the internet
self.send_response(200)
self.send_header('Content-type', 'text/html')
self.end_headers()
self.wfile.write(f.read())
f.close()
return
except IOError:
self.send_error(404,'File Not Found: %s' % self.path)
def main():
try:
server = HTTPServer(('', 80), MyHandler)
print 'started httpserver...'
server.serve_forever()
except KeyboardInterrupt:
print '^C received, shutting down server'
server.socket.close()
if __name__ == '__main__':
main()
What if I want to server a ZIP file also... how would I do that? I don't think this line would work right?
self.wfile.write(f.read())
Upvotes: 12
Views: 29759
Reputation: 1793
If you want to share files in a folder of any type, then you can also try typing the command
python -m SimpleHTTPServer
This will start the server at port 8000 and you can browse the files (via directory listing)
Upvotes: 3
Reputation: 4128
Pass binary as a parameter to open(). This:
f = open(curdir + sep + self.path, 'rb')
Instead of this:
f = open(curdir + sep + self.path)
UNIX doesn't distinguish between binary and text, but windows does. But if the script executes on UNIX, the "b" will just be ignored so you're safe.
Upvotes: 10
Reputation: 193121
Your line would work just fine. The problem would be setting the Content-type
appropriately. You'd want to set it to application/zip
instead of text/html
.
Upvotes: 4