dokuwiki

import requests
import sys
url = sys.argv[1]
for i in open(sys.argv[2],"r"):
  #print i.rstrip() # removes newline character 
  r = requests.get("http://{}/{}".format(url,i.rstrip()))
  if r.status_code == 200:
      print "http://{}/{}".format(url,i.rstrip())

##to run
python wfuzz_1.py 127.0.0.1 common.txt