python - How to find the position of specific element in a list? -

i have list this:

 website =    ['http://freshtutorial.com/install-xamp-ubuntu/', 'http://linuxg.net/how-to-install-xampp-on-ubuntu-13-04-12-10-12-04/', 'http://ubuntuforums.org/showthread.php?t=2149654', 'http://andyhat.co.uk/2012/07/installing-xampp-32bit-ubuntu-11-10-12-04/', 'http://askubuntu.com/questions/303068/error-with-tar-command-cannot-install-xampp-1-8-1-on-ubuntu-13-04', 'http://askubuntu.com/questions/73541/how-to-install-xampp'] 

i want search if following list contain url or not.

url in format : url = 'http://freshtutorial.com'

the website 1st element of list. thus, want print 1 not 0.

i want in loop if there's no website url, go again , dynamically generate list , again search website.

i have done upto now:

for in website:     if url in website:         print "true" 

i can't seem print position , wrap in loop. also, better use regex or if in that syntax. thanks

here complete program:

def search(li,ur):     u in li:         if u.startswith(ur):             return li.index(u)+1             return 0  def main():     website = ['http://freshtutorial.com/install-xamp-ubuntu/', 'http://linuxg.net/how-to-install-xampp-on-ubuntu-13-04-12-10-12-04/', 'http://ubuntuforums.org/showthread.php?t=2149654', 'http://andyhat.co.uk/2012/07/installing-xampp-32bit-ubuntu-11-10-12-04/', 'http://askubuntu.com/questions/303068/error-with-tar-command-cannot-install-xampp-1-8-1-on-ubuntu-13-04', 'http://askubuntu.com/questions/73541/how-to-install-xampp']     url = 'http://freshtutorial.com'     print search(website,url)  if __name__ == '__main__':     main()