How would you simplify this program? Python
Question: I wrote this program, which purpose is to visit the 18th link on the list of
links and then on the new page visit the 18th link again.

This program works as intended, but it's a little repetitive and inelegant.

I was wondering if you have any ideas on how to make it simpler, without using
any functions. If I wanted to repeat the process 10 or 100 times, this would
become very long.

Thanks for any suggestions!

    
    
    # Note - this code must run in Python 2.x and you must download
    # http://www.pythonlearn.com/code/BeautifulSoup.py
    # Into the same folder as this program
    
    import urllib
    from BeautifulSoup import *
    
    url = raw_input('Enter - ')
    if len(url) < 1 :
        url='http://python-data.dr-chuck.net/known_by_Oluwanifemi.html'
    html = urllib.urlopen(url).read()
    soup = BeautifulSoup(html)
    
    # Retrieve all of the anchor tags
    tags = soup('a')
    urllist = list()
    count = 0
    loopcount = 0
    for tag in tags:
        count = count + 1
        tg = tag.get('href', None)
        if count == 18:
            print count, tg
            urllist.append(tg)
    
    
    url2 = (urllist[0])
    html2 = urllib.urlopen(url2).read()
    soup2 = BeautifulSoup(html2)
    
    tags2 = soup2('a')
    count2 = 0
    for tag2 in tags2:
        count2 = count2 + 1
        tg2 = tag2.get('href', None)
        if count2 == 18:
            print count2, tg2
            urllist.append(tg2)
    


Answer: This is what you could do.

    
    
    import urllib
    from BeautifulSoup import *
    
    url_1 = input('') or 'http://python-data.dr-chuck.net/known_by_Oluwanifemi.html'
    
    html_1 = urllib.urlopen(url_1).read()
    soup_1 = BeautifulSoup(html_1)
    
    tags = soup('a')
    url_retr1 = tags[17].get('href', None)
    
    html_2 = urllib.urlopen(url_retr1).read()
    soup_2 = BeautifulSoup(html_2)
    
    tags_2 = soup_2('a')
    url_retr1 = tags_2[17].get('href', None)
    

