你如何简化这个程序？Python

Question

我写了这个程序，目的是访问链接列表上的第18个链接，然后在新的页面上再次访问第18个链接。你知道吗

这个程序按预期工作，但有点重复和不雅。你知道吗

我想知道你是否有什么想法，如何使它更简单，不使用任何函数。如果我想重复这个过程10或100次，这将变得很长。你知道吗

谢谢你的建议！你知道吗

# 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 1

这就是你能做的。你知道吗

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)