为什么 listfunction 不会遍历 url？

Question

I am trying to write a program that opens a url, finds a name in a certain line, and saves it. Then it should find the url in the same line as the name, open it, and find the name + url in the same line # as the previous page. It should do this 4 times.

I can't get it to iterate through the new url parameter. It keeps returning the same name and url. What is going wrong here? Thanks.

from bs4 import BeautifulSoup
from urllib.request import urlopen
import re
import ssl
linklist = list()
namelist = list()
linelist = list()
count = 0
listposition = int(input("Please enter list position: "))
goodnamelist = list(["Fikret"])
nexturl = "http://py4e-data.dr-chuck.net/known_by_Fikret.html"
def listfunction(url):
    ctx = ssl.create_default_context()
    #Allows reading of HTTPS pages
    ctx.check_hostname = False
    ctx.verify_mode = ssl.CERT_NONE
    html = urlopen(url, context=ctx).read()
    soup = BeautifulSoup(html, "html.parser")
    linelist = soup('a')
    for line in linelist:
        #Creates list of lines in webpage:
        linklist.append(re.findall("(http://.+)\"", str(line)))
        #Creates list of names in line:
        namelist.append(re.findall(">(.+)</a>", str(line)))
    #Creates list of names in the designated user-input position:
    goodnamelist.append(namelist[listposition][0])
    nexturl = linklist[listposition][0]
    return nexturl
while (count < 4):
    nexturl = listfunction(nexturl)
    print(listfunction(nexturl))
    count += 1
    print(nexturl)
    continue
print(linelist)
print(linklist)
print(namelist)
print(nexturl)
print(goodnamelist)
print(listfunction(nexturl))

Answer 1

You do not actually set nexturl in listfunction() . Therefore the method just returns the same initial global variable every time.