使用python 3.x 对pythonchallenge-----4的解答过程

pythonchallenge-4地址 : http://www.pythonchallenge.com/pc/def/linkedlist.php
图片如下:

题目解析:通过页面源代码解析,要打开链接http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=12345,然后获取nothing值,一直循环直到得出答案
解题过程:

from urllib import request,parse
import re

url = r'http://www.pythonchallenge.com/pc/def/linkedlist.php?'

headers = {
    'User-Agent': r'Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) '
                    r'Chrome/45.0.2454.85 Safari/537.36 115Browser/6.0.3',
    'Connection': 'keep-alive'
}
data = {}
data['nothing'] = "12345"

def getnothing(url,data,headers):
    data = parse.urlencode(data)
    url = url + data
    req = request.Request(url,headers=headers)
    page = request.urlopen(req).read().decode('utf-8')
    return page

for i in range(251):
    htmlstr = getnothing(url=url,data=data,headers=headers)
    pattern = re.compile('d+')
    nothingnum = re.findall(pattern,htmlstr)
    if nothingnum:
        data['nothing'] = nothingnum[0]
        if len(nothingnum)>1:
            data['nothing'] = nothingnum[1]
    else:
        data['nothing'] = str(int(data['nothing'])/2)
    print(str(i) + " ---- " + htmlstr)
    print(data['nothing'])

 答案:

250 ---- peak.html

 心得:在第85次和第140次的时候分别有个小坑

85 ---- Yes. Divide by two and keep going
...
...
...
140 ---- There maybe misleading numbers in the 
text. One example is 82683. Look only for the next nothing and the next nothing is 63579
 
 
原文地址:https://www.cnblogs.com/yinsjun/p/7465908.html