用python脚本24小时刷浏览器的访问量方法

如下所示：

# -*- coding=utf-8 -*-
import urllib2
import socket
import time

urls = raw_input("Please enter a web address: \n> ")
print "\nAccess web page start..."
brushNum = 3600
for i in range(brushNum):
 url = urls
 socket.setdefaulttimeout
 req_header = {'User-Agent':'Mozilla/5.0 (Windows NT 6.1) AppleWebKit/537.11 (KHTML, like Gecko) Chrome/23.0.1271.64 Safari/537.11',
 'Accept':'text/html;q=0.9,*/*;q=0.8',
 'Accept-Charset':'ISO-8859-1,utf-8;q=0.7,*;q=0.3',
 'Accept-Encoding':'gzip',
 'Connection':'close',
 'Referer':None
 }
 req_timeout = 60
 req = urllib2.Request(url,None,req_header)
 resp = urllib2.urlopen(req,None,req_timeout)
 html = resp.read()
 print "Success!\t",i + 1
 print "Rest 10 seconds to continue...\n"
 time.sleep(10)

这次重构了代码，运行脚本会提示输入需要刷新的网页，刷新次数默认为3600，超时时间改为120秒，这样效果好一点。

来源：https://blog.csdn.net/happy_jijiawei/article/details/53606468