利用Python检测URL状态

yipeiwu_com6年前Python基础

需求:Python检测URL状态,并追加保存200的URL

代码一:

#! /usr/bin/env python
#coding=utf-8
import sys
import requests
def getHttpStatusCode(url):
  try:
    request = requests.get(url)
    httpStatusCode = request.status_code
    return httpStatusCode
  except requests.exceptions.HTTPError as e:
    return e
 
if __name__ == "__main__":
  with open('1.txt', 'r') as f:
    for line in f:
      try:
        status = getHttpStatusCode(line.strip('\n'))#换行符
        if status == 200:
          with open('200.txt','a') as f:
            f.write(line + '\n')
            print line
        else:
          print 'no 200 code'
      except Exception as e:
        print e

代码二:

#! /usr/bin/env python
# -*--coding:utf-8*-

import requests

def request_status(line):
  conn = requests.get(line)
  if conn.status_code == 200:
    with open('url_200.txt', 'a') as f:
      f.write(line + '\n')
    return line13   else:
    return None


if __name__ == '__main__':
  with open('/1.txt', 'rb') as f:
    for line in f:
      try:
        purge_url = request_status(line.strip('\n'))
      except Exception as e:
        pass

代码三:

#! /usr/bin/env python
#coding:utf-8
import os,urllib,linecache
import sys
result = list()

for x in linecache.updatecache(r'1.txt'):
  try:
    a = urllib.urlopen(x.replace('/n','')).getcode()
    #print x,a
  except Exception,e:
    print e
  if a == 200:
    #result.append(x)               #保存
    #result.sort()                    #排序结果
    #open('2.txt', 'w').write('%s' % '\n'.join(result)) #保存入结果文件
    with open ('200urllib.txt','a') as f: ## r只读,w可写,a追加
      f.write(x + '\n')
  else:
    print 'error'

总结

以上所述是小编给大家介绍的python 检测url 状态,希望对大家有所帮助,如果大家有任何疑问请给我留言,小编会及时回复大家的。在此也非常感谢大家对【听图阁-专注于Python设计】网站的支持!
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