利用Python检测URL状态

Python  /  管理员 发布于 7年前   144

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-8import sysimport requestsdef 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 requestsdef 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 Noneif __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-8import os,urllib,linecacheimport sysresult = 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

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