利用Python检测URL状态

最近小编接到这样的需求，Python检测URL状态，并追加保存200的URL。接下来通过实例代码给大家分析讲解，需要的朋友跟随小编一起看看吧

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编程之家网站的支持！

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