发布于2019-08-05 11:54 阅读(1225) 评论(0) 点赞(4) 收藏(2)
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'
作者:mimi
链接:https://www.pythonheidong.com/blog/article/4449/a70087dd958ee476fedd/
来源:python黑洞网
任何形式的转载都请注明出处,如有侵权 一经发现 必将追究其法律责任
昵称:
评论内容:(最多支持255个字符)
---无人问津也好,技不如人也罢,你都要试着安静下来,去做自己该做的事,而不是让内心的烦躁、焦虑,坏掉你本来就不多的热情和定力
Copyright © 2018-2021 python黑洞网 All Rights Reserved 版权所有,并保留所有权利。 京ICP备18063182号-1
投诉与举报,广告合作请联系vgs_info@163.com或QQ3083709327
免责声明:网站文章均由用户上传,仅供读者学习交流使用,禁止用做商业用途。若文章涉及色情,反动,侵权等违法信息,请向我们举报,一经核实我们会立即删除!