Posted in Python onDecember 18, 2018
应用场景:
状态不是200的URL重试多次
代码比较简单还有部分注释
python2.7实现:
# -*-coding:utf-8-*- """ ayou """ import requests def url_retry(url,num_retries=3): print("access!") try: request = requests.get(url,timeout=60) #raise_for_status(),如果不是200会抛出HTTPError错误 request.raise_for_status() html = request.content except requests.HTTPError as e: html=None if num_retries>0: #如果不是200就重试,每次递减重试次数 return url_retry(url,num_retries-1) #如果url不存在会抛出ConnectionError错误,这个情况不做重试 except requests.exceptions.ConnectionError as e: return return html url_retry("http://httpbin.org/status/404")
python3.5实现:
# -*-coding:utf-8-*- """ ayou """ import asyncio import aiohttp async def print_page(url,num_retries=3): async with aiohttp.ClientSession() as session: try: async with session.get(url,timeout=60) as response: print("access!") #raise_for_status(),如果不是200会抛出HttpProcessingError错误 response.raise_for_status() body = await response.text() except aiohttp.errors.HttpProcessingError as e: body = None if num_retries > 0: #如果不是200就重试,每次递减重试次数 return await print_page(url, num_retries - 1) #不存在URL会抛出ClientResponseError错误 except aiohttp.errors.ClientResponseError as e: return e session.close() print(body) return body def main(): #这是一个不存在URL # url = 'http://httpbin.org/status/404111' #这是一个404的URL url = 'http://httpbin.org/status/404' loop = asyncio.get_event_loop() loop.run_until_complete(print_page(url)) loop.close() if __name__ == '__main__': main()
爬虫URL重试机制封装成修饰器(python2.7以及python3.5以上)
python2.7版本:
# -*-coding:utf-8-*- """ ayou """ import requests #定义一个重试修饰器,默认重试一次 def retry(num_retries=1): #用来接收函数 def wrapper(func): #用来接收函数的参数 def wrapper(*args,**kwargs): #为了方便看抛出什么错误定义一个错误变量 last_exception =None #循环执行包装的函数 for _ in range(num_retries): try: #如果没有错误就返回包装的函数,这样跳出循环 return func(*args, **kwargs) except Exception as e: #捕捉到错误不要return,不然就不会循环了 last_exception = e #如果要看抛出错误就可以抛出 # raise last_exception return wrapper return wrapper if __name__=="__main__": @retry(5) def url_retry(url): request = requests.get(url, timeout=60) print("access!") request.raise_for_status() html = request.content print(html) return html url_retry("http://httpbin.org/status/404") # url_retry("http://httpbin.org/status/404111") # url_retry("http://www.baidu.com")
python3.5以上版本:
# -*-coding:utf-8-*- """ ayou """ import aiohttp,asyncio #定义一个重试修饰器,默认重试一次 def retry(num_retries=1): #用来接收函数 def wrapper(func): #用来接收函数的参数 def wrapper(*args,**kwargs): #为了方便看抛出什么错误定义一个错误变量 last_exception =None #循环执行包装的函数 for _ in range(num_retries): try: #如果没有错误就返回包装的函数,这样跳出循环 return func(*args, **kwargs) except Exception as e: #捕捉到错误不要return,不然就不会循环了 last_exception = e #如果要看抛出错误就可以抛出 # raise last_exception return wrapper return wrapper async def print_page(url): async with aiohttp.ClientSession() as session: async with session.get(url,timeout=60) as response: print("access!") #raise_for_status(),如果不是200会抛出HttpProcessingError错误 response.raise_for_status() body = await response.text() session.close() print(body) return body @retry(5) def loop_get(): # url = "http://www.baidu.com" # url = 'http://httpbin.org/status/404111' url = 'http://httpbin.org/status/404' loop = asyncio.get_event_loop() loop.run_until_complete(print_page(url)) loop.close() if __name__ == '__main__': loop_get()
以上这篇python爬虫URL重试机制的实现方法(python2.7以及python3.5)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持三水点靠木。
python爬虫URL重试机制的实现方法(python2.7以及python3.5)
- Author -
ZJL-阿友声明:登载此文出于传递更多信息之目的,并不意味着赞同其观点或证实其描述。
Reply on: @reply_date@
@reply_contents@