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-*
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脚本之家。
python 爬虫 URL 重试机制