python之requests库请求错误重试机制写法最优解用tenacity库

云霄IT

已于 2024-09-20 17:22:51 修改

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文章标签： python 开发语言

于 2024-09-20 17:22:02 首次发布

本文链接：https://blog.youkuaiyun.com/weixin_51111267/article/details/142390830

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代码参考如下：

from tenacity import retry, stop_after_attempt,wait_exponential
import requests
import re


@retry(stop=stop_after_attempt(3),wait=wait_exponential(multiplier=1, min=1, max=5), retry_error_callback=lambda x:None)
def requests_get(url, params=None, **kwargs):
    print('retry...')
    response = requests.get(url, params=params, **kwargs)
    response.raise_for_status()   # If response status code is 4xx or 5xx, raise HTTPError
    return response

def send():
    # url = 'https://www.runoob.com/'
    url = 'https://www.runoob.com/123456'
    headers = {
        "user-agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/111.0.0.0 Safari/537.36"
    }
    resp = requests_get(url,headers=headers)
    if not resp:
        print("Retry to max!")
        return
    title = re.findall("<title>(.*?)</title>",resp.text)[0]
    print(title)


if __name__ == '__main__':
    send()