我是Python的新手,正在尝试从csv导入urls列表并检查每个urls的报头状态。到目前为止,我得到了以下输出urls列表的命令:
import csv
from requests_html import HTMLSession
from bs4 import BeautifulSoup
import requests
with open('list.csv', 'r') as f:
reader = csv.reader(f)
for row in reader:
url = " ".join(row)
print(url)我现在想在每个url旁边打印标题状态代码,所以我尝试了一下,但不起作用。如有任何帮助,我们将不胜感激:
with open('list.csv', 'r') as f:
reader = csv.reader(f)
for row in reader:
url = " ".join(row)
headers = requests.head(url.get('href'))
print(url, (headers.status_code))我得到的错误是:
headers = requests.head(url.get('href')) AttributeError:'str‘对象没有'get’属性
我需要输出如下内容:
https://www.domainone.com 200
https://www.domaintwo.com 200
https://www.domainthree.com 404发布于 2020-03-18 00:49:06
您已经有了url和响应。只需将它们打印出来:
resp = requests.head(url)
print('%s %d' % (url, resp.status_code))https://stackoverflow.com/questions/60726629
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