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[英]how to get the content of a title using BeautifulSoup4 and requests
[英]How do I get the “title” of an anchor tag using BeautifulSoup4?
我不知道如何获得标题上的标题。 这是我的代码:
from flask import Flask
import requests
from bs4 import BeautifulSoup
laptops = 'http://webscraper.io/test-sites/e-commerce/allinone/computers/laptops'
def scrape():
page = requests.get('http://webscraper.io/test-sites/e-commerce/allinone/computers/laptops')
soup = BeautifulSoup(page.content, "lxml")
links = soup("a", {"class":"title"})
for link in links:
print(link.prettify())
scrape()
结果示例:
<a class="title" href="/test-sites/e-commerce/allinone/product/251" title="Asus VivoBook X441NA-GA190">
Asus VivoBook X4...
</a>
<a class="title" href="/test-sites/e-commerce/allinone/product/252" title="Prestigio SmartBook 133S Dark Grey">
Prestigio SmartB...
</a>
<a class="title" href="/test-sites/e-commerce/allinone/product/253" title="Prestigio SmartBook 133S Gold">
Prestigio SmartB...
</a>
我如何获得“头衔”?
可以通过订阅或元素上的.attrs
字典访问诸如title
类的属性:
for link in links:
print(link['title'])
请参阅Attributes上的BeautifulSoup文档 。
对于给定的URL,将产生:
Asus VivoBook X441NA-GA190
Prestigio SmartBook 133S Dark Grey
Prestigio SmartBook 133S Gold
Aspire E1-510
Lenovo V110-15IAP
Lenovo V110-15IAP
Hewlett Packard 250 G6 Dark Ash Silver
# ... etc
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