Python BeautifulSoup:get_text（）从bs4标记返回空字符串

我正试图从中提取信息

首先，我解析页面：

import requests
from bs4 import BeautifulSoup
page = requests.get("https://www.theguardian.com/politics/2019/oct/20/boris-johnson-could-be-held-in-contempt-of-court-over-brexit-letter")
soup = BeautifulSoup(page.content, 'html.parser')

然后我从标题开始：

title = soup.find('meta', property="og:title")

如果我把它打印出来，我会得到：

<meta content="Boris Johnson could be held in contempt of court over Brexit letter" property="og:title"/>

但是，当我运行

title.get_text（）

时，结果是一个空字符串：

'

---

我的错误在哪里？

那是因为标签实际上没有定义任何文本。在本例中，您要查找的“文本”包含在带有属性

content

的

标记中。因此，您需要提取

内容的值

：

import requests
from bs4 import BeautifulSoup
page = requests.get("https://www.theguardian.com/politics/2019/oct/20/boris-johnson-could-be-held-in-contempt-of-court-over-brexit-letter")
soup = BeautifulSoup(page.content, 'html.parser')

title = soup.find('meta', property="og:title")['content']

输出：

print (title)
Boris Johnson could be held in contempt of court over Brexit letter

print (title.attrs)
{'property': 'og:title', 'content': 'Boris Johnson could be held in contempt of court over Brexit letter'}

您可以使用

.attrs

获取所有属性和值。这将返回给定标记中属性和值的字典（键：值对）：

title = soup.find('meta', property="og:title")

print (title.attrs)

输出：

print (title)
Boris Johnson could be held in contempt of court over Brexit letter

print (title.attrs)
{'property': 'og:title', 'content': 'Boris Johnson could be held in contempt of court over Brexit letter'}

---

这是因为实际上没有任何由标记定义的文本。在本例中，您要查找的“文本”包含在带有属性

content

的

标记中。因此，您需要提取

内容的值

：

import requests
from bs4 import BeautifulSoup
page = requests.get("https://www.theguardian.com/politics/2019/oct/20/boris-johnson-could-be-held-in-contempt-of-court-over-brexit-letter")
soup = BeautifulSoup(page.content, 'html.parser')

title = soup.find('meta', property="og:title")['content']

输出：

print (title)
Boris Johnson could be held in contempt of court over Brexit letter

print (title.attrs)
{'property': 'og:title', 'content': 'Boris Johnson could be held in contempt of court over Brexit letter'}

您可以使用

.attrs

获取所有属性和值。这将返回给定标记中属性和值的字典（键：值对）：

title = soup.find('meta', property="og:title")

print (title.attrs)

输出：

print (title)
Boris Johnson could be held in contempt of court over Brexit letter

print (title.attrs)
{'property': 'og:title', 'content': 'Boris Johnson could be held in contempt of court over Brexit letter'}

---

谢谢，效果很好。有可能得到整个元素列表吗？不仅仅是

['content']

，而是想看到所有这些内容吗？谢谢，效果很好。有可能得到整个元素列表吗？不仅仅是

['content']

，而是要看到它们？