Question

我只希望跨越范围之外的文本，而不是跨越范围内的任何内容。我目前的代码给了我所有这些：

birthday = bsObj.find( "div", {"class":"age"} )
# <div class="age"><span class="category">Age:</span> 23 (10/21/1992)</div>
birthday.get_text()
birthplace = bsObj.find( "div", {"class":"hometown"} )
# <div class="hometown"><span class="category">Birthplace:</span> Barranquilla, Colombia</div>
birthplace.get_text()

结果：

"Age: 24 (04/21/1991)","Birthplace: Barranquilla, Colombia"

期望的结果：

"24 (04/21/1991)","Barranquilla, Colombia"

Answer 1

在get_text（）

之前清除范围

from bs4 import BeautifulSoup

html_doc ='<html><body><div class="age"><span class="category">Age:</span> 23 (10/21/1992)</div><div class="hometown"><span class="category">Birthplace:</span> Barranquilla, Colombia</div></body></html>'

bsObj = BeautifulSoup(html_doc, 'html.parser')

# <div class="age"><span class="category">Age:</span> 23 (10/21/1992)</div>
birthday = bsObj.find( "div", {"class":"age"} )
birthday.span.clear()
print(birthday.get_text()) # 23 (10/21/1992)

# <div class="hometown"><span class="category">Birthplace:</span> Barranquilla, Colombia</div>
birthplace = bsObj.find( "div", {"class":"hometown"} )
birthplace.span.clear()
print(birthplace.get_text()) # Barranquilla, Colombia

Answer 2

使用span

clear()

使用strip()

from bs4 import BeautifulSoup

soup = BeautifulSoup('<div class="age"><span class="category">Age:</span> 23 (10/21/1992)</div>', 'html.parser')
soup.span.clear()
print(soup.get_text().strip())

输出：

23 (10/21/1992)

如何使用BeautifulSoup和Python获取文本后面的文本？

2 个答案: