使用BeautifulSoup从html中提取除script标签内容之外的文本

Question

我有这样的html

<span class="age">

    Ages 15

    <span class="loc" id="loc_loads1">

     </span>

     <script>

        getCurrentLocationVal("loc_loads1",29.45218856,59.38139268,1);

     </script>

</span>

我正在尝试Age 15使用BeautifulSoup

所以我写了python代码如下

代码：

from bs4 import BeautifulSoup as bs

import urllib3

URL = 'html file'

http = urllib3.PoolManager()

page = http.request('GET', URL)

soup = bs(page.data, 'html.parser')

age = soup.find("span", {"class": "age"})

print(age.text)

输出：

Age 15 getCurrentLocationVal("loc_loads1",29.45218856,59.38139268,1);

我只想要标签Age 15内的功能script。有没有办法只获取 text: Age 15？或者有什么方法可以排除script标签的内容？

PS：script标签太多，URL不同。我不喜欢从输出中替换文本。

Answer 1

用 .find(text=True)前任：from bs4 import BeautifulSouphtml = """<span class="age">&nbsp; &nbsp; Ages 15&nbsp; &nbsp; <span class="loc" id="loc_loads1">&nbsp; &nbsp; &nbsp;</span>&nbsp; &nbsp; &nbsp;<script>&nbsp; &nbsp; &nbsp; &nbsp; getCurrentLocationVal("loc_loads1",29.45218856,59.38139268,1);&nbsp; &nbsp; &nbsp;</script></span>"""soup = BeautifulSoup(html, "html.parser")print(soup.find("span", {"class": "age"}).find(text=True).strip())输出：Ages 15

Answer 2

迟到的答案，但为了将来参考，您还可以使用分解（）从 中删除所有script元素html，即：soup = BeautifulSoup(html, "html.parser")&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;# remove script and style elements&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;for script in soup(["script", "style"]):&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp; script.decompose()&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;print(soup.find("span", {"class": "age"}).text.strip())&nbsp; &nbsp;&nbsp;# Ages 15