如何从BeautifulSoup中的span标签获取文本

Question

我的链接看起来像这样

<div class="systemRequirementsMainBox">
<div class="systemRequirementsRamContent">
<span title="000 Plus Minimum RAM Requirement">1 GB</span> </div>

我想1 GB从那里开始.我试过了

tt  = [a['title'] for a in soup.select(".systemRequirementsRamContent span")]
for ram in tt:
    if "RAM" in ram.split():
        print (soup.string)

它输出None.

我试过a['text']但它给了我KeyError.我怎么能解决这个问题,我的错误是什么？

Answer 1

您可以使用css选择器,使用标题文本拉出所需的范围:

soup = BeautifulSoup("""<div class="systemRequirementsMainBox">
<div class="systemRequirementsRamContent">
<span title="000 Plus Minimum RAM Requirement">1 GB</span> </div>""", "xml")

print(soup.select_one("span[title*=RAM]").text)

找到包含RAM的title属性的span,它相当于在python中说,.if "RAM" in span["title"]

或者使用find with re.compile

import re
print(soup.find("span", title=re.compile("RAM")).text)

获取所有数据:

from bs4 import BeautifulSoup 
r  = requests.get("http://www.game-debate.com/games/index.php?g_id=21580&game=000%20Plus").content

soup = BeautifulSoup(r,"lxml")
cont = soup.select_one("div.systemRequirementsRamContent")
ram = cont.select_one("span")
print(ram["title"], ram.text)
for span in soup.select("div.systemRequirementsSmallerBox.sysReqGameSmallBox span"):
        print(span["title"],span.text)

哪个会给你:

000 Plus Minimum RAM Requirement 1 GB
000 Plus Minimum Operating System Requirement Win Xp 32
000 Plus Minimum Direct X Requirement DX 9
000 Plus Minimum Hard Disk Drive Space Requirement 500 MB
000 Plus GD Adjusted Operating System Requirement Win Xp 32
000 Plus GD Adjusted Direct X Requirement DX 9
000 Plus GD Adjusted Hard Disk Drive Space Requirement 500 MB
000 Plus Recommended Operating System Requirement Win Xp 32
000 Plus Recommended Hard Disk Drive Space Requirement 500 MB

Answer 2

find_all()我尝试使用 (BeautifulSoup) 中的函数提取 HTML 文档内所有 span 标记内的文本bs4：

from bs4 import BeautifulSoup
import requests
url="YOUR_URL_HERE"
response=requests.get(url)
soup=BeautifulSoup(response.content,html5lib)
spans=soup.find_all('span',"ENTER_Css_CLASS_HERE")
for span in spans:
  print(span.text)