Python抓取(Beautiful Soup)从此HTML获取数据

时间:2015-01-21 19:21:16

标签: python html web-scraping beautifulsoup html-parsing

 <ul>
  <li>
    <div class="c_logo_box">
     <a href="money-transfer-companies/ria-money-transfer/"><img src="http://www.compareremit.com/uploads/ria-logo11.png" style="height:57px;width:147px;" alt="RIA Money Transfer"></a>
     <span class="rs"> <span class="txt13">&#8377;</span> 61.24</span>
       </div>
  </li>
 ...

我想从&#39; alt = Ria Money Transfer&#39;中删除名称。来自61.24的评分

到目前为止,我有这个Python代码:

#!/usr/bin/python

import requests
import re
from bs4 import BeautifulSoup

r = requests.get('http://www.compareremit.com')
data = r.text

soup = BeautifulSoup(data)
for rate in soup.find_all('li', re.compile('money')):
print rate.text

它什么都没给我。有人能告诉我我错过了什么吗?另外,我在查看我在for循环搜索中支持哪个元素时遇到问题,你能否澄清一下在这种情况下如何知道在for循环中指定什么条件?

2 个答案:

答案 0 :(得分:1)

有多种方法可以达到元素。一种选择是依赖a标记,href包含ria-money-transfer部分,然后获取包含费率的following span element

import re

from bs4 import BeautifulSoup
import requests

response = requests.get('http://www.compareremit.com')
soup = BeautifulSoup(response.content)

link = soup.find('div', class_='c_logo_box').find('a', href=re.compile(r'ria-money-transfer'))
print(link.img.get('alt'))

rate = link.find_next_sibling('span').text.split(' ')[-1]
print(rate)

打印:

RIA Money Transfer
61.24

答案 1 :(得分:1)

您的代码在逻辑上不正确。您可以通过多种方式执行此操作,请尝试此代码

#!的/ usr / bin中/ Python的

import requests
import re
from bs4 import BeautifulSoup

r = requests.get('http://www.compareremit.com')
data = r.text

soup = BeautifulSoup(data)
for rate in soup.find_all('div',{"class":"c_logo_box"}):
    print rate.a.img['alt'] 
    print rate.span.text