如何使用美丽汤从xml标签中提取属性？

Question

我想在Django中使用美丽的汤来提取xml标签。这是我正在使用的标记示例：

<item> 
<title> 
Title goes here 
</title> 
<link> 
Link1 goes here 
</link> 
<description> 
Description goes here 
</description> 
<media:thumbnail url="Image URL goes here" height="222" width="300"/> 
<pubDate>Thu, 15 Sep 2016 13:24:48 EDT</pubDate> 
<guid isPermaLink="true"> 
Link2 goes here 
</guid> 
</item> 

我已经获得标题，链接和描述标记的字符串。但是我无法从media:thumbnail标签获取网址。

这是我得到的标签的其余部分的值的片段：

soup=BeautifulSoup(urlopen(xmllink),'xml') 
for items in soup.find_all('item'): 
    listTitle.append(items.title.get_text()) 
    listURL.append(items.link.get_text()) 
    listDescription.append(items.description.get_text()) 

帮助

Answer 1

的问题是，因为不是每一个项目都有一个媒体：缩略图所以你需要检查第一：

In [60]: import requests 

In [61]: from bs4 import BeautifulSoup 

In [62]: soup = BeautifulSoup(requests.get("https://rss.sciencedaily.com/computers_math/computer_programming.xml").content, "xml") 

In [63]: 

In [63]: for item in soup.find_all("item"): 
    ....:   thumb = item.find("thumbnail") 
    ....:   if thumb: 
    ....:     print(thumb["url"]) 
    ....:   
https://images.sciencedaily.com/2016/09/160915132448.jpg 
https://images.sciencedaily.com/2016/09/160915090018.jpg 
https://images.sciencedaily.com/2016/09/160914090327.jpg 
https://images.sciencedaily.com/2016/09/160913134149.jpg 
https://images.sciencedaily.com/2016/09/160909094844.jpg 
https://images.sciencedaily.com/2016/09/160907125004.jpg 
https://images.sciencedaily.com/2016/09/160906085157.jpg 
https://images.sciencedaily.com/2016/08/160831085055.jpg 
https://images.sciencedaily.com/2016/08/160822181811.jpg 
https://images.sciencedaily.com/2016/08/160815134941.jpg 
https://images.sciencedaily.com/2016/08/160815134817.jpg 
https://images.sciencedaily.com/2016/08/160809095640.jpg 
https://images.sciencedaily.com/2016/08/160803140137.jpg 
https://images.sciencedaily.com/2016/07/160722104135.jpg 
https://images.sciencedaily.com/2016/07/160721144139.jpg 
https://images.sciencedaily.com/2016/07/160721103855.jpg 
https://images.sciencedaily.com/2016/07/160720094641.jpg 
https://images.sciencedaily.com/2016/07/160718133206.jpg 
https://images.sciencedaily.com/2016/07/160713105850.jpg 
https://images.sciencedaily.com/2016/07/160711151055.jpg 
https://images.sciencedaily.com/2016/07/160707083258.jpg 
https://images.sciencedaily.com/2016/06/160629125823.jpg 
https://images.sciencedaily.com/2016/06/160627125140.jpg 
https://images.sciencedaily.com/2016/06/160624101050.jpg 
https://images.sciencedaily.com/2016/06/160622104810.jpg 

更快的替代方案将是使用LXML：

from lxml import etree 

for item in tree.findall(".//item/media:thumbnail",tree.nsmap): 
    parent = item.getparent() 
    print(parent.xpath("title/text()")[0]) 
    print(parent.xpath("link/text()")[0]) 
    print(item.get("url"))