BeautifulSoup 打印多个标签/属性

Question

首先，这是我第一次尝试 Python，到目前为止它看起来很容易使用，尽管我仍然遇到了问题..

我正在尝试将 XML 文件更改为 rss-XML 原始的 xml 源看起来像这样：

<news title="Random Title" date="Date and Time" subtitle="The article txt"></news>

它最终应该看起来像这样：

<item>
<pubDate>Date and Time</pubDate>
<title>Random Title</title>
<content:encoded>The article txt</content:encoded>
</item>

我正在尝试使用 python 和 BeautifulSoup 来执行此操作，使用以下脚本

from BeautifulSoup import BeautifulSoup
import re

doc = [
'<news post_title="Random Title" post_date="Date and Time" post_content="The article txt">''</news></p>'
    ]
soup = BeautifulSoup(''.join(doc))

print soup.prettify()

posttitle = soup.news['post_title']
postdate = soup.news['post_date']
postcontent = soup.news['post_content']

print "<item>"
print "<pubDate>"
print postdate
print "</pubDate>"
print "<title>"
print posttitle
print "</title>"
print "<content:encoded>"
print postcontent
print "</content:encoded>"
print "</item>"

这里的问题是，它只检索最上面的字符串 XML，而不是其他字符串。 有人能给我一些解决这个问题的指导吗？

干杯:)

Answer 1

您的示例 doc 变量仅包含一个 元素。

但一般来说，您需要循环浏览新闻元素，

例如

for news in soup.findAll('news'):
    posttitle = news['post_title']
    postdate = news['post_date']
    postcontent = news['post_content']
    print "<item>"
    print "<pubDate>"
    print postdate
    print "</pubDate>"
    print "<title>"
    print posttitle
    print "</title>"
    print "<content:encoded>"
    print postcontent
    print "</content:encoded>"
    print "</item>"

Answer 2

窃取代码并更正它：

for news in soup.findAll('news'):
    posttitle = news['post_title']
    postdate = news['post_date']
    postcontent = news['post_content']
    print "<item>"
    print "<pubDate>"
    print postdate
    print "</pubDate>"
    print "<title>"
    print posttitle
    print "</title>"
    print "<content:encoded>"
    print postcontent
    print "</content:encoded>"
    print "</item>"