Python BeautifulSoup XML解析

Question

我编写了一个简单的脚本来使用BeautifulSoup模块解析XML聊天记录.标准的soup.prettify()工作正常,除了聊天记录中有很多毛病.您可以看到脚本代码和我正在使用的一些XML输入文件:

码

import sys
from BeautifulSoup import BeautifulSoup as Soup

def parseLog(file):
    file = sys.argv[1]
    handler = open(file).read()
    soup = Soup(handler)
    print soup.prettify()

if __name__ == "__main__":
    parseLog(sys.argv[1])

测试XML输入

<?xml version="1.0"?>
<?xml-stylesheet type='text/xsl' href='MessageLog.xsl'?>
<Log FirstSessionID="1" LastSessionID="2"><Message Date="10/31/2010" Time="3:43:48 PM"     DateTime="2010-10-31T20:43:48.937Z" SessionID="1"><From><User FriendlyName="Jon"/></From>    <To><User FriendlyName="Bill"/></To><Text Style="font-family:Segoe UI; color:#000000; ">hey, what's up?</Text></Message>
<Message Date="10/31/2010" Time="3:44:03 PM" DateTime="2010-10-15T20:44:03.421Z" SessionID="1"><From><User FriendlyName="Jon"/></From><To><User FriendlyName="Bill"/></To><Text Style="font-family:Segoe UI; color:#000000; ">Got your message</Text></Message> 
<Message Date="10/31/2010" Time="3:44:31 PM" DateTime="2010-10-15T20:44:31.390Z" SessionID="2"><From><User FriendlyName="Bill"/></From><To><User FriendlyName="Jon"/></To><Text Style="font-family:Segoe UI; color:#000000; ">oh, great</Text></Message>
<Message Date="10/31/2010" Time="3:44:59 PM" DateTime="2010-10-15T20:44:59.281Z" SessionID="2"><From><User FriendlyName="Bill"/></From><To><User FriendlyName="Jon"/></To><Text Style="font-family:Segoe UI; color:#000000; ">hey, i gotta run</Text></Message>

我希望能够将其输出为如下格式或至少比纯XML更具可读性的格式:

乔恩:嘿,怎么了？[10/31/10 @ 3:43p]

乔恩:收到你的留言[10/31/10 @ 3:44p]

比尔:哦,太棒了[10/31/10 @ 3:44p]

我听说过有关PyParsing模块的一些不错的东西,也许是时候试一试了.

Answer 1

BeautifulSoup使得获取xml中的属性和值非常简单.我调整了你的示例函数来使用这些功能.

import sys
from BeautifulSoup import BeautifulSoup as Soup

def parseLog(file):
    file = sys.argv[1]
    handler = open(file).read()
    soup = Soup(handler)
    for message in soup.findAll('message'):
        msg_attrs = dict(message.attrs)
        f_user = message.find('from').user
        f_user_dict = dict(f_user.attrs)
        print "%s: %s [%s @ %s]" % (f_user_dict[u'friendlyname'],
                                    message.find('text').decodeContents(),
                                    msg_attrs[u'date'],
                                    msg_attrs[u'time'])


if __name__ == "__main__":
    parseLog(sys.argv[1])

Answer 2

我建议使用内置的ElementTree模块.BeautifulSoup旨在处理不良格式的代码,如被黑客攻击的HTML,而XML格式良好,并且意图由XML库读取.

更新:我最近阅读的一些内容建议将lxml作为构建的库并增强标准ElementTree.