关于读取xml的问题

[ft1834334]

owners.xml
<owners>
<owner name="jack">
<filename>looting</filename>
<filename>RaceConfig</filename>
</owner>

<owner name="rose">
<filename>NBACourt</filename>
<filename>StreetCourt</filename>
</owner>

<owner name="mike">
<filename>ControlGrade</filename>
<filename>Talent</filename>
</owner>
</owners>

需求是这样的，我会传入文件名
比如：looting和talent2个文件名
然后去搜索所有xml的节点的文本信息
如果搜索到了，就返回上一个节点值。

例子：传入looting，talent2个文件名
之后返回jack和talent2个name值。。。。
求完整代码。。。小弟想了很多版本都没能实现。。。求指教

[libralibra]

s = '''<owners>
<owner name="jack">
<filename>looting</filename>
<filename>RaceConfig</filename>
</owner>

<owner name="rose">
<filename>NBACourt</filename>
<filename>StreetCourt</filename>
</owner>

<owner name="mike">
<filename>ControlGrade</filename>
<filename>Talent</filename>
</owner>
</owners>'''

import re,string
res = r'<owner name=\"(.*?)\"(.*?)<\/owner>'
m = re.findall(res,s,re.S)

fname = raw_input("Please input the file name<ENTER for exit>: ")

while len(fname.strip())>0:
    for item in m:
        if string.lower(fname.strip()) in string.lower(item[1].strip()):
            print "Input:",fname,"Name:",item[0]
            break
    fname = raw_input("Please input the file name<ENTER for exit>: ")

结果

>>> 
Input: looting Name: jack
Input: talent Name: mike
>>> 

[angel_su]

看看手册，内建就有几个解析html/xml模块，或者试试第三方模块BeautifulSoup，蛮好用的，代码类似：

from bs4 import BeautifulSoup

soup = BeautifulSoup('owners.xml')
files = ['looting', 'talent']
results = soup.find_all(text=lambda s: s.lower() in files)
names = [each.find_parent('owner')['name'] for each in results]