Текст в XML в Python

Question

Я пытаюсь преобразовать текст в xml формат. И я использую LXML Библиотека. Но я получаю сообщение об ошибке. Пожалуйста, помогите мне. Спасибо

import re
from lxml import etree
import urllib,urllib2
def get_movie_info(movie_id):
    URL = "http://www.raaga.com/a/rss.asp?%s"%(movie_id)
    f = urllib.urlopen(URL)
    movie_info = f.read()
    rss = "".join([ line.strip() for line in movie_info ])
    mi_tree = etree.fromstring(rss)
    #mi_title = self._parse_movie_title(mi_tree.xpath("/rss/channel/title/text()")[0])
    #mi_tracks = mi_tree.xpath("/rss/channel/item")
    return mi_tree

get_movie_info('A0000102')

Вот мой след

    Traceback (most recent call last):
  File "py1.py", line 14, in <module>
    get_movie_info('A0000102')
  File "py1.py", line 9, in get_movie_info
    mi_tree = etree.fromstring(rss)
  File "lxml.etree.pyx", line 2743, in lxml.etree.fromstring (src/lxml\lxml.etre
e.c:52665)
  File "parser.pxi", line 1573, in lxml.etree._parseMemoryDocument (src/lxml\lxm
l.etree.c:79932)
  File "parser.pxi", line 1452, in lxml.etree._parseDoc (src/lxml\lxml.etree.c:7
8774)
  File "parser.pxi", line 960, in lxml.etree._BaseParser._parseDoc (src/lxml\lxm
l.etree.c:75389)
  File "parser.pxi", line 564, in lxml.etree._ParserContext._handleParseResultDo
c (src/lxml\lxml.etree.c:71739)
  File "parser.pxi", line 645, in lxml.etree._handleParseResult (src/lxml\lxml.e
tree.c:72614)
  File "parser.pxi", line 585, in lxml.etree._raiseParseError (src/lxml\lxml.etr
ee.c:71955)
lxml.etree.XMLSyntaxError: xmlParsePITarget: invalid name prefix 'xml', line 1,
column 13

Answer 1

У меня отлично работает без этой строки:

rss = "".join([ line.strip() for line in movie_info ])

Примерно так:

mi_tree = etree.fromstring(movie_info)