python比较2个xml内容的方法

from xml.etree import ElementTree 
OK=True 
main_pid = 10000 
loop_depth = 0 
def compare_xml(left, right, key_info='.'):
  global loop_depth 
  loop_depth += 1 
  if loop_depth == 1: print 
  if left.tag != right.tag: 
    print_diff(main_pid, key_info, 'difftag', left.tag, right.tag)
    return 
  if left.text != right.text: 
    print_diff(main_pid, key_info, 'difftext', left.text, right.text)
    return 
  leftitems = dict(left.items()) 
  rightitems = dict(right.items()) 
  for k,v in leftitems.items(): 
    if k not in rightitems: 
      s = '%s/%s' % (key_info, left.tag) 
      print_diff(main_pid, s, 'lostattr', k, "")
  for k,v in rightitems.items(): 
    if k not in leftitems: 
      s = '%s/%s' % (key_info, right.tag) 
      print_diff(main_pid, s, 'extraattr', "", k)
  leftnodes = left.getchildren() 
  rightnodes = right.getchildren() 
  leftlen = len(leftnodes) 
  rightlen = len(rightnodes) 
  if leftlen != rightlen: 
    s = '%s/%s' % (key_info, right.tag)
    print_diff(main_pid, s, 'difflen', leftlen, rightlen)
    return 
  l = leftlen<rightlen and leftlen or rightlen
  d = {} 
  for i in xrange(l):     
    node=leftnodes[i] 
    if node.tag not in d: 
      d[node.tag] = 1 
      tag = node.tag 
    else: 
      tag = node.tag + str(d[node.tag])
      d[node.tag] += 1 
    s = '%s/%s' % (key_info, tag) 
    compare_xml(leftnodes[i], rightnodes[i], s)
def print_diff(main_pid, key_info, msg, base_type, test_type):
  global OK 
  info = u'[ %-5s ] %s -> %-40s [ %s != %s ]'%(msg.upper(), main_pid, key_info.strip('./'), base_type, test_type)
  print info.encode('gbk')
  OK = False

if __name__ == '__main__': 
  s1 = '''''<?xml version="1.0" encoding="UTF-8"?> \
     <employees> \ 
     <employee id = '1'> \ 
      <name>linux</name>\ 
      <age>30</age>\ 
     </employee>\ 
     <employee id = '2'> \ 
      <name>windows</name>\ 
      <age>20</age>\ 
     </employee>\ 
     </employees>'''  
  s2 = '''''<?xml version="1.0" encoding="UTF-8"?> \
     <employees> \ 
     <employee id = '3'> \ 
      <name>windows</name>\ 
      <age>20</age>\ 
     </employee>\ 
     <employee id = '4'> \ 
      <name>linux</name>\ 
      <age>30</age>\ 
     </employee>\ 
     </employees>'''  
  lroot = ElementTree.fromstring(s1) 
  rroot = ElementTree.fromstring(s2) 
  compare_xml(lroot, rroot)