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,'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,'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,'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,msg,base_type,test_type):
  global OK 
  info = u'[ %-5s ] %s -> %-40s [ %s != %s ]'%(msg.upper(),main_pid,key_info.strip('./'),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)