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 %-40s [ %s != %s ]'%(msg.upper(), main_pid, key_info.strip('./'), base_type, test_type)
  print info.encode('gbk')
  OK = False

调用：

if __name__ == '__main__': 
  s1 = '''''<&＃63;xml version="1.0" encoding="UTF-8"&＃63;> \
      \ 
      \ 
      linux\ 
      30\ 
     \ 
      \ 
      windows\ 
      20\ 
     \ 
     '''  
  s2 = '''''<&＃63;xml version="1.0" encoding="UTF-8"&＃63;> \
      \ 
      \ 
      windows\ 
      20\ 
     \ 
      \ 
      linux\ 
      30\ 
     \ 
     '''  
  lroot = ElementTree.fromstring(s1) 
  rroot = ElementTree.fromstring(s2) 
  compare_xml(lroot, rroot)

希望本文所述对大家的Python程序设计有所帮助。