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)