python比较2个xml内容

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from xml.etree import ElementTreeOK=Truemain_pid = 10000loop_depth = 0def 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

调用：

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)

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