python比较2个xml内容

1. from xml.etree import ElementTree
3. OK=True
4. main_pid = 10000
5. loop_depth = 0
6. def compare_xml(left, right, key_info='.'):
7. global loop_depth
8. loop_depth += 1
9. if loop_depth == 1: print
10. if left.tag != right.tag:
11. print_diff(main_pid, key_info, 'difftag', left.tag, right.tag)
12. return
13. if left.text != right.text:
14. print_diff(main_pid, key_info, 'difftext', left.text, right.text)
15. return
16. leftitems = dict(left.items())
17. rightitems = dict(right.items())
18. for k,v in leftitems.items():
19. if k not in rightitems:
20. s = '%s/%s' % (key_info, left.tag)
21. print_diff(main_pid, s, 'lostattr', k, "")
22. for k,v in rightitems.items():
23. if k not in leftitems:
24. s = '%s/%s' % (key_info, right.tag)
25. print_diff(main_pid, s, 'extraattr', "", k)
26. leftnodes = left.getchildren()
27. rightnodes = right.getchildren()
28. leftlen = len(leftnodes)
29. rightlen = len(rightnodes)
30. if leftlen != rightlen:
31. s = '%s/%s' % (key_info, right.tag)
32. print_diff(main_pid, s, 'difflen', leftlen, rightlen)
33. return
34. l = leftlen<rightlen and leftlen or rightlen
35. d = {}
36. for i in xrange(l):
37. node=leftnodes[i]
38. if node.tag not in d:
39. d[node.tag] = 1
40. tag = node.tag
41. else:
42. tag = node.tag + str(d[node.tag])
43. d[node.tag] += 1
44. s = '%s/%s' % (key_info, tag)
45. compare_xml(leftnodes[i], rightnodes[i], s)
48. def print_diff(main_pid, key_info, msg, base_type, test_type):
49. global OK
50. info = u'[ %-5s ] %s -> %-40s [ %s != %s ]'%(msg.upper(), main_pid, key_info.strip('./'), base_type, test_type)
51. print info.encode('gbk')
52. OK = False

调用：

1. 
   
2. if __name__ == '__main__':
3. s1 = '''<?xml version="1.0" encoding="UTF-8"?> \
4. <employees> \
5. <employee id = '1'> \
6. <name>linux</name>\
7. <age>30</age>\
8. </employee>\
9. <employee id = '2'> \
10. <name>windows</name>\
11. <age>20</age>\
12. </employee>\
13. </employees>'''
15. s2 = '''<?xml version="1.0" encoding="UTF-8"?> \
16. <employees> \
17. <employee id = '3'> \
18. <name>windows</name>\
19. <age>20</age>\
20. </employee>\
21. <employee id = '4'> \
22. <name>linux</name>\
23. <age>30</age>\
24. </employee>\
25. </employees>'''
26. lroot = ElementTree.fromstring(s1)
27. rroot = ElementTree.fromstring(s2)
28. compare_xml(lroot, rroot)