这篇文章主要介绍了python比较2个xml内容的方法,涉及Python操作XML文件的相关技巧,需要的朋友可以参考下
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 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调用：
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 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)希望本文所述对大家的Python程序设计有所帮助。
