发布于 2015-10-04 09:26:38 | 241 次阅读 | 评论: 0 | 来源: 网友投递
Python编程语言
Python 是一种面向对象、解释型计算机程序设计语言，由Guido van Rossum于1989年底发明，第一个公开发行版发行于1991年。Python语法简洁而清晰，具有丰富和强大的类库。它常被昵称为胶水语言，它能够把用其他语言制作的各种模块（尤其是C/C++）很轻松地联结在一起。
本文实例讲述了python比较2个xml内容的方法。分享给大家供大家参考。具体分析如下：
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：
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)