python自定义解析简单xml格式文件的方法

因为公司内部的接口返回的字串支持2种形式：php数组，xml；结果php数组python不能直接用，而xml字符串的格式不是标准的，所以也不能用标准模块解析。【不标准的地方是某些节点会的名称是以数字开头的】，所以写个简单的脚步来解析一下文件，用来做接口测试。

#!/usr/bin/env python
#encoding: utf-8
import re
class xmlparse:
  def __init__(self, xmlstr):
    self.xmlstr = xmlstr
    self.xmldom = self.__convet2utf8()
    self.xmlnodelist = []
    self.xpath = ''
  def __convet2utf8(self):
    headstr = self.__get_head()
    xmldomstr = self.xmlstr.replace(headstr, '')
    if 'gbk' in headstr: 
      xmldomstr = xmldomstr.decode('gbk').encode('utf-8')
    elif 'gb2312' in headstr:
      xmldomstr = self.xmlstr.decode('gb2312').encode('utf-8')
    return xmldomstr
  def __get_head(self):
    headpat = r'<\&＃63;xml.*\&＃63;>'
    headpatobj = re.compile(headpat)
    headregobj = headpatobj.match(self.xmlstr)
    if headregobj:
      headstr = headregobj.group()
      return headstr
    else:
      return ''
  def parse(self, xpath):
    self.xpath = xpath
    xpatlist = []
    xpatharr = self.xpath.split('/')
    for xnode in xpatharr:
      if xnode:
        spcindex = xnode.find('[')
        if spcindex > -1:
          index = int(xnode[spcindex+1:-1])
          xnode = xnode[:spcindex]
        else:
          index = 0;
        temppat = ('<%s>(.*&＃63;)' % (xnode, xnode),index)
        xpatlist.append(temppat)
    xmlnodestr = self.xmldom
    for xpat,index in xpatlist:
      xmlnodelist = re.findall(xpat,xmlnodestr)
      xmlnodestr = xmlnodelist[index]
      if xmlnodestr.startswith(r'aaaaabbbbb'
  xpath1 = '/product_id'
  xpath2 = '/product_id[1]'
  xpath3 = '/a/product_id'
  xp = xmlparse(xmlstr)
  print 'xmlstr:',xp.xmlstr
  print 'xmldom:',xp.xmldom
  print '------------------------------'
  getstr = xp.parse(xpath1)
  print 'xpath:',xp.xpath
  print 'get list:',xp.xmlnodelist
  print 'get string:', getstr
  print '------------------------------'
  getstr = xp.parse(xpath2)
  print 'xpath:',xp.xpath
  print 'get list:',xp.xmlnodelist
  print 'get string:', getstr
  print '------------------------------'
  getstr = xp.parse(xpath3)
  print 'xpath:',xp.xpath
  print 'get list:',xp.xmlnodelist
  print 'get string:', getstr

运行结果：

xmlstr: <&＃63;xml version="1.0" encoding="utf-8" standalOne="yes" &＃63;>aaaaabbbbb
xmldom: aaaaabbbbb
------------------------------
xpath: /product_id
get list: ['aaaaa', 'bbbbb']
get string: aaaaa
------------------------------
xpath: /product_id[1] 
get list: ['aaaaa', 'bbbbb']
get string: bbbbb
------------------------------
xpath: /a/product_id
get list: ['aaaaa']
get string: aaaaa

因为返回的xml格式比较简单，没有带属性的节点，所以处理起来就比较简单了。但测试还是发现有一个bug。即当相同节点嵌套时会出现正则匹配出问题，该问题的可以通过避免在xpath中出现有嵌套节点的名称来解决，否则只有重写复杂的机制了。

希望本文所述对大家的Python程序设计有所帮助。