python自定义解析简单xml格式文件的方法

本文实例讲述了python自定义解析简单xml格式文件的方法。分享给大家供大家参考。具体分析如下：

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'<\?xml.*\?>'
        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>(.*?)</%s>' % (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'<![CDATA['):
            xmlnodestr = xmlnodestr.replace(r'<![CDATA[','')[:-3]
        self.xmlnodelist = xmlnodelist
        return xmlnodestr
    if '__main__' == __name__:
      xmlstr = '<?xml version="1.0" encoding="utf-8" standalone="yes" ?><resultObject><a><product_id>aaaaa</product_id><product_name><![CDATA[bbbbb]]></a><b><product_id>bbbbb</product_id><product_name><![CDATA[bbbbb]]></b></product_name></resultObject>'
      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: <?xml version="1.0" encoding="utf-8" standalone="yes" ?><resultObject><a><product_id>aaaaa</product_id><product_name><![CDATA[bbbbb]]></a><b><product_id>bbbbb</product_id><product_name><![CDATA[bbbbb]]></b></product_name></resultObject>
    xmldom: <resultObject><a><product_id>aaaaa</product_id><product_name><![CDATA[bbbbb]]></a><b><product_id>bbbbb</product_id><product_name><![CDATA[bbbbb]]></b></product_name></resultObject>
    ------------------------------
    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。