python自定义解析简单xml格式文件的方法

本文实例讲述了python/' target='_blank'>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-8import reclass 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 xmlnodestrif '__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_idget list: ['aaaaa', 'bbbbb']get string: aaaaa------------------------------xpath: /product_id[1] get list: ['aaaaa', 'bbbbb']get string: bbbbb------------------------------xpath: /a/product_idget 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。