python自定义解析简单xml格式文件

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 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_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。