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 = '<"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: <"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。
【python自定义解析简单xml格式文件的方法】相关文章：