python自訂解析簡單xml格式檔案的方法

本文執行個體講述了python自訂解析簡單xml格式檔案的方法。分享給大家供大家參考。具體分析如下：

因為公司內部的介面返回的字串支援2種形式：php數組，xml；結果php數組python不能直接用，而xml字串的格式不是標準的，所以也不能用標準模組解析。【不標準的地方是某些節點會的名稱是以數字開頭的】，所以寫個簡單的腳步來解析一下檔案，用來做介面測試。

#!/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>(.*?)' % (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''):        xmlnodestr = xmlnodestr.replace(r'<![CDATA[','')[:-3]    self.xmlnodelist = xmlnodelist    return xmlnodestrif '__main__' == __name__:  xmlstr = '<&#63;xml version="1.0" encoding="utf-8" standalone="yes" &#63;><resultObject><product_id>aaaaa</product_id><product_name><![CDATA[bbbbbbbbbbbbbbb'  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

運行結果：

xmlstr: <?xml version="1.0" encoding="utf-8" standalone="yes" ?>aaaaabbbbbbbbbbbbbbbxmldom: aaaaabbbbbbbbbbbbbbb------------------------------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

因為返回的xml格式比較簡單，沒有帶屬性的節點，所以處理起來就比較簡單了。但測試還是發現有一個bug。即當相同節點嵌套時會出現正則匹配出問題，該問題的可以通過避免在xpath中出現有嵌套節點的名稱來解決，否則只有重寫複雜的機制了。

希望本文所述對大家的Python程式設計有所協助。

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