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python自定義解析簡單xml格式檔案

因為公司內部的介面返回的字串支援2種形式:php陣列,xml;結果php陣列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-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  
                  

執行結果:

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。即當相同節點巢狀時會出現正則匹配出問題,該問題的可以通過避免在xpath中出現有巢狀節點的名稱來解決,否則只有重寫複雜的機制了。