python自定义解析简单xml格式文件的方法
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 = '<?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_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。