开发者

Python 基于xml.etree.ElementTree实现XML对比示例详解

目录
  • 测试环境
  • 代码实现
  • 测试运行
    • 运行结果

测试环境

python 3.6

Win10

代码实现

#!/usr/bin/env python 3.4.0
#-*- encoding:utf-8 -*-

__author__ = 'shouke'

import XML.etree.ElementTree as ET

def compare_xml_node_attributes(xml_node1, xml_node2):
    result = []
    node1_attributes_dict = xml_node1.attrib
    node2_attributes_dict = xml_node2.attrib
    for attrib1, value in node1_attributes_dict.items():
        value2 =  node2_attributes_dict.get(attrib1)
        if value == value2:
            node2_attributes_dict.pop(attrib1)
        else:
            if value2:
                attrib2 = attrib1
                node2_attributes_dict.pop(attrib2)
            else:
                attrib2 = '不存在'
            result.append('结点1属性:{attrib1} 值:{value1},结点2属性:{attrib1} 值:{value2}'.format(attrib1=attrib1 or '不存在',
                                    fRgZP                                                     value1=value or '不存在',
                                                                                         attrib2=attrib2,
                                                                                         value2=value2 or '不存在'))

    for attrib2, value2 in node2_attributes_dict.items():
        result.append('结点1属性:{attrib1} 值:{value1},结点2属性:{attrib1} 值:{value2}'.format(attrib1='不存在',
                                                                                         value1='不存在',
                                                                                         attrib2=attrib2,
                                                                                         value2=value2))
    return result


def compare_xml_node_children(xml_javascriptnode1, xml_node2, node1_xpath, node2_xpath):
    def get_node_children(xml_node, node_xpath):
        result = {}
        for child in list(xml_node):
            if child.tag not in result:
                result[child.tag] = [{'node':child, 'xpath': '%s/%s[%s]' % (node_xpath, child.tag, 1)}]
            else:
                result[child.tag].append({'node':child, 'xpath': '%s/%s[%s]' % (node_xpath, child.tag, len(result[child.tag])+1)})
        return result

    result = []
    children_of_node1_dict = get_node_children(xml_node1, node1_xpath)
    children_of_node2_dict = get_node_children(xml_node2, node2_xpath)

    temp_list1 = []
    temp_list2 = []
    for child_tag, child_node_list in children_of_node1_dict.items():
        second_child_node_list = children_of_node2_dict.get(child_tag, [])
        if not second_child_node_list:
            # 获取xml1中比xml2中多出的子结点
            for i in range(0, len(child_node_list)):
                temp_list1.append('%s/%s[%s]' % (node1_xpath, child_node_list[i]['node'].tag, i+1))
            continue

        for first_child, second_child in zip(child_node_list, second_child_node_list):
            result.extend(compare_xml_nodes(first_child['node'], second_child['node'], first_child['xpath'], second_child['xpath']))

        # 获取xml2中对应结点比xml1中对应结点多出的同名子结点
        for i in range(l开发者_C学习en(child_node_list), len(second_child_node_list)):
            temp_list2.append('%s/%s[%s]' % (node2_xpathttp://www.devze.comh, second_child_node_list[i]['node'].tag, i+1))
        children_of_node2_dict.pop(child_tag)

    if temp_list1:
        result.append('子结点不一样:xml1结点(xpath:{xpath1})比xml2结点(xpath:{xpath2})多了以下子结点:\n{differences}'.format (xpath1=node1_xpath,
                                                                                                  xpath2=node2_xpath,
                                                                                                  differences='\n'.join(temp_list1)))
    # 获取xml2比xml1中多出的子结点
    for child_tag, child_node_list in children_of_node2_dict.items():
        for i in range(0, len(child_node_list)):
            temp_list2.append('%s/%s[%s]' % (node1_xpath, child_node_list[i]['node'].tag, i+1))

    if temp_list2:
        result.append('子结点不一样:xml1结点(xpath:{xpath1})比xml2结点(xpath:{xpath2})少了以下子结点:\n{differences}'.format (xpath1=node1_xpath,
                                                                                                  xpath2=node2_xpath,
                                                                                                  differences='\n'.join(temp_list2)))
    return result


def compare_xml_nodes(xml_node1, xml_node2, node1_xpath='', node2_xpath=''):
    result = []
    # 比较标签
    if xml_node1.tag !=  xml_node2.tag:
        result.append('标签不一样:xml1结点(xpath:{xpath1}):{tag1},xml2结点(xpath:{xpath2}):{tag2}'.format (xpath1=node1_xpath,
                                                       编程客栈                                           tag1=xml_node1.tag,
                                                                                                  xpath2=node2_xpath,
                                                                                                  tag2=xml_node2.tag))

    # 比较文本
    if xml_node1.text !=  xml_node2.text:
        result.append('文本不一样:xml1结点(xpath:{xpath1}):{text1},xml2结点(xpath:{xpath2}):{text2}'.format (xpath1=node1_xpath,
                                                                                                  tag1=xml_node1.text or '',
                                                                        python                          xpath2=node2_xpath,
                                                                                                  tag2=xml_node2.text or ''))

    # 比较属性
    res = compare_xml_node_attributes(xml_node1, xml_node2)
    if res:
        result.append('属性不一样:xml1结点(xpath:{xpath1}),xml2结点(xpath:{xpath2}):\n{differences}'.format (xpath1=node1_xpath,
                                                                                                  xpath2=node2_xpath,
                                                                                                  differences='\n'.join(res)))
    # 比较子结点
    res = compare_xml_node_children(xml_node1, xml_node2, node1_xpath, node2_xpath)
    if res:
        result.extend(res)

    return result

def compare_xml_strs(xml1_str, xml2_str, mode=3):
    '''
    @param: mode 比较模式,预留,暂时没用。目前默认 xml 子元素如果为列表,则列表有序列表,按序比较
    '''
    root1 = ET.fromstring(xml1_str.strip())
    root2 = ET.fromstring(xml2_str.strip())

    return compare_xml_nodes(root1, root2, '/%s' % root1.tag, '/%s' % root2.tag)

测试运行

xml_str1 = '''
<?xml version = "1.0" encoding="utf-8" ?>
<data>
    <country name="Liechtenstein">
        <rangk>1</rangk>
        <year>2008</year>
        <gdppc>141100</gdppc>
        <neighbor name="Austria" direction="E" ></neighbor>
        <neighbor name="Switzerland" direction="W" ></neighbor>
    </country>
    <country name="Singpore">
        <rank>4</rank>
        <year>2011</year>
        <gdppc>59900</gdppc>
        <neighbor name="Malaysia" direction="N" ></neighbor>
    </country>
    <country name="Panama">
        <rank>68</rank>
        <year>2011</year>
        <gdppc>13600</gdppc>
        <neighbor name="Costa Rica" direction="W" ></neighbor>
        <neighbor name="Colombia" direction="W" ></neighbor>
    </country>
</data>
'''
xml_str2 = '''
<?xml version = "1.0" encoding="utf-8" ?>
<data>
    <country name="Liechtenstein">
        <rangk>1</rangk>
        <year>2008</year>
        <gdppc>141100</gdppc>
        <neighbor name="Austria" direction="E" ></neighbor>
        <neighbor name="Switzerland" direction="W" ></neighbor>
    </country>
    <country name="Singpore">
        <rank>4</rank>
        <year>2011</year>
        <gdppc>59900</gdppc>
        <neighbor name="Malaysia" direction="N" ></neighbor>
    </country>
    <country name="Panama">
        <rank>68</rank>
        <year>2011</year>
        <gdppc>13600</gdppc>
        <neighbor name="Costa Rica" direction="W" ></neighbor>
        <neighbor name="Colombia" direction="W" ></neighbor>
    </country>
</data>
'''

xml_str3 = '''
<?xml version = "1.0" encoding="utf-8" ?>
<data>
    <class name="computer">
        <rangk>1</rangk>
        <year>unknow</year>
        <addr>sz</addr>
        <book name="Java programming" price="10" ></book>
        <book name="python programming" price="10" ></book>
    </class>
    <class name="philosophy">
        <rangk>2</rangk>
        <year>unknown</year>
        <book name="A little history of philosophy" price="15" ></book>
        <book name="contemporary introduction" price="15" ></book>
    </class>
    <class name="history">
        <rangk>3</rangk>
        <year>unknown</year>
        <addr>other addr</addr>
        <book name="The South China Sea" price="10" ></book>
        <book name="Chinese Among Others" price="10" ></book>
    </class>
</data>
'''
xml_str4 = '''
<?xml version = "1.0" encoding="utf-8" ?>
<data>
    <class name="computer">
        <year>unknow</year>
        <addr>sz</addr>
        <book name="java programming" price="10" ></book>
        <book name="python programming" price="10" ></book>
    </class>
    <class name="philosophy">
        <year>unknown</year>
        <addr>other addr</addr>
        <book name="A little history of philosophy" price="15" ></book>
        <book name="contemporary introduction" price="16" ></book>
    </class>
</data>
'''
if __name__ == '__main__':
    res_list = compare_xml_strs(xml_str1, xml_str2)
    if res_list:
        print('xml1和xml2不一样:\n%s' % '\n'.join(res_list))
    else:
        print('xml1和xml2一样')

    res_list = compare_xml_strs(xml_str3, xml_str4)
    if res_list:
        print('xml3和xml4不一样:\n%s' % '\n'.join(res_list))
    else:
        print('xml3和xml4一样')

运行结果

xml1和xml2一样 xml3和xml4不一样: 子结点不一样:xml1结点(xpath:/data/class[1])比xml2结点(xpath:/data/class[1])多了以下子结点: /data/class[1]/rangk[1] 属性不一样:xml1结点(xpath:/data/class[2]/book[2]),xml2结点(xpath:/data/class[2]/book[2]): 结点1属性:price 值:15,结点2属性:price 值:16 子结点不一样:xml1结点(xpath:/data/class[2])比xml2结点(xpath:/data/class[2])多了以下子结点: /data/class[2]/rangk[1] 子结点不一样:xml1结点(xpath:/data/class[2])比xml2结点(xpath:/data/class[2])少了以下子结点: /data/class[2]/addr[1]

到此这篇关于Python 基于xml.etree.ElementTree实现XML对比的文章就介绍到这了,更多相关Python 实现XML对比内容请搜索我们以前的文章或继续浏览下面的相关文章希望大家以后多多支持我们!

0

上一篇:

下一篇:

精彩评论

暂无评论...
验证码 换一张
取 消

最新开发

开发排行榜