python中两个xml文件的比较

Question

我是用python编程的新手，我在理解这个概念时遇到了一些困难。我想比较两个xml文件。这些xml文件相当大。 我将给出一个我想要比较的文件类型的例子。

XML文件1：

<xml>
    <property1>
          <property2>    
               <property3>

               </property3>
          </property2>    
    </property1>    
</xml>

xml文件2：

<xml>
    <property1>
        <property2>    
            <property3> 
                <property4>

                </property4>    
            </property3>
        </property2>    
    </property1>

</xml>

我命名的property1，property2与文件中实际的property2不同。xml文件中有很多属性。 我想比较一下这两个xml文件。

我正在使用一个lxml解析器来尝试比较这两个文件并打印出它们之间的差异。

我不知道如何分析和比较它自动。

我试图阅读lxml解析器，但我不知道如何使用它来解决我的问题。

有人能告诉我该如何处理这个问题吗。

代码片段非常有用

还有一个问题，我是在遵循正确的概念，还是我遗漏了其他东西？请纠正我你不知道的任何新概念

Answer 1

另一个使用xml.etree的脚本。很糟糕，但很有效：）

#!/usr/bin/env python

import sys
import xml.etree.ElementTree as ET

from termcolor import colored

tree1 = ET.parse(sys.argv[1])
root1 = tree1.getroot()

tree2 = ET.parse(sys.argv[2])
root2 = tree2.getroot()

class Element:
    def __init__(self,e):
        self.name = e.tag
        self.subs = {}
        self.atts = {}
        for child in e:
            self.subs[child.tag] = Element(child)

        for att in e.attrib.keys():
            self.atts[att] = e.attrib[att]

        print "name: %s, len(subs) = %d, len(atts) = %d" % ( self.name, len(self.subs), len(self.atts) )

    def compare(self,el):
        if self.name!=el.name:
            raise RuntimeError("Two names are not the same")
        print "----------------------------------------------------------------"
        print self.name
        print "----------------------------------------------------------------"
        for att in self.atts.keys():
            v1 = self.atts[att]
            if att not in el.atts.keys():
                v2 = '[NA]'
                color = 'yellow'
            else:
                v2 = el.atts[att]
                if v2==v1:
                    color = 'green'
                else:
                    color = 'red'
            print colored("first:\t%s = %s" % ( att, v1 ), color)
            print colored("second:\t%s = %s" % ( att, v2 ), color)

        for subName in self.subs.keys():
            if subName not in el.subs.keys():
                print colored("first:\thas got %s" % ( subName), 'purple')
                print colored("second:\thasn't got %s" % ( subName), 'purple')
            else:
                self.subs[subName].compare( el.subs[subName] )



e1 = Element(root1)
e2 = Element(root2)

e1.compare(e2)

Answer 2

我解决这个问题的方法是将每个XML转换成一个xml.etree.ElementTree并遍历每个层。 我还提供了在进行比较时忽略属性列表的功能。

第一段代码包含使用的类：

import xml.etree.ElementTree as ET
import logging

class XmlTree():

    def __init__(self):
        self.hdlr = logging.FileHandler('xml-comparison.log')
        self.formatter = logging.Formatter('%(asctime)s %(levelname)s %(message)s')

    @staticmethod
    def convert_string_to_tree( xmlString):

        return ET.fromstring(xmlString)

    def xml_compare(self, x1, x2, excludes=[]):
        """
        Compares two xml etrees
        :param x1: the first tree
        :param x2: the second tree
        :param excludes: list of string of attributes to exclude from comparison
        :return:
            True if both files match
        """

        if x1.tag != x2.tag:
            self.logger.debug('Tags do not match: %s and %s' % (x1.tag, x2.tag))
            return False
        for name, value in x1.attrib.items():
            if not name in excludes:
                if x2.attrib.get(name) != value:
                    self.logger.debug('Attributes do not match: %s=%r, %s=%r'
                                 % (name, value, name, x2.attrib.get(name)))
                    return False
        for name in x2.attrib.keys():
            if not name in excludes:
                if name not in x1.attrib:
                    self.logger.debug('x2 has an attribute x1 is missing: %s'
                                 % name)
                    return False
        if not self.text_compare(x1.text, x2.text):
            self.logger.debug('text: %r != %r' % (x1.text, x2.text))
            return False
        if not self.text_compare(x1.tail, x2.tail):
            self.logger.debug('tail: %r != %r' % (x1.tail, x2.tail))
            return False
        cl1 = x1.getchildren()
        cl2 = x2.getchildren()
        if len(cl1) != len(cl2):
            self.logger.debug('children length differs, %i != %i'
                         % (len(cl1), len(cl2)))
            return False
        i = 0
        for c1, c2 in zip(cl1, cl2):
            i += 1
            if not c1.tag in excludes:
                if not self.xml_compare(c1, c2, excludes):
                    self.logger.debug('children %i do not match: %s'
                                 % (i, c1.tag))
                    return False
        return True

    def text_compare(self, t1, t2):
        """
        Compare two text strings
        :param t1: text one
        :param t2: text two
        :return:
            True if a match
        """
        if not t1 and not t2:
            return True
        if t1 == '*' or t2 == '*':
            return True
        return (t1 or '').strip() == (t2 or '').strip()

第二段代码包含两个XML示例及其比较：

xml1 = "<note><to>Tove</to><from>Jani</from><heading>Reminder</heading><body>Don't forget me this weekend!</body></note>"

xml2 = "<note><to>Tove</to><from>Daniel</from><heading>Reminder</heading><body>Don't forget me this weekend!</body></note>"

tree1 = XmlTree.convert_string_to_tree(xml1)
tree2 = XmlTree.convert_string_to_tree(xml2)

comparator = XmlTree()

if comparator.xml_compare(tree1, tree2, ["from"]):
    print "XMLs match"
else:
    print "XMLs don't match"

此代码的大部分功劳必须给予syawar

Answer 3

这实际上是一个相当具有挑战性的问题（由于“差异”在这里通常是指旁观者的眼睛，因为在语义上会有“等价”的信息，您可能不希望将其标记为差异）。

您可以尝试使用xmldiff，这是基于论文中的工作Change Detection in Hierarchically Structured Information。