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比较XML代码段?

diff xml python

Gregg Lind · 技术社区 · 15 年前

建立在 another SO question ,如何检查两个格式良好的XML片段在语义上是否相等。我只需要“相等”或不相等,因为我在单元测试中使用它。

在我想要的系统中,它们是相等的(注意“开始”的顺序和'结束'):

<?xml version='1.0' encoding='utf-8' standalone='yes'?>
<Stats start="1275955200" end="1276041599">
</Stats>

# Reordered start and end

<?xml version='1.0' encoding='utf-8' standalone='yes'?>
<Stats end="1276041599" start="1275955200" >
</Stats>

我可以使用lmxl和其他工具,而且一个只允许重新排序属性的简单函数也可以很好地工作!

基于IANB答案的工作代码段:

from formencode.doctest_xml_compare import xml_compare
# have to strip these or fromstring carps
xml1 = """    <?xml version='1.0' encoding='utf-8' standalone='yes'?>
    <Stats start="1275955200" end="1276041599"></Stats>"""
xml2 = """     <?xml version='1.0' encoding='utf-8' standalone='yes'?>
    <Stats end="1276041599" start="1275955200"></Stats>"""
xml3 = """ <?xml version='1.0' encoding='utf-8' standalone='yes'?>
    <Stats start="1275955200"></Stats>"""

from lxml import etree
tree1 = etree.fromstring(xml1.strip())
tree2 = etree.fromstring(xml2.strip())
tree3 = etree.fromstring(xml3.strip())

import sys
reporter = lambda x: sys.stdout.write(x + "\n")

assert xml_compare(tree1,tree2,reporter)
assert xml_compare(tree1,tree3,reporter) is False

10 回复 | 直到 8 年前

Thomas Grainger 11 年前

你可以使用 formencode.doctest_xml_compare --xml_compare函数比较两个elementtree或lxml树。

Anentropic 11 年前

元素的顺序在XML中可能很重要,这可能是为什么大多数建议的其他方法在顺序不同时会比较不相等…即使元素具有相同的属性和文本内容。

但是我也想要一个顺序不敏感的比较,所以我想到了:

from lxml import etree
import xmltodict  # pip install xmltodict


def normalise_dict(d):
    """
    Recursively convert dict-like object (eg OrderedDict) into plain dict.
    Sorts list values.
    """
    out = {}
    for k, v in dict(d).iteritems():
        if hasattr(v, 'iteritems'):
            out[k] = normalise_dict(v)
        elif isinstance(v, list):
            out[k] = []
            for item in sorted(v):
                if hasattr(item, 'iteritems'):
                    out[k].append(normalise_dict(item))
                else:
                    out[k].append(item)
        else:
            out[k] = v
    return out


def xml_compare(a, b):
    """
    Compares two XML documents (as string or etree)

    Does not care about element order
    """
    if not isinstance(a, basestring):
        a = etree.tostring(a)
    if not isinstance(b, basestring):
        b = etree.tostring(b)
    a = normalise_dict(xmltodict.parse(a))
    b = normalise_dict(xmltodict.parse(b))
    return a == b

Mark E. Haase 12 年前

我有同样的问题:我想比较两个具有相同属性但顺序不同的文档。

似乎LXML中的XML规范化(C14N)可以很好地解决这个问题,但我绝对不是XML专家。我很好奇是否有人能指出这种方法的缺点。

parser = etree.XMLParser(remove_blank_text=True)

xml1 = etree.fromstring(xml_string1, parser)
xml2 = etree.fromstring(xml_string2, parser)

print "xml1 == xml2: " + str(xml1 == xml2)

ppxml1 = etree.tostring(xml1, pretty_print=True)
ppxml2 = etree.tostring(xml2, pretty_print=True)

print "pretty(xml1) == pretty(xml2): " + str(ppxml1 == ppxml2)

xml_string_io1 = StringIO()
xml1.getroottree().write_c14n(xml_string_io1)
cxml1 = xml_string_io1.getvalue()

xml_string_io2 = StringIO()
xml2.getroottree().write_c14n(xml_string_io2)
cxml2 = xml_string_io2.getvalue()

print "canonicalize(xml1) == canonicalize(xml2): " + str(cxml1 == cxml2)

运行这个可以让我:

$ python test.py 
xml1 == xml2: false
pretty(xml1) == pretty(xml2): false
canonicalize(xml1) == canonicalize(xml2): true

Guillaume Vincent 10 年前

这里是一个简单的解决方案,将XML转换为字典(使用 XMLtoDICT )把字典放在一起比较

import json
import xmltodict

class XmlDiff(object):
    def __init__(self, xml1, xml2):
        self.dict1 = json.loads(json.dumps((xmltodict.parse(xml1))))
        self.dict2 = json.loads(json.dumps((xmltodict.parse(xml2))))

    def equal(self):
        return self.dict1 == self.dict2

单元测试

import unittest

class XMLDiffTestCase(unittest.TestCase):

    def test_xml_equal(self):
        xml1 = """<?xml version='1.0' encoding='utf-8' standalone='yes'?>
        <Stats start="1275955200" end="1276041599">
        </Stats>"""
        xml2 = """<?xml version='1.0' encoding='utf-8' standalone='yes'?>
        <Stats end="1276041599" start="1275955200" >
        </Stats>"""
        self.assertTrue(XmlDiff(xml1, xml2).equal())

    def test_xml_not_equal(self):
        xml1 = """<?xml version='1.0' encoding='utf-8' standalone='yes'?>
        <Stats start="1275955200">
        </Stats>"""
        xml2 = """<?xml version='1.0' encoding='utf-8' standalone='yes'?>
        <Stats end="1276041599" start="1275955200" >
        </Stats>"""
        self.assertFalse(XmlDiff(xml1, xml2).equal())

或者在简单的python方法中:

import json
import xmltodict

def xml_equal(a, b):
    """
    Compares two XML documents (as string or etree)

    Does not care about element order
    """
    return json.loads(json.dumps((xmltodict.parse(a)))) == json.loads(json.dumps((xmltodict.parse(b))))

Jeremy Brown 15 年前

如果采用DOM方法,则可以在比较节点(节点类型、文本、属性)的同时同时遍历两个树。

递归解决方案将是最优雅的——只要在一对节点不“相等”时进行短路进一步比较,或者当您检测到一棵树中的一片叶子是另一棵树中的一个分支时进行短路进一步比较,等等。

user3116268 11 年前

考虑到这个问题,我提出了以下解决方案,使XML元素具有可比性和可排序性:

import xml.etree.ElementTree as ET
def cmpElement(x, y):
    # compare type
    r = cmp(type(x), type(y))
    if r: return r 
    # compare tag
    r = cmp(x.tag, y.tag)
    if r: return r
    # compare tag attributes
    r = cmp(x.attrib, y.attrib)
    if r: return r
    # compare stripped text content
    xtext = (x.text and x.text.strip()) or None
    ytext = (y.text and y.text.strip()) or None
    r = cmp(xtext, ytext)
    if r: return r
    # compare sorted children
    if len(x) or len(y):
        return cmp(sorted(x.getchildren()), sorted(y.getchildren()))
    return 0

ET._ElementInterface.__lt__ = lambda self, other: cmpElement(self, other) == -1
ET._ElementInterface.__gt__ = lambda self, other: cmpElement(self, other) == 1
ET._ElementInterface.__le__ = lambda self, other: cmpElement(self, other) <= 0
ET._ElementInterface.__ge__ = lambda self, other: cmpElement(self, other) >= 0
ET._ElementInterface.__eq__ = lambda self, other: cmpElement(self, other) == 0
ET._ElementInterface.__ne__ = lambda self, other: cmpElement(self, other) != 0

Community CDub 8 年前

适应 Anentropic's great answer 到python 3(基本上,更改 iteritems() 到 items() 和 basestring 到 string ):

from lxml import etree
import xmltodict  # pip install xmltodict

def normalise_dict(d):
    """
    Recursively convert dict-like object (eg OrderedDict) into plain dict.
    Sorts list values.
    """
    out = {}
    for k, v in dict(d).items():
        if hasattr(v, 'iteritems'):
            out[k] = normalise_dict(v)
        elif isinstance(v, list):
            out[k] = []
            for item in sorted(v):
                if hasattr(item, 'iteritems'):
                    out[k].append(normalise_dict(item))
                else:
                    out[k].append(item)
        else:
            out[k] = v
    return out


def xml_compare(a, b):
    """
    Compares two XML documents (as string or etree)

    Does not care about element order
    """
    if not isinstance(a, str):
        a = etree.tostring(a)
    if not isinstance(b, str):
        b = etree.tostring(b)
    a = normalise_dict(xmltodict.parse(a))
    b = normalise_dict(xmltodict.parse(b))
    return a == b

maxschlepzig 10 年前

自从 order of attributes is not significant in XML ,您希望忽略由于不同的属性顺序和 XML canonicalization (C14N) 确定性排序属性,您可以使用该方法测试相等性:

xml1 = b'''    <?xml version='1.0' encoding='utf-8' standalone='yes'?>
    <Stats start="1275955200" end="1276041599"></Stats>'''
xml2 = b'''     <?xml version='1.0' encoding='utf-8' standalone='yes'?>
    <Stats end="1276041599" start="1275955200"></Stats>'''
xml3 = b''' <?xml version='1.0' encoding='utf-8' standalone='yes'?>
    <Stats start="1275955200"></Stats>'''

import lxml.etree

tree1 = lxml.etree.fromstring(xml1.strip())
tree2 = lxml.etree.fromstring(xml2.strip())
tree3 = lxml.etree.fromstring(xml3.strip())

import io

b1 = io.BytesIO()
b2 = io.BytesIO()
b3 = io.BytesIO()

tree1.getroottree().write_c14n(b1)
tree2.getroottree().write_c14n(b2)
tree3.getroottree().write_c14n(b3)

assert b1.getvalue() == b2.getvalue()
assert b1.getvalue() != b3.getvalue()

注意,这个例子假设使用python 3。对于python 3,使用 b'''...''' 弦乐和 io.BytesIO 是必需的,而对于python 2,此方法也适用于普通字符串和 io.StringIO .

janbrohl 9 年前

Simpletal使用自定义xml.sax处理程序比较XML文档 https://github.com/janbrohl/SimpleTAL/blob/python2/tests/TALTests/XMLTests/TALAttributeTestCases.py#L47-L112 (比较getxmlchesum的结果) 但是我更喜欢生成一个列表而不是MD5哈希

Pankaj Raheja 8 年前

下面的代码段怎么样?可以很容易地增强以包括属性:

def separator(self):
    return "!@#$%^&*" # Very ugly separator

def _traverseXML(self, xmlElem, tags, xpaths):
    tags.append(xmlElem.tag)
    for e in xmlElem:
        self._traverseXML(e, tags, xpaths)

    text = ''
    if (xmlElem.text):
        text = xmlElem.text.strip()

    xpaths.add("/".join(tags) + self.separator() + text)
    tags.pop()

def _xmlToSet(self, xml):
    xpaths = set() # output
    tags = list()
    root = ET.fromstring(xml)
    self._traverseXML(root, tags, xpaths)

    return xpaths

def _areXMLsAlike(self, xml1, xml2):
    xpaths1 = self._xmlToSet(xml1)
    xpaths2 = self._xmlToSet(xml2)`enter code here`

    return xpaths1 == xpaths2