比较XML片段？

时间：2010-06-09 15:33:25

标签： python xml diff

在another SO question的基础上，如何检查两个结构良好的XML片段在语义上是否相等。我所需要的只是“平等”与否，因为我正在使用它进行单元测试。

在我想要的系统中，这些是相同的（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，只需要重新排序属性的简单功能也可以正常工作！

基于IanB答案的工作片段：

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

9 个答案:

答案 0 :(得分：25)

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

答案 1 :(得分：14)

元素的顺序在XML中可能很重要，这可能就是为什么大多数其他建议的方法会在顺序不同时进行比较...即使元素具有相同的属性和文本内容。

但我也想要一个对顺序不敏感的比较，所以我提出了这个：

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

答案 2 :(得分：5)

我遇到了同样的问题：我想要比较两个具有相同属性但是顺序不同的文档。

似乎lxml中的XML Canonicalization（C14N）适用于此，但我绝对不是XML专家。我很想知道其他人是否可以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)

运行它给了我：

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

答案 3 :(得分：5)

这是一个简单的解决方案，将XML转换为字典（使用 xmltodict ）并将字典一起比较

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())

或简单的python方法：

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))))

答案 4 :(得分：2)

如果采用DOM方法，您可以同时遍历两棵树，同时比较节点（节点类型，文本，属性）。

递归解决方案将是最优雅的 - 只要一对节点不是“相等”，或者一旦你在一棵树中检测到一个树叶，当它是另一棵树中的树枝时，就会进行短路进一步比较等。

答案 5 :(得分：2)

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

答案 6 :(得分：0)

将Anentropic's great answer改编为Python 3（基本上，将iteritems()更改为items()，将basestring更改为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

答案 7 :(得分：0)

SimpleTAL使用自定义xml.sax处理程序来比较xml文档 https://github.com/janbrohl/SimpleTAL/blob/python2/tests/TALTests/XMLTests/TALAttributeTestCases.py#L47-L112 （比较getXMLChecksum的结果） 但我更喜欢生成列表而不是md5-hash

答案 8 :(得分：0)

以下代码段如何？可以轻松增强，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
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