在Java中比较两个XML文档的最佳方法。

I'm trying to write an automated test of an application that basically translates a custom message format into an XML message and sends it out the other end. I've got a good set of input/output message pairs so all I need to do is send the input messages in and listen for the XML message to come out the other end.

我正在尝试编写一个应用程序的自动化测试，它基本上将自定义消息格式转换为XML消息，并将其发送到另一端。我有一组很好的输入/输出消息对，所以我需要做的就是发送输入消息，并侦听XML消息从另一端出来。

When it comes time to compare the actual output to the expected output I'm running into some problems. My first thought was just to do string comparisons on the expected and actual messages. This doens't work very well because the example data we have isn't always formatted consistently and there are often times different aliases used for the XML namespace (and sometimes namespaces aren't used at all.)

当需要将实际输出与预期输出进行比较时，我遇到了一些问题。我的第一个想法是对预期的和实际的消息进行字符串比较。这不是很好，因为我们所使用的示例数据并不总是一致的格式，而且常常会有不同的别名用于XML名称空间(有时甚至不使用名称空间)。

I know I can parse both strings and then walk through each element and compare them myself and this wouldn't be too difficult to do, but I get the feeling there's a better way or a library I could leverage.

我知道我可以对两个字符串进行分析然后遍历每个元素并将它们进行比较，这不会太难，但是我感觉有更好的方法或者是一个可以利用的库。

So, boiled down, the question is:

所以，问题是:

Given two Java Strings which both contain valid XML how would you go about determining if they are semantically equivalent? Bonus points if you have a way to determine what the differences are.

给定两个包含有效XML的Java字符串，您将如何确定它们是否具有语义等价性?如果你有一种方法来确定这些差异是什么，那就加分了。

14 个解决方案

#1

Sounds like a job for XMLUnit

听起来像是XMLUnit的工作。

• http://www.xmlunit.org/
• http://www.xmlunit.org/
• https://github.com/xmlunit
• https://github.com/xmlunit

Example:

例子:

public class SomeTest extends XMLTestCase {
  @Test
  public void test() {
    String xml1 = ...
    String xml2 = ...

    XMLUnit.setIgnoreWhitespace(true); // ignore whitespace differences

    // can also compare xml Documents, InputSources, Readers, Diffs
    assertXMLEquals(xml1, xml2);  // assertXMLEquals comes from XMLTestCase
  }
}

#2

The following will check if the documents are equal using standard JDK libraries.

下面将检查这些文档是否与使用标准JDK库相同。

DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
dbf.setNamespaceAware(true);
dbf.setCoalescing(true);
dbf.setIgnoringElementContentWhitespace(true);
dbf.setIgnoringComments(true);
DocumentBuilder db = dbf.newDocumentBuilder();

Document doc1 = db.parse(new File("file1.xml"));
doc1.normalizeDocument();

Document doc2 = db.parse(new File("file2.xml"));
doc2.normalizeDocument();

Assert.assertTrue(doc1.isEqualNode(doc2));

normalize() is there to make sure there are no cycles (there technically wouldn't be any)

规范化()是为了确保没有循环(技术上不会有任何周期)

The above code will require the white spaces to be the same within the elements though, because it preserves and evaluates it. The standard XML parser that comes with Java does not allow you to set a feature to provide a canonical version or understand xml:space if that is going to be a problem then you may need a replacement XML parser such as xerces or use JDOM.

上面的代码将要求在元素中保持相同的空白，因为它保存并计算了它。使用Java的标准XML解析器不允许您设置一个特性来提供规范版本或理解XML:如果这将是一个问题，那么您可能需要一个替换的XML解析器，例如xerces或使用JDOM。

#3

Xom has a Canonicalizer utility which turns your DOMs into a regular form, which you can then stringify and compare. So regardless of whitespace irregularities or attribute ordering, you can get regular, predictable comparisons of your documents.

Xom有一个规范化的实用程序，它将您的dom转换成一个规则的形式，您可以将其stringify和compare。因此，无论是否存在空白或属性排序，都可以对文档进行常规的、可预测的比较。

This works especially well in IDEs that have dedicated visual String comparators, like Eclipse. You get a visual representation of the semantic differences between the documents.

这在具有专用可视字符串比较器(如Eclipse)的ide中尤其有效。您可以看到文档之间语义差异的可视化表示。

#4

The latest version of XMLUnit can help the job of asserting two XML are equal. Also XMLUnit.setIgnoreWhitespace() and XMLUnit.setIgnoreAttributeOrder() may be necessary to the case in question.

XMLUnit的最新版本可以帮助断言两个XML是相等的。还有XMLUnit.setIgnoreWhitespace()和XMLUnit.setIgnoreAttributeOrder()可能对该案例有必要。

See working code of a simple example of XML Unit use below.

参见下面的一个简单的XML单元示例的工作代码。

import org.custommonkey.xmlunit.DetailedDiff;
import org.custommonkey.xmlunit.XMLUnit;
import org.junit.Assert;

public class TestXml {

    public static void main(String[] args) throws Exception {
        String result = "<abc             attr=\"value1\"                title=\"something\">            </abc>";
        // will be ok
        assertXMLEquals("<abc attr=\"value1\" title=\"something\"></abc>", result);
    }

    public static void assertXMLEquals(String expectedXML, String actualXML) throws Exception {
        XMLUnit.setIgnoreWhitespace(true);
        XMLUnit.setIgnoreAttributeOrder(true);

        DetailedDiff diff = new DetailedDiff(XMLUnit.compareXML(expectedXML, actualXML));

        List<?> allDifferences = diff.getAllDifferences();
        Assert.assertEquals("Differences found: "+ diff.toString(), 0, allDifferences.size());
    }

}

If using Maven, add this to your pom.xml:

如果使用Maven，将其添加到您的pom.xml:

<dependency>
    <groupId>xmlunit</groupId>
    <artifactId>xmlunit</artifactId>
    <version>1.4</version>
</dependency>

#5

Thanks, I extended this, try this ...

谢谢，我扩展了这个，试试这个…

import java.io.ByteArrayInputStream;
import java.util.LinkedHashMap;
import java.util.List;
import java.util.Map;

import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;

import org.w3c.dom.Document;
import org.w3c.dom.NamedNodeMap;
import org.w3c.dom.Node;

public class XmlDiff 
{
    private boolean nodeTypeDiff = true;
    private boolean nodeValueDiff = true;

    public boolean diff( String xml1, String xml2, List<String> diffs ) throws Exception
    {
        DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
        dbf.setNamespaceAware(true);
        dbf.setCoalescing(true);
        dbf.setIgnoringElementContentWhitespace(true);
        dbf.setIgnoringComments(true);
        DocumentBuilder db = dbf.newDocumentBuilder();


        Document doc1 = db.parse(new ByteArrayInputStream(xml1.getBytes()));
        Document doc2 = db.parse(new ByteArrayInputStream(xml2.getBytes()));

        doc1.normalizeDocument();
        doc2.normalizeDocument();

        return diff( doc1, doc2, diffs );

    }

    /**
     * Diff 2 nodes and put the diffs in the list 
     */
    public boolean diff( Node node1, Node node2, List<String> diffs ) throws Exception
    {
        if( diffNodeExists( node1, node2, diffs ) )
        {
            return true;
        }

        if( nodeTypeDiff )
        {
            diffNodeType(node1, node2, diffs );
        }

        if( nodeValueDiff )
        {
            diffNodeValue(node1, node2, diffs );
        }


        System.out.println(node1.getNodeName() + "/" + node2.getNodeName());

        diffAttributes( node1, node2, diffs );
        diffNodes( node1, node2, diffs );

        return diffs.size() > 0;
    }

    /**
     * Diff the nodes
     */
    public boolean diffNodes( Node node1, Node node2, List<String> diffs ) throws Exception
    {
        //Sort by Name
        Map<String,Node> children1 = new LinkedHashMap<String,Node>();      
        for( Node child1 = node1.getFirstChild(); child1 != null; child1 = child1.getNextSibling() )
        {
            children1.put( child1.getNodeName(), child1 );
        }

        //Sort by Name
        Map<String,Node> children2 = new LinkedHashMap<String,Node>();      
        for( Node child2 = node2.getFirstChild(); child2!= null; child2 = child2.getNextSibling() )
        {
            children2.put( child2.getNodeName(), child2 );
        }

        //Diff all the children1
        for( Node child1 : children1.values() )
        {
            Node child2 = children2.remove( child1.getNodeName() );
            diff( child1, child2, diffs );
        }

        //Diff all the children2 left over
        for( Node child2 : children2.values() )
        {
            Node child1 = children1.get( child2.getNodeName() );
            diff( child1, child2, diffs );
        }

        return diffs.size() > 0;
    }


    /**
     * Diff the nodes
     */
    public boolean diffAttributes( Node node1, Node node2, List<String> diffs ) throws Exception
    {        
        //Sort by Name
        NamedNodeMap nodeMap1 = node1.getAttributes();
        Map<String,Node> attributes1 = new LinkedHashMap<String,Node>();        
        for( int index = 0; nodeMap1 != null && index < nodeMap1.getLength(); index++ )
        {
            attributes1.put( nodeMap1.item(index).getNodeName(), nodeMap1.item(index) );
        }

        //Sort by Name
        NamedNodeMap nodeMap2 = node2.getAttributes();
        Map<String,Node> attributes2 = new LinkedHashMap<String,Node>();        
        for( int index = 0; nodeMap2 != null && index < nodeMap2.getLength(); index++ )
        {
            attributes2.put( nodeMap2.item(index).getNodeName(), nodeMap2.item(index) );

        }

        //Diff all the attributes1
        for( Node attribute1 : attributes1.values() )
        {
            Node attribute2 = attributes2.remove( attribute1.getNodeName() );
            diff( attribute1, attribute2, diffs );
        }

        //Diff all the attributes2 left over
        for( Node attribute2 : attributes2.values() )
        {
            Node attribute1 = attributes1.get( attribute2.getNodeName() );
            diff( attribute1, attribute2, diffs );
        }

        return diffs.size() > 0;
    }
    /**
     * Check that the nodes exist
     */
    public boolean diffNodeExists( Node node1, Node node2, List<String> diffs ) throws Exception
    {
        if( node1 == null && node2 == null )
        {
            diffs.add( getPath(node2) + ":node " + node1 + "!=" + node2 + "\n" );
            return true;
        }

        if( node1 == null && node2 != null )
        {
            diffs.add( getPath(node2) + ":node " + node1 + "!=" + node2.getNodeName() );
            return true;
        }

        if( node1 != null && node2 == null )
        {
            diffs.add( getPath(node1) + ":node " + node1.getNodeName() + "!=" + node2 );
            return true;
        }

        return false;
    }

    /**
     * Diff the Node Type
     */
    public boolean diffNodeType( Node node1, Node node2, List<String> diffs ) throws Exception
    {       
        if( node1.getNodeType() != node2.getNodeType() ) 
        {
            diffs.add( getPath(node1) + ":type " + node1.getNodeType() + "!=" + node2.getNodeType() );
            return true;
        }

        return false;
    }

    /**
     * Diff the Node Value
     */
    public boolean diffNodeValue( Node node1, Node node2, List<String> diffs ) throws Exception
    {       
        if( node1.getNodeValue() == null && node2.getNodeValue() == null )
        {
            return false;
        }

        if( node1.getNodeValue() == null && node2.getNodeValue() != null )
        {
            diffs.add( getPath(node1) + ":type " + node1 + "!=" + node2.getNodeValue() );
            return true;
        }

        if( node1.getNodeValue() != null && node2.getNodeValue() == null )
        {
            diffs.add( getPath(node1) + ":type " + node1.getNodeValue() + "!=" + node2 );
            return true;
        }

        if( !node1.getNodeValue().equals( node2.getNodeValue() ) )
        {
            diffs.add( getPath(node1) + ":type " + node1.getNodeValue() + "!=" + node2.getNodeValue() );
            return true;
        }

        return false;
    }


    /**
     * Get the node path
     */
    public String getPath( Node node )
    {
        StringBuilder path = new StringBuilder();

        do
        {           
            path.insert(0, node.getNodeName() );
            path.insert( 0, "/" );
        }
        while( ( node = node.getParentNode() ) != null );

        return path.toString();
    }
}

#6

skaffman seems to be giving a good answer.

斯卡弗曼似乎给出了一个很好的答案。

another way is probably to format the XML using a commmand line utility like xmlstarlet(http://xmlstar.sourceforge.net/) and then format both the strings and then use any diff utility(library) to diff the resulting output files. I don't know if this is a good solution when issues are with namespaces.

另一种方法可能是使用像xmlstarlet(http://xmlstar.sourceforge.net/)这样的命令行实用程序格式化XML，然后格式化字符串，然后使用任何diff实用程序(库)来对生成的输出文件进行diff处理。我不知道这是不是一个好的解决方案，当问题有名称空间的时候。

#7

Building on Tom's answer, here's an example using XMLUnit v2.

根据Tom的答案，这里有一个使用XMLUnit v2的例子。

It uses these maven dependencies

它使用这些maven依赖项。

    <dependency>
        <groupId>org.xmlunit</groupId>
        <artifactId>xmlunit-core</artifactId>
        <version>2.0.0</version>
        <scope>test</scope>
    </dependency>
    <dependency>
        <groupId>org.xmlunit</groupId>
        <artifactId>xmlunit-matchers</artifactId>
        <version>2.0.0</version>
        <scope>test</scope>
    </dependency>

..and here's the test code

. .这是测试代码。

import static org.junit.Assert.assertThat;
import static org.xmlunit.matchers.CompareMatcher.isIdenticalTo;
import org.xmlunit.builder.Input;
import org.xmlunit.input.WhitespaceStrippedSource;

public class SomeTest extends XMLTestCase {
    @Test
    public void test() {
        String result = "<root></root>";
        String expected = "<root>  </root>";

        // ignore whitespace differences
        // https://github.com/xmlunit/user-guide/wiki/Providing-Input-to-XMLUnit#whitespacestrippedsource
        assertThat(result, isIdenticalTo(new WhitespaceStrippedSource(Input.from(expected).build())));

        assertThat(result, isIdenticalTo(Input.from(expected).build())); // will fail due to whitespace differences
    }
}

The documentation that outlines this is https://github.com/xmlunit/xmlunit#comparing-two-documents

概述这一点的文档是https://github.com/xmlunit/xmlunit#比较-两个文档。

#8

I'm using Altova DiffDog which has options to compare XML files structurally (ignoring string data).

我使用的是Altova DiffDog，它可以在结构上比较XML文件(忽略字符串数据)。

This means that (if checking the 'ignore text' option):

这意味着(如果检查“忽略文本”选项):

<foo a="xxx" b="xxx">xxx</foo>

and

和

<foo b="yyy" a="yyy">yyy</foo> 

are equal in the sense that they have structural equality. This is handy if you have example files that differ in data, but not structure!

它们在结构上是平等的。如果您有不同于数据的示例文件，而不是结构，这很方便。

#9

This will compare full string XMLs (reformatting them on the way). It makes it easy to work with your IDE (IntelliJ, Eclipse), cos you just click and visually see the difference in the XML files.

这将比较完整的字符串XMLs(在路上重新格式化它们)。它使使用IDE (IntelliJ, Eclipse)变得容易，因为您只需单击并直观地看到XML文件中的差异。

import org.apache.xml.security.c14n.CanonicalizationException;
import org.apache.xml.security.c14n.Canonicalizer;
import org.apache.xml.security.c14n.InvalidCanonicalizerException;
import org.w3c.dom.Element;
import org.w3c.dom.bootstrap.DOMImplementationRegistry;
import org.w3c.dom.ls.DOMImplementationLS;
import org.w3c.dom.ls.LSSerializer;
import org.xml.sax.InputSource;
import org.xml.sax.SAXException;

import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import javax.xml.transform.TransformerException;
import java.io.IOException;
import java.io.StringReader;

import static org.apache.xml.security.Init.init;
import static org.junit.Assert.assertEquals;

public class XmlUtils {
    static {
        init();
    }

    public static String toCanonicalXml(String xml) throws InvalidCanonicalizerException, ParserConfigurationException, SAXException, CanonicalizationException, IOException {
        Canonicalizer canon = Canonicalizer.getInstance(Canonicalizer.ALGO_ID_C14N_OMIT_COMMENTS);
        byte canonXmlBytes[] = canon.canonicalize(xml.getBytes());
        return new String(canonXmlBytes);
    }

    public static String prettyFormat(String input) throws TransformerException, ParserConfigurationException, IOException, SAXException, InstantiationException, IllegalAccessException, ClassNotFoundException {
        InputSource src = new InputSource(new StringReader(input));
        Element document = DocumentBuilderFactory.newInstance().newDocumentBuilder().parse(src).getDocumentElement();
        Boolean keepDeclaration = input.startsWith("<?xml");
        DOMImplementationRegistry registry = DOMImplementationRegistry.newInstance();
        DOMImplementationLS impl = (DOMImplementationLS) registry.getDOMImplementation("LS");
        LSSerializer writer = impl.createLSSerializer();
        writer.getDomConfig().setParameter("format-pretty-print", Boolean.TRUE);
        writer.getDomConfig().setParameter("xml-declaration", keepDeclaration);
        return writer.writeToString(document);
    }

    public static void assertXMLEqual(String expected, String actual) throws ParserConfigurationException, IOException, SAXException, CanonicalizationException, InvalidCanonicalizerException, TransformerException, IllegalAccessException, ClassNotFoundException, InstantiationException {
        String canonicalExpected = prettyFormat(toCanonicalXml(expected));
        String canonicalActual = prettyFormat(toCanonicalXml(actual));
        assertEquals(canonicalExpected, canonicalActual);
    }
}

I prefer this to XmlUnit because the client code (test code) is cleaner.

我更倾向于使用XmlUnit，因为客户机代码(测试代码)更干净。

#10

AssertJ 1.4+ has specific assertions to compare XML content:

AssertJ 1.4+有特定的断言来比较XML内容:

String expectedXml = "<foo />";
String actualXml = "<bar />";
assertThat(actualXml).isXmlEqualTo(expectedXml);

Here is the Documentation

这是文档

#11

Using JExamXML with java application

使用java应用程序使用JExamXML。

    import com.a7soft.examxml.ExamXML;
    import com.a7soft.examxml.Options;

       .................

       // Reads two XML files into two strings
       String s1 = readFile("orders1.xml");
       String s2 = readFile("orders.xml");

       // Loads options saved in a property file
       Options.loadOptions("options");

       // Compares two Strings representing XML entities
       System.out.println( ExamXML.compareXMLString( s1, s2 ) );

#12

I required the same functionality as requested in the main question. As I was not allowed to use any 3rd party libraries, I have created my own solution basing on @Archimedes Trajano solution.

我需要在主要问题中要求相同的功能。由于我不允许使用任何第三方库，所以我基于@Archimedes Trajano解决方案创建了自己的解决方案。

Following is my solution.

以下是我的解决方案。

import java.io.ByteArrayInputStream;
import java.nio.charset.Charset;
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;

import org.junit.Assert;
import org.w3c.dom.Document;

/**
 * Asserts for asserting XML strings.
 */
public final class AssertXml {

    private AssertXml() {
    }

    private static Pattern NAMESPACE_PATTERN = Pattern.compile("xmlns:(ns\\d+)=\"(.*?)\"");

    /**
     * Asserts that two XML are of identical content (namespace aliases are ignored).
     * 
     * @param expectedXml expected XML
     * @param actualXml actual XML
     * @throws Exception thrown if XML parsing fails
     */
    public static void assertEqualXmls(String expectedXml, String actualXml) throws Exception {
        // Find all namespace mappings
        Map<String, String> fullnamespace2newAlias = new HashMap<String, String>();
        generateNewAliasesForNamespacesFromXml(expectedXml, fullnamespace2newAlias);
        generateNewAliasesForNamespacesFromXml(actualXml, fullnamespace2newAlias);

        for (Entry<String, String> entry : fullnamespace2newAlias.entrySet()) {
            String newAlias = entry.getValue();
            String namespace = entry.getKey();
            Pattern nsReplacePattern = Pattern.compile("xmlns:(ns\\d+)=\"" + namespace + "\"");
            expectedXml = transletaNamespaceAliasesToNewAlias(expectedXml, newAlias, nsReplacePattern);
            actualXml = transletaNamespaceAliasesToNewAlias(actualXml, newAlias, nsReplacePattern);
        }

        // nomralize namespaces accoring to given mapping

        DocumentBuilder db = initDocumentParserFactory();

        Document expectedDocuemnt = db.parse(new ByteArrayInputStream(expectedXml.getBytes(Charset.forName("UTF-8"))));
        expectedDocuemnt.normalizeDocument();

        Document actualDocument = db.parse(new ByteArrayInputStream(actualXml.getBytes(Charset.forName("UTF-8"))));
        actualDocument.normalizeDocument();

        if (!expectedDocuemnt.isEqualNode(actualDocument)) {
            Assert.assertEquals(expectedXml, actualXml); //just to better visualize the diffeences i.e. in eclipse
        }
    }


    private static DocumentBuilder initDocumentParserFactory() throws ParserConfigurationException {
        DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
        dbf.setNamespaceAware(false);
        dbf.setCoalescing(true);
        dbf.setIgnoringElementContentWhitespace(true);
        dbf.setIgnoringComments(true);
        DocumentBuilder db = dbf.newDocumentBuilder();
        return db;
    }

    private static String transletaNamespaceAliasesToNewAlias(String xml, String newAlias, Pattern namespacePattern) {
        Matcher nsMatcherExp = namespacePattern.matcher(xml);
        if (nsMatcherExp.find()) {
            xml = xml.replaceAll(nsMatcherExp.group(1) + "[:]", newAlias + ":");
            xml = xml.replaceAll(nsMatcherExp.group(1) + "=", newAlias + "=");
        }
        return xml;
    }

    private static void generateNewAliasesForNamespacesFromXml(String xml, Map<String, String> fullnamespace2newAlias) {
        Matcher nsMatcher = NAMESPACE_PATTERN.matcher(xml);
        while (nsMatcher.find()) {
            if (!fullnamespace2newAlias.containsKey(nsMatcher.group(2))) {
                fullnamespace2newAlias.put(nsMatcher.group(2), "nsTr" + (fullnamespace2newAlias.size() + 1));
            }
        }
    }

}

It compares two XML strings and takes care of any mismatching namespace mappings by translating them to unique values in both input strings.

它将两个XML字符串进行比较，并通过将它们转换为输入字符串中的惟一值来处理任何不匹配的名称空间映射。

Can be fine tuned i.e. in case of translation of namespaces. But for my requirements just does the job.

可以进行微调，例如在名称空间转换的情况下。但就我的要求而言，这就是工作。

#13

Below code works for me

下面的代码对我有用。

String xml1 = ...
String xml2 = ...
XMLUnit.setIgnoreWhitespace(true);
XMLUnit.setIgnoreAttributeOrder(true);
XMLAssert.assertXMLEqual(actualxml, xmlInDb);

#14

Since you say "semantically equivalent" I assume you mean that you want to do more than just literally verify that the xml outputs are (string) equals, and that you'd want something like

既然您说“语义等价”，我假设您的意思是您想要做的不仅仅是验证xml输出是(string)等于，并且您想要类似的东西。

<foo> some stuff here</foo></code>

< foo >一些东西< / foo > < /代码>

and

和

<foo>some stuff here</foo></code>

< foo >一些东西< / foo > < /代码>

do read as equivalent. Ultimately it's going to matter how you're defining "semantically equivalent" on whatever object you're reconstituting the message from. Simply build that object from the messages and use a custom equals() to define what you're looking for.

读是等价的。最终，无论你要从哪个对象上重新定义信息，你都要定义“语义对等”。只需从消息中构建该对象并使用自定义equals()来定义您要查找的内容。