Comment valider un fichier XML en utilisant Java avec un XSD ayant une include?

je suis à l'aide de Java 5 javax.XML.validation.Validateur pour valider le fichier XML. Je l'ai fait pour un schéma qui utilise uniquement des importations et tout fonctionne bien. Maintenant j'essaie de valider avec un autre schéma qui utilise import Et one include. Le problème que j'ai est que les éléments du schéma principal sont ignorés, la validation dit qu'il ne peut pas trouver leur déclaration.

<!-Voici comment je construis le schéma:

InputStream includeInputStream = getClass().getClassLoader().getResource("include.xsd").openStream();
InputStream importInputStream = getClass().getClassLoader().getResource("import.xsd").openStream();
InputStream mainInputStream = getClass().getClassLoader().getResource("main.xsd").openStream();
Source[] sourceSchema = new SAXSource[]{includeInputStream , importInputStream, 
mainInputStream };
Schema schema = factory.newSchema(sourceSchema);

Maintenant, voici l'extrait de la déclaration en principal.xsd

<xsd:schema xmlns="http://schema.omg.org/spec/BPMN/2.0" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:import="http://www.foo.com/import" targetNamespace="http://main/namespace" elementFormDefault="qualified" attributeFormDefault="unqualified">
    <xsd:import namespace="http://www.foo.com/import" schemaLocation="import.xsd"/>
    <xsd:include schemaLocation="include.xsd"/>
    <xsd:element name="element" type="tElement"/>
    <...>
</xsd:schema>

si je copie le code de mon xsd inclus dans le main.xsd, il fonctionne très bien. Si Je ne le fais pas, la validation ne trouve pas la déclaration de "Element".

25
demandé sur Line 2010-02-26 18:49:50

7 réponses

vous devez utiliser un Isresourceresolver pour que cela fonctionne. veuillez jeter un oeil à l'exemple de code ci-dessous.

une méthode de validation:

// note that if your XML already declares the XSD to which it has to conform, then there's no need to declare the schemaName here
void validate(String xml, String schemaName) throws Exception {

    DocumentBuilderFactory builderFactory = DocumentBuilderFactory.newInstance();
    builderFactory.setNamespaceAware(true);

    DocumentBuilder parser = builderFactory
            .newDocumentBuilder();

    // parse the XML into a document object
    Document document = parser.parse(new StringInputStream(xml));

    SchemaFactory factory = SchemaFactory
            .newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);

    // associate the schema factory with the resource resolver, which is responsible for resolving the imported XSD's
    factory.setResourceResolver(new ResourceResolver());

            // note that if your XML already declares the XSD to which it has to conform, then there's no need to create a validator from a Schema object
    Source schemaFile = new StreamSource(getClass().getClassLoader()
            .getResourceAsStream(schemaName));
    Schema schema = factory.newSchema(schemaFile);

    Validator validator = schema.newValidator();
    validator.validate(new DOMSource(document));
}

la ressource résolveur de mise en œuvre:

public class ResourceResolver  implements LSResourceResolver {

public LSInput resolveResource(String type, String namespaceURI,
        String publicId, String systemId, String baseURI) {

     // note: in this sample, the XSD's are expected to be in the root of the classpath
    InputStream resourceAsStream = this.getClass().getClassLoader()
            .getResourceAsStream(systemId);
    return new Input(publicId, systemId, resourceAsStream);
}

 }

la implémentation D'entrée renvoyée par le résolveur de ressources:

public class Input implements LSInput {

private String publicId;

private String systemId;

public String getPublicId() {
    return publicId;
}

public void setPublicId(String publicId) {
    this.publicId = publicId;
}

public String getBaseURI() {
    return null;
}

public InputStream getByteStream() {
    return null;
}

public boolean getCertifiedText() {
    return false;
}

public Reader getCharacterStream() {
    return null;
}

public String getEncoding() {
    return null;
}

public String getStringData() {
    synchronized (inputStream) {
        try {
            byte[] input = new byte[inputStream.available()];
            inputStream.read(input);
            String contents = new String(input);
            return contents;
        } catch (IOException e) {
            e.printStackTrace();
            System.out.println("Exception " + e);
            return null;
        }
    }
}

public void setBaseURI(String baseURI) {
}

public void setByteStream(InputStream byteStream) {
}

public void setCertifiedText(boolean certifiedText) {
}

public void setCharacterStream(Reader characterStream) {
}

public void setEncoding(String encoding) {
}

public void setStringData(String stringData) {
}

public String getSystemId() {
    return systemId;
}

public void setSystemId(String systemId) {
    this.systemId = systemId;
}

public BufferedInputStream getInputStream() {
    return inputStream;
}

public void setInputStream(BufferedInputStream inputStream) {
    this.inputStream = inputStream;
}

private BufferedInputStream inputStream;

public Input(String publicId, String sysId, InputStream input) {
    this.publicId = publicId;
    this.systemId = sysId;
    this.inputStream = new BufferedInputStream(input);
}
}
56
répondu Stefan De Boey 2010-02-26 16:12:18

j'ai du faire quelques modifications à ce post by AMegmondoEmber

mon fichier de schéma principal avait quelques includes provenant de dossiers frères, et les fichiers inclus avaient aussi quelques includes provenant de leurs dossiers locaux. J'ai aussi dû retracer le chemin de la ressource de base et le chemin relatif de la ressource actuelle. Ce code fonctionne pour moi savoir, mais s'il vous plaît garder à l'esprit qu'il suppose que tous les fichiers xsd ont un nom unique. Si vous avez des fichiers xsd avec le même nom, mais un contenu différent sur différents chemins, cela vous posera probablement des problèmes.

import java.io.ByteArrayInputStream;
import java.io.InputStream;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
import org.w3c.dom.ls.LSInput;
import org.w3c.dom.ls.LSResourceResolver;

/**
 * The Class ResourceResolver.
 */
public class ResourceResolver implements LSResourceResolver {

    /** The logger. */
    private final Logger logger = LoggerFactory.getLogger(this.getClass());

    /** The schema base path. */
    private final String schemaBasePath;

    /** The path map. */
    private Map<String, String> pathMap = new HashMap<String, String>();

    /**
     * Instantiates a new resource resolver.
     *
     * @param schemaBasePath the schema base path
     */
    public ResourceResolver(String schemaBasePath) {
        this.schemaBasePath = schemaBasePath;
        logger.warn("This LSResourceResolver implementation assumes that all XSD files have a unique name. "
                + "If you have some XSD files with same name but different content (at different paths) in your schema structure, "
                + "this resolver will fail to include the other XSD files except the first one found.");
    }

    /* (non-Javadoc)
     * @see org.w3c.dom.ls.LSResourceResolver#resolveResource(java.lang.String, java.lang.String, java.lang.String, java.lang.String, java.lang.String)
     */
    @Override
    public LSInput resolveResource(String type, String namespaceURI,
            String publicId, String systemId, String baseURI) {
        // The base resource that includes this current resource
        String baseResourceName = null;
        String baseResourcePath = null;
        // Extract the current resource name
        String currentResourceName = systemId.substring(systemId
                .lastIndexOf("/") + 1);

        // If this resource hasn't been added yet
        if (!pathMap.containsKey(currentResourceName)) {
            if (baseURI != null) {
                baseResourceName = baseURI
                        .substring(baseURI.lastIndexOf("/") + 1);
            }

            // we dont need "./" since getResourceAsStream cannot understand it
            if (systemId.startsWith("./")) {
                systemId = systemId.substring(2, systemId.length());
            }

            // If the baseResourcePath has already been discovered, get that
            // from pathMap
            if (pathMap.containsKey(baseResourceName)) {
                baseResourcePath = pathMap.get(baseResourceName);
            } else {
                // The baseResourcePath should be the schemaBasePath
                baseResourcePath = schemaBasePath;
            }

            // Read the resource as input stream
            String normalizedPath = getNormalizedPath(baseResourcePath, systemId);
            InputStream resourceAsStream = this.getClass().getClassLoader()
                    .getResourceAsStream(normalizedPath);

            // if the current resource is not in the same path with base
            // resource, add current resource's path to pathMap
            if (systemId.contains("/")) {
                pathMap.put(currentResourceName, normalizedPath.substring(0,normalizedPath.lastIndexOf("/")+1));
            } else {
                // The current resource should be at the same path as the base
                // resource
                pathMap.put(systemId, baseResourcePath);
            }
            Scanner s = new Scanner(resourceAsStream).useDelimiter("\A");
            String s1 = s.next().replaceAll("\n", " ") // the parser cannot understand elements broken down multiple lines e.g. (<xs:element \n name="buxing">)
                    .replace("\t", " ") // these two about whitespaces is only for decoration
                    .replaceAll("\s+", " ").replaceAll("[^\x20-\x7e]", ""); // some files has a special character as a first character indicating utf-8 file
            InputStream is = new ByteArrayInputStream(s1.getBytes());

            return new LSInputImpl(publicId, systemId, is); // same as Input class
        }

        // If this resource has already been added, do not add the same resource again. It throws
        // "org.xml.sax.SAXParseException: sch-props-correct.2: A schema cannot contain two global components with the same name; this schema contains two occurrences of ..."
        // return null instead.
        return null;
    }

    /**
     * Gets the normalized path.
     *
     * @param basePath the base path
     * @param relativePath the relative path
     * @return the normalized path
     */
    private String getNormalizedPath(String basePath, String relativePath){
        if(!relativePath.startsWith("../")){
            return basePath + relativePath;
        }
        else{
            while(relativePath.startsWith("../")){
                basePath = basePath.substring(0,basePath.substring(0, basePath.length()-1).lastIndexOf("/")+1);
                relativePath = relativePath.substring(3);
            }
            return basePath+relativePath;
        }
    }
}
3
répondu burcakulug 2017-05-23 10:29:47

la réponse acceptée est parfaitement ok, mais ne fonctionne pas avec Java 8 sans quelques modifications. Il serait également utile de pouvoir spécifier un chemin de base à partir duquel les schémas importés sont lus.

j'ai utilisé dans mon Java 8 le code suivant qui permet de spécifier un chemin de schéma intégré autre que le chemin racine:

import com.sun.org.apache.xerces.internal.dom.DOMInputImpl;
import org.w3c.dom.ls.LSInput;
import org.w3c.dom.ls.LSResourceResolver;

import java.io.InputStream;
import java.util.Objects;

public class ResourceResolver implements LSResourceResolver {

    private String basePath;

    public ResourceResolver(String basePath) {
        this.basePath = basePath;
    }

    @Override
    public LSInput resolveResource(String type, String namespaceURI, String publicId, String systemId, String baseURI) {
        // note: in this sample, the XSD's are expected to be in the root of the classpath
        InputStream resourceAsStream = this.getClass().getClassLoader()
                .getResourceAsStream(buildPath(systemId));
        Objects.requireNonNull(resourceAsStream, String.format("Could not find the specified xsd file: %s", systemId));
        return new DOMInputImpl(publicId, systemId, baseURI, resourceAsStream, "UTF-8");
    }

    private String buildPath(String systemId) {
        return basePath == null ? systemId : String.format("%s/%s", basePath, systemId);
    }
}

cette implémentation donne aussi à l'utilisateur un message significatif dans le cas où le schéma ne peut pas être lu.

3
répondu gil.fernandes 2017-09-18 08:22:43

pour nous, la source de résolution ressemblait à ceci. Après quelques exceptions prolog et bizarre Le type d'élément "xs: schema" doit être suivi des spécifications de l'attribut " > "ou"/>". Le type d'élément "xs: element" doit être suivi des spécifications d'attribut " > "ou"/>". (en raison de la répartition des lignes multiples)

l'historique du chemin était nécessaire en raison de la structure de includes

main.xsd (this has include "includes/subPart.xsd")
/includes/subPart.xsd (this has include "./subSubPart.xsd")
/includes/subSubPart.xsd

alors le code ressemble à:

String pathHistory = "";

@Override
public LSInput resolveResource(String type, String namespaceURI, String publicId, String systemId, String baseURI) {
    systemId = systemId.replace("./", "");// we dont need this since getResourceAsStream cannot understand it
    InputStream resourceAsStream = Message.class.getClassLoader().getResourceAsStream(systemId);
    if (resourceAsStream == null) {
        resourceAsStream = Message.class.getClassLoader().getResourceAsStream(pathHistory + systemId);
    } else {
        pathHistory = getNormalizedPath(systemId);
    }
    Scanner s = new Scanner(resourceAsStream).useDelimiter("\A");
    String s1 = s.next()
            .replaceAll("\n"," ") //the parser cannot understand elements broken down multiple lines e.g. (<xs:element \n name="buxing">) 
            .replace("\t", " ") //these two about whitespaces is only for decoration
            .replaceAll("\s+", " ") 
            .replaceAll("[^\x20-\x7e]", ""); //some files has a special character as a first character indicating utf-8 file
    InputStream is = new ByteArrayInputStream(s1.getBytes());

    return new LSInputImpl(publicId, systemId, is);
}

private String getNormalizedPath(String baseURI) {
    return baseURI.substring(0, baseURI.lastIndexOf(System.getProperty("file.separator"))+ 1) ;
}
0
répondu AMegmondoEmber 2014-01-16 04:53:06

si vous ne trouvez pas un élément dans xml, vous obtiendrez l'exception xml:lang. Les éléments sont sensibles à la casse

-1
répondu Ramakrishna 2014-02-24 10:06:00

la réponse acceptée est très verbeuse, et construit un DOM en mémoire d'abord, inclut semble fonctionner hors de la boîte pour moi, y compris les références relatives.

    SchemaFactory schemaFactory = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
    Schema schema = schemaFactory.newSchema(new File("../foo.xsd"));
    Validator validator = schema.newValidator();
    validator.validate(new StreamSource(new File("./foo.xml")));
-1
répondu teknopaul 2016-06-23 12:52:40
SchemaFactory schemaFactory = SchemaFactory
                                .newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
Source schemaFile = new StreamSource(getClass().getClassLoader()
                                .getResourceAsStream("cars-fleet.xsd"));
Schema schema = schemaFactory.newSchema(schemaFile);
Validator validator = schema.newValidator();
StreamSource source = new StreamSource(xml);
validator.validate(source);
-4
répondu valerian 2013-04-18 14:35:29