Java – How to validate a xml file against a given xsd file while parsing it with a sax parser

javaxmlxsd

I want to parse a xml file using a SAXParser or XMLReader and verify that the file conforms to a specific xsd file (new File( "example.xsd" )).

It's easy to

do the validation against a xsd file in an extra step using a Validator like in this SO answer.
to validate while parsing by specifying the name of the xsd as "http://apache.org/xml/properties/schema/external-noNamespaceSchemaLocation" like in this SO answer.

But how can I validate against a new File( "example.xsd" ) while parsing?

Best Answer

Assuming Java 5 or above, set the schema on the SAXParserFactory:

SchemaFactory schemaFactory = SchemaFactory
    .newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
Schema schema = schemaFactory.newSchema(new File("myschema.xsd"));
SAXParserFactory saxFactory = SAXParserFactory.newInstance();
saxFactory.setSchema(schema);
SAXParser parser = saxFactory.newSAXParser();
parser.parse("data.xml", new DefaultHandler() {
  // TODO: other handler methods
  @Override
  public void error(SAXParseException e) throws SAXException {
    throw e;
  }
});

You handle validation errors by overriding the error method on your handler and acting as you see fit.

Related Solutions

Java – What’s the best way to validate an XML file against an XSD file

The Java runtime library supports validation. Last time I checked this was the Apache Xerces parser under the covers. You should probably use a javax.xml.validation.Validator.

import javax.xml.XMLConstants;
import javax.xml.transform.Source;
import javax.xml.transform.stream.StreamSource;
import javax.xml.validation.*;
import java.net.URL;
import org.xml.sax.SAXException;
//import java.io.File; // if you use File
import java.io.IOException;
...
URL schemaFile = new URL("http://host:port/filename.xsd");
// webapp example xsd: 
// URL schemaFile = new URL("http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd");
// local file example:
// File schemaFile = new File("/location/to/localfile.xsd"); // etc.
Source xmlFile = new StreamSource(new File("web.xml"));
SchemaFactory schemaFactory = SchemaFactory
    .newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
try {
  Schema schema = schemaFactory.newSchema(schemaFile);
  Validator validator = schema.newValidator();
  validator.validate(xmlFile);
  System.out.println(xmlFile.getSystemId() + " is valid");
} catch (SAXException e) {
  System.out.println(xmlFile.getSystemId() + " is NOT valid reason:" + e);
} catch (IOException e) {}

The schema factory constant is the string http://www.w3.org/2001/XMLSchema which defines XSDs. The above code validates a WAR deployment descriptor against the URL http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd but you could just as easily validate against a local file.

You should not use the DOMParser to validate a document (unless your goal is to create a document object model anyway). This will start creating DOM objects as it parses the document - wasteful if you aren't going to use them.

C# – The most performant way to validate XML against XSD

Others have already mentioned the XmlReader class for doing the validation, and I wont elaborate further into that.

Your question does not specify much context. Will you be doing this validation repeatedly for several xml documents, or just once? I'm reading a scenario where you are just validating a lot of xml documents (from a third party system?) and storing them for future use.

My contribution to the performance hunt would be to use a compiled XmlSchemaSet which would be thread safe, so several threads can reuse it without needing to parse the xsd document again.

var xmlSchema = XmlSchema.Read(stream, null);
var xmlSchemaSet = new XmlSchemaSet();
xmlSchemaSet.Add(xmlSchema);
xmlSchemaSet.Compile();

CachedSchemas.Add(name, xmlSchemaSet);

Best Answer

Related Solutions

Java – What’s the best way to validate an XML file against an XSD file

C# – The most performant way to validate XML against XSD

Related Topic