Hi,
I want to validate an xml file against an xsd schema. The xml files root element does not have any namespace or xsi details. It has no attributes so just .
I have tried the following code from http://www.ibm.com/developerworks/xml/library/x-javaxmlvalidapi.html with no luck as I receive
cvc-elt.1: Cannot find the declaration of element 'root'
SchemaFactory factory = SchemaFactory.newInstance("http://www.w3.org/2001/XMLSchema");
File schemaFile = new File("schema.xsd");
Schema xsdScheme = factory.newSchema(schemaFile);
Validator validator = xsdScheme.newValidator();
Source source = new StreamSource(xmlfile);
validator.validate(source);
The xml validates fine with the namespace headers included etc (added via xmlspy), but I would have thought the xml namespace could be declared without having to manually edit the source file?
Edit and Solution:
public static void validateAgainstXSD(File file) {
try {
SchemaFactory factory = SchemaFactory.newInstance("http://www.w3.org/2001/XMLSchema");
File schemaFile = new File("path/to/xsd");
Schema xsdScheme = factory.newSchema(schemaFile);
Validator validator = xsdScheme.newValidator();
SAXSource source = new SAXSource(
new NamespaceFilter(XMLReaderFactory.createXMLReader()),
new InputSource(new FileInputStream(file)));
validator.validate(source,null);
} catch (Exception e) {
e.printStackTrace();
}
}
protected static class NamespaceFilter extends XMLFilterImpl {
String requiredNamespace = "namespace";
public NamespaceFilter(XMLReader parent) {
super(parent);
}
@Override
public void startElement(String arg0, String arg1, String arg2, Attributes arg3) throws SAXException {
if(arg0 != requiredNamespace) arg0 = requiredNamespace;
super.startElement(arg0, arg1, arg2, arg3);
}
}