如何在没有本地XSD文件的情况下根据XML模式验证XML？

Question

This is my XML:

<?xml version="1.0"?>
<root xmlns="http://example.com/first-schema.xsd"
  xmlns:f="http://example.com/second-schema.xsd">
  <f:foo>test</f:foo>
</root>

Now I want to validate whether this XML is XMLSchema-valid. I don't have these first-schema.xsd and second-schema.xsd files locally. Moreover, I don't know anything about them. I just want to make sure that my XML document is valid against its schemas. Is it possible to do in Java?

Answer 1

You can use the javax.xml.validation APIs for this.

SchemaFactory sf = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI); 
URL schemaURL = // The URL to your XML Schema; 
Schema schema = sf.newSchema(schemaURL); 
Validator validator = schema.newValidator();
DOMSource source = new DOMSource(xmlDOM);
validator.validate(source);

The example below demonstrates how to validate a JAXB object model against a schema, but you'll see it's easy to replace the JAXBSource with a DOMSource for DOM:

• http://bdoughan.blogspot.com/2010/11/validate-jaxb-object-model-with-xml.html