对xsd执行xml验证

时间:2011-03-18 12:16:31

标签: java xml saxparser

我将XML作为字符串,将XSD作为文件,我需要使用XSD验证XML。我怎么能这样做?

2 个答案:

答案 0 :(得分:11)

您可以使用javax.xml.validation API执行此操作。

public boolean validate(String inputXml, String schemaLocation)
  throws SAXException, IOException {
  // build the schema
  SchemaFactory factory = SchemaFactory.newInstance("http://www.w3.org/2001/XMLSchema");
  File schemaFile = new File(schemaLocation);
  Schema schema = factory.newSchema(schemaFile);
  Validator validator = schema.newValidator();

  // create a source from a string
  Source source = new StreamSource(new StringReader(inputXml));

  // check input
  boolean isValid = true;
  try  {

    validator.validate(source);
  } 
  catch (SAXException e) {

    System.err.println("Not valid");
    isValid = false;
  }

  return isValid;
}

答案 1 :(得分:1)

您可以使用javax.xml.validation API:

String xml = "<root/>";  // XML as String
File xsd = new File("schema.xsd");  // XSD as File

SchemaFactory sf = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
Schema schema = sf.newSchema(xsd); 

SAXParserFactory spf = SAXParserFactory.newInstance();
spf.setSchema(schema);
SAXParser sp = spf.newSAXParser();
XMLReader xr = sp.getXMLReader();
xr.parse(new InputSource(new StringReader(xml)));