开发者

performing xml validation against xsd

开发者 https://www.devze.com 2023-02-17 10:48 出处:网络
I have XML as a string and an XSD as a file, and I nee开发者_StackOverflowd to validate the XML with the XSD. How can I do this?You can use javax.xml.validation API to do this.

I have XML as a string and an XSD as a file, and I nee开发者_StackOverflowd to validate the XML with the XSD. How can I do this?


You can use javax.xml.validation API to do this.

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;
}


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

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)));
0

精彩评论

暂无评论...
验证码 换一张
取 消

关注公众号