如何使用公共Digester解析xml文件并使其填充java.util.Date对象？

Question

示例XML.

<person>
  <name>Joe Dirt</name>
  <ssn>123-45-6789</ssn>
  <dob>07/04/1981</dob>
</person>

示例Java类

public class Person(){
  private String name;
  private String ssn;
  private java.util.Date dob;
.....
}

Sample Digester规则

<?xml version="1.0"?>
<digester-rules>
  <pattern value="message">
      <object-create-rule classname="Person" />
      <bean-property-setter-rule pattern="name" propertyname="name" />
      <bean-property-setter-rule pattern="ssn" propertyname="ssn" />
      <bean-property-setter-rule pattern="dob" propertyname="dob" />
 </pattern>
</digester-rules>

Answer 1

应该能够注册一个DateConverter,所有应该工作:

import org.apache.commons.beanutils.Converter;

class MyDateConverter implements Converter {
    @Override
    public Object convert(Class clazz, Object value) {
        if (clazz.equals(Date.class)) {
            SimpleDateFormat sdf = new SimpleDateFormat("dd/MM/yyyy");
            try {
                return sdf.parse((String) value);
            } catch (ParseException pe) {
                throw new IllegalArgumentException(pe);
            } catch (ClassCastException cce) {
                throw new IllegalArgumentException(cce);
            }
        } else {
            throw new IllegalArgumentException("Expected Date class");
        }
    }
}

......

import org.apache.commons.beanutils.ConvertUtils;

@Test
public void testXmlRules() throws Exception {
    ConvertUtils.register(new MyDateConverter(), Date.class);
    Digester digester = DigesterLoader.createDigester(new InputSource(this
            .getClass().getResourceAsStream("rules.xml")));
    Person person = (Person) digester.parse(this.getClass()
            .getResourceAsStream("person.xml"));
    Assert.assertEquals("Joe Dirt", person.getName());
    Assert.assertEquals("123-45-6789", person.getSsn());
    Assert.assertEquals(new SimpleDateFormat("dd/MM/yyyy")
            .parse("07/04/1981"), person.getDob());
}