如何在JAXB中添加多個值？

Question

import javax.xml.bind.annotation.XmlElement; 

     public class Test1 { 
      private String Type; 
      private String value; 
      @XmlElement(name="type") 
      public final String getType(){ 
       return Type; 
      } 
      public final void setType(final String type){ 
       this.Type=type; 
      } 
      @XmlElement(name="value") 
      public final String getValue(){ 
       return value; 
      } 
      public final void setValue(final String value) { 
       this.value = value; 
      } 
     } 
    import java.util.ArrayList; 

    import javax.xml.bind.annotation.XmlElement; 
    import javax.xml.bind.annotation.XmlElementWrapper; 
    import javax.xml.bind.annotation.XmlRootElement; 

    @XmlRootElement(name="test") 
    public class Test2 { 
     ArrayList<Test1> test1; 
     @XmlElementWrapper(name="users") 
     @XmlElement(name="user") 
     public final ArrayList<Test1> getUser(){ 

      return test1; 
     } 
     public final void setUser(final ArrayList<Test1> mm){ 
      this.test1=mm; 
     } 
    } 
import java.util.ArrayList; 

import javax.xml.bind.JAXBContext; 
import javax.xml.bind.JAXBException; 
import javax.xml.bind.Marshaller; 

import com.covidien.laptopagent.xml.XMLUserLogin; 



public class TEST { 

    public static void main(String agrs[]){ 
     Test1 test1=new Test1(); 
     test1.setType("100"); 
     test1.setValue("none"); 
     test1.setType("101"); 
     test1.setValue("ok"); 
     ArrayList<Test1> permfile=new ArrayList<Test1>(); 
     permfile.add(test1); 


     ArrayList<Test2>test2 =new ArrayList<Test2>(); 
     Test2 test2obj=new Test2(); 
     test2obj.setuser(permfile); 




     JAXBContext context; 
     try { 
      context = JAXBContext.newInstance(XMLUserLogin.class); 
      Marshaller m = context.createMarshaller(); 
      m.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, Boolean.TRUE); 
      m.marshal(test2, System.out); 
      System.out.println(); 
     } catch (JAXBException e) { 
      // TODO Auto-generated catch block 
      e.printStackTrace(); 
     } 
    } 

} 

除外輸出：

<?xml version="1.0" encoding="UTF-8" standalone="yes"?> 
<test> 
    ...... 
    <users> 
     <user> 
       <type>100</type> 
       <value>none</value> 
     </user> 
     <user> 
       <type>101</type> 
       <value>ok</value> 
      </user> 
</users> 
............... 
............... 
</test> 

但它返回只有一個用戶的價值：

<?xml version="1.0" encoding="UTF-8" standalone="yes"?> 
<test> 
    ...... 
    <users> 
     <user> 
       <type>100</type> 
       <value>none</value> 
     </user> 
</users> 
............... 
............... 
</test> 

如何解決這個問題？有人能幫助我，我做錯了什麼？

感謝

Answer 1

修復這樣的代碼

Test1 user1=new Test1(); 
    user1.setType("100"); 
    user1.setValue("none"); 

    Test1 user2=new Test1(); 
    user2.setType("101"); 
    user2.setValue("ok"); 

    ArrayList<Test1> permfile=new ArrayList<Test1>(); 
    permfile.add(user1); 
    permfile.add(user2);