Question

我正在使用lwuit使用j2me我有一个问题就是

当我在midlet中startApp()时，我首先设置了Display.init(this)

并且运行应用程序lwuit工作正常，但是当我在midlet中使用Form startApp()事件时它工作得很好但是在这种形式的actionevent中我称之为新形式并且以这种新形式当我按下它时，我放了一个后退命令它不会在主midlet上移动

请帮助知道如何使用

import javax.microedition.MIDlet;

import  some lwuit UILibrary

public class mainMiddlet extends MIDlet implement ActionListner
{
      public mainMiddlet(){
                  try{

                       Display.init(this);
                       //somthing is here 
                       form=new Form();

                       form.addActionListener(this);

                     }catch(Exception e){}
       }
       public void actionperformed(ActionEven ae){
                //here i call new form 
                //in action event of this form 
                new form().show();
        }
       //here some middlet default method 


}
public class newForm extends Form {

    //in this form I am put one command back and when i am pressed it 
    // I call mainMiddlet but it throw error internal application java.lang.nullpointer
   // can I back on mainmiddlet from on form to another form 
   // my main problem is I am not move on mainmiddlet for exit middlet because destoryall()
   // is method of middlet 

}

Answer 1

它很简单。您可以在下一个form back命令中调用show()方法。例如，

<强> MainMidlet.java

// create the midlet and write inside of the midlet
final Form form = new Form();

form.addCommand(new Command("Next") {

    public void actionPerformed(ActionEvent evt) {
            new NewForm(form).show();
       }
    });

<强> NewForm.java

   // create the NewForm class and write inside of the class

        public NewForm(final Form form) {
   // Constructor
        addCommand(new Command("Back") {

            public void actionPerformed(ActionEvent evt) {
                    form.show();
               }
            });
    }

使用lwuit UI库我不破坏j2me应用程序

1 个答案: