Java JTable不会加载

时间:2018-05-12 15:14:09

标签: java swing hashmap jtable

我有一个HashMap<Person, ArrayList<Account>>,我想要一个包含来自HashMap的所有键(人)的表。问题是如果我尝试在视图中创建它,表是空的,但如果我在main中创建它(使用完全相同的代码)它可以工作并填充数据。我不知道为什么。

public class App 
{
public static void main( String[] args )
{

    View view = new View();
    Bank b = new Bank();
    Person p = new Person("Name","Prenume", "id");
    Person p1 = new Person("Name2","Prenume2", "id2");
    b.addPerson(p);
    b.addPerson(p1);

    view.personTbl = view.tablePers(b.mapper);
    view.sp = new JScrollPane(view.personTbl);
    view.personFrame.add(view.sp,BorderLayout.CENTER);
    view.viewPers.setVisible(true);
}
}

如果我像这样创建它不起作用

 public View() {
    personFrame = new JFrame("Person");
    personFrame.setLayout(new BorderLayout());
    personFrame.setSize(400, 400);
    persBtnsPan = new JPanel();
    viewPers = new JFrame("View Pers");
    viewPers.setLayout(new BorderLayout());
    viewPers.setSize(400, 400);
    viewPersPan = new JPanel();
    addPerson = new JButton("add");
    editPerson = new JButton("edit");
    viewP = new JButton("view");

    personTbl = tablePers(b.mapper);
    sp = new JScrollPane(personTbl);

    persBtnsPan.add(addPerson);
    persBtnsPan.add(editPerson);
    persBtnsPan.add(viewP);
    personFrame.add(persBtnsPan, BorderLayout.NORTH);
    personFrame.add(sp,BorderLayout.CENTER);
    }

这是创建表的方法

    public static <T> JTable tablePers(HashMap<Person, ArrayList<Account>> map) {
    DefaultTableModel model = new DefaultTableModel(
        new Object[] { "Nume", "Prenume", "CNP" }, 0);
    for (HashMap.Entry<Person, ArrayList<Account>> entry : map.entrySet()) {
        model.addRow(new Object[] { entry.getKey().getNume(), entry.getKey().getPrenume(), entry.getKey().getCnp() });
    }
    tab = new JTable(model);
    return tab;
}

1 个答案:

答案 0 :(得分:1)

您有参考问题,因为视图中的b.mapper与主视图中的public class App { Bank b = new Bank(); // the model //... do things with b here View view = new View(b); // pass the model in // and then in View's constructor use the model } 不同,但也不应该。但是,View处理这样的数据是没有意义的。您需要直接或间接将模型传递给View,然后让视图显示模型的状态。

如,

public static void main(String[] args) {
    BankModel model = new BankModel();
    model.addPerson(new BankPerson("LastName1", "FirstName1", "id1"));
    model.addPerson(new BankPerson("LastName2", "FirstName2", "id2"));
    BankView view = new BankView();
    BankController controller = new BankController(model, view);
    controller.displayGui();
}

此外,如果这是一个M-V-C程序,那么您的Controller将监听视图和模型的更改,并将通知相关实体更改,更新模型和视图。

如,

jQuery