java - 如何重新绘制我的 JTable

Question

所以我有一个带有 JTable 的应用程序。 JTable 位于 JScrollPane 内部，而 JScrollPane 绘制在 JFrame 上。

现在，在我的应用程序中，我打开一个新窗口以向该表添加新行，然后单击按钮保存更改后，新窗口将关闭。

现在我尝试在新窗口关闭后添加这些行:

askTableInfo(); //a method to save the info in database to table and then save the table to variable 'table'
table.repaint();
scrollPane.repaint(); 

当然还有 repaint();通过它自己。但它似乎仍然没有更新我在 JFrame 中的表。

这里可能出现什么问题？

public class AppWindow extends JFrame implements ActionListener {
    String user = "";
    JLabel greetText = new JLabel();
    JPanel panel = new JPanel();
    JPanel panel2 = new JPanel(new GridLayout(1, 3));
    JScrollPane scrollPane;
    JTable tabel;
    JButton newBook = new JButton("Add a book");
    JButton deleteBook = new JButton("Remove a book");
    JButton changeBook = new JButton("Change a book");
    int ID;

    public AppWindow(String user, int ID) {
        this.ID = ID;
        this.user = user;
        setSize(500, 500);
        setTitle("Books");

        setLayout(new BorderLayout());
        greetText.setText("Hi "+user+" here are your books:");
        add(greetText, BorderLayout.NORTH); 

        askData();  
        panel.add(scrollPane);
        add(panel, BorderLayout.CENTER);

        panel2.add(newBook);
        panel2.add(deleteBook);
        panel2.add(changeBook);
        add(paneel2, BorderLayout.SOUTH);

        newBook.addActionListener(this);

        setVisible(true);
    }

    private void askData() {
        DataAsker asker = null;
        try {
            asker = new AndmeKysija(ID);
        } catch (SQLException e) {
            e.printStackTrace();
        }
        table = asker.giveTable();
        scrollPane = new JScrollPane(tabel);
    }

    public static void main(String[] args){
        AppWindow window = new AppWindow("Name", 2);
    }

    @Override
    public void actionPerformed(ActionEvent e) {
        if(e.getSource() == newBook){
                new BookAdded(ID);
                panel.revalidate();
                panel.add(scrollPane);
                panel.repaint();
                repaint();
        }   
    }
}

Answer 1

不是最好的方法，但它会让你越过界限......

public AppWindow(String user, int ID) {
    this.ID = ID;
    this.user = user;
    setSize(500, 500);
    setTitle("Books");

    setLayout(new BorderLayout());
    greetText.setText("Hi "+user+" here are your books:");
    add(greetText, BorderLayout.NORTH); 

    JTable table = askData();  
    scrollPane.setViewportView(table);
    panel.add(scrollPane);
    add(panel, BorderLayout.CENTER);

    panel2.add(newBook);
    panel2.add(deleteBook);
    panel2.add(changeBook);
    add(paneel2, BorderLayout.SOUTH);

    newBook.addActionListener(this);

    setVisible(true);
}

private JTable askData() {
    DataAsker asker = null;
    try {
        asker = new AndmeKysija(ID);
    } catch (SQLException e) {
        e.printStackTrace();
    }
    return asker.giveTable();
}

public static void main(String[] args){
    AppWindow window = new AppWindow("Name", 2);
}

@Override
public void actionPerformed(ActionEvent e) {
    if(e.getSource() == newBook){
        new BookAdded(ID);
        JTable table = askData();  
        scrollPane.setViewportView(table);
    }   
}

您应该做的是根据 AndmeKysija 的结果创建一个新的 TableModel，AndmeKysija 应该有 UI 的想法或概念。然后，您只需使用 JTable#setModel 来更新 View ...

Swing 使用 Model-View-Controller 的变体范例