函数通过将参数作为组件名称返回JTable

时间:2013-05-17 10:56:05

标签: java swing jtable components

public JTable getTable(String Component_name)
    {
        JTable table=new JTable();
    //in this function i want to search all the JTables that have been created on runtime! 
    //and then i want to return one JTable by the name "Component_Name"//

        return table;
    }

这就是我想要的功能;我想出了一个创建一个具有私有JTable表和私有String名称的新类组件的解决方案,但是在按名称搜索JTable时仍然会出现问题。

3 个答案:

答案 0 :(得分:2)

结合@Alican Ozgoren,@ mKorbel和answer的建议,您可以构建Map<String, NamedModel>,以便按名称快速访问TableModel

public TableModel getTable(String name) {
    return map.get(name);
}

以下示例替换单个JTable的模型,而另一个example显示多个表。

Test image

import java.awt.BorderLayout;
import java.awt.EventQueue;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.util.HashMap;
import java.util.Map;
import javax.swing.DefaultComboBoxModel;
import javax.swing.JComboBox;
import javax.swing.JFrame;
import javax.swing.JTable;
import javax.swing.table.DefaultTableModel;
import javax.swing.table.TableModel;

/**
 * @see https://stackoverflow.com/a/16611982/230513
 * @see https://stackoverflow.com/a/10623134/230513
 */
public class Test {

    private static final int N = 25;
    private DefaultComboBoxModel dcbm = new DefaultComboBoxModel();
    private JComboBox combo = new JComboBox(dcbm);
    private JTable table = new JTable(1, 1);
    private Map<String, NamedModel> map = new HashMap<String, NamedModel>();

    public TableModel getTable(String name) {
        return map.get(name);
    }

    private void display() {
        JFrame f = new JFrame("Test");
        f.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
        for (int i = 0; i < N; i++) {
            String name = "Table " + String.valueOf(i);
            NamedModel model = new NamedModel(name);
            map.put(name, model);
            dcbm.addElement(model);
        }
        combo.setSelectedIndex(-1);
        combo.addActionListener(new ActionListener() {
            @Override
            public void actionPerformed(ActionEvent e) {
                TableModel model = (TableModel) combo.getSelectedItem();
                table.setModel(model);
            }
        });
        f.add(combo, BorderLayout.NORTH);
        f.add(table);
        f.pack();
        f.setLocationRelativeTo(null);
        f.setVisible(true);
    }

    private static class NamedModel extends DefaultTableModel {

        private String name;

        public NamedModel(String name) {
            super(1, 1);
            this.name = name;
        }

        @Override
        public Object getValueAt(int row, int col) {
            return name + ", " + row + ", " + col;
        }

        @Override
        public String toString() {
            return name;
        }
    }

    public static void main(String[] args) {
        EventQueue.invokeLater(new Runnable() {
            @Override
            public void run() {
                new Test().display();
            }
        });
    }
}

答案 1 :(得分:1)

  

这就是我想要的功能;我想出了一个解决方案   创建一个具有私有JTable表的新组件   私有字符串名称,但在搜索JTable时仍然会遇到问题   按名称。

同意,非常好的设计,让我在这种情况下... ...

public JTable getTable(String Component_name)

可能是

public JTable getTable(myTableModel, arrays implemented in JTables API)

不要声明(重载)组件名称,这可能是内部方法返回JTable,但我仍然找不到原因

答案 2 :(得分:0)

成功回答!!

我创造了         private ArrayList List = new ArrayList();

和我的ActionListener ::

 AddTableMenuItem.addActionListener(new ActionListener() 
    {
                //This method will be called whenever you click the button. 
            public void actionPerformed(ActionEvent e){
               Table table=new Table();

              JTable.setName("Table"+count);

              AddT(table);                               // Table List 
    }

其中

  public ArrayList<JTable> getTable()
  {
        return List;
  }

 public void AddT(JTable tl)
 {
     if(tl.getName()!=null)
     {getTable().add(tl);}
     else return;
 }

现在我有一个所有JTable的列表,现在我可以通过名称实现获取如下::

JTable one= new JTable();
  JTable two= new JTable();
  ComboBoxModel combo = comboBox.getModel();

 for(JPanel t: getTable())
  {  
       if(combo.getSelectedItem().equals(t.getName()))
       {
          one=t;
       }
  }