批量生产JTable

我想制作25个JTable。 我通过这样做来生成表名：

for(int i=0; i < 26; i++) { TableNames[i] = "Table" + i + ""; ... 

我怎样才能在数组中使用这些String名称作为新的JTable名称？ 即

 TableNames[i] = new JTable(model){ ... 

Answer 1

考虑使用List或List而不是数组。 将名称传递给表的构造函数或工厂方法。 下面的示例使用Component的名称，但JComponent的客户端属性可能更通用。

更新 ：下面修订的Java 8示例说明了如何动态添加新表。

 import java.awt.BorderLayout; import java.awt.EventQueue; import java.awt.GridLayout; import java.awt.event.ActionEvent; import java.util.ArrayList; import java.util.List; import javax.swing.AbstractAction; import javax.swing.JButton; import javax.swing.JFrame; import javax.swing.JPanel; import javax.swing.JScrollBar; import javax.swing.JScrollPane; import javax.swing.JTable; import javax.swing.event.ListSelectionEvent; import javax.swing.table.DefaultTableModel; /** * @see http://stackoverflow.com/a/10623134/230513 */ public class NamedTableTest extends JPanel { private static final int N = 25; private final List list = new ArrayList<>(); public NamedTableTest() { super(new GridLayout(0, 1)); for (int i = 0; i < N; i++) { list.add(new NamedTable("Table " + String.valueOf(i))); this.add(list.get(i)); } } private static class NamedTable extends JTable { public NamedTable(final String name) { super(new DefaultTableModel(1, 1) { @Override public Object getValueAt(int row, int col) { return name + ", " + row + ", " + col; } }); this.setName(name); } @Override public void valueChanged(ListSelectionEvent e) { if (!e.getValueIsAdjusting()) { System.out.println(NamedTable.this.getName()); } } } private void display() { JFrame f = new JFrame("NamedTable"); f.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); final JScrollPane jsp = new JScrollPane(this); f.add(jsp); f.add(new JButton(new AbstractAction("Add") { @Override public void actionPerformed(ActionEvent e) { list.add(new NamedTable("Table " + String.valueOf(list.size()))); NamedTableTest.this.add(list.get(list.size() - 1)); NamedTableTest.this.validate(); EventQueue.invokeLater(new Runnable() { @Override public void run() { JScrollBar sb = jsp.getVerticalScrollBar(); sb.setValue(sb.getMaximum()); } }); } }), BorderLayout.SOUTH); f.pack(); f.setLocationRelativeTo(null); f.setVisible(true); } public static void main(String[] args) { EventQueue.invokeLater(() -> { new NamedTableTest().display(); }); } } 

Answer 2

 TableNames[i] = new JTable(model) 

放入for循环我同意先前的声明make error试试这个

 for(i=0;i