Question

所以这是一个家伙如何做到的例子

import java.awt.BorderLayout;
import javax.swing.JButton;
import javax.swing.JFrame;
import javax.swing.JLabel;
import javax.swing.JPanel;
import javax.swing.JScrollPane;
import javax.swing.JTable;
import javax.swing.JTextField;
import javax.swing.RowFilter;
import javax.swing.SwingUtilities;
import javax.swing.event.DocumentEvent;
import javax.swing.event.DocumentListener;
import javax.swing.table.DefaultTableModel;
import javax.swing.table.TableModel;
import javax.swing.table.TableRowSorter;

public class TestTableSortFilter extends JPanel {

private String[] columnNames
        = {"Country", "Capital", "Population in Millions", "Democracy"};

private Object[][] data = {
    {"USA", "Washington DC", 280, true},
    {"Canada", "Ottawa", 32, true},
    {"United Kingdom", "London", 60, true},
    {"Germany", "Berlin", 83, true},
    {"France", "Paris", 60, true},
    {"Norway", "Oslo", 4.5, true},
    {"India", "New Delhi", 1046, true}
};

private DefaultTableModel model = new DefaultTableModel(data, columnNames);
private JTable jTable = new JTable(model);

private TableRowSorter<TableModel> rowSorter
        = new TableRowSorter<>(jTable.getModel());

private JTextField jtfFilter = new JTextField();
private JButton jbtFilter = new JButton("Filter");

public TestTableSortFilter() {
    jTable.setRowSorter(rowSorter);

    JPanel panel = new JPanel(new BorderLayout());
    panel.add(new JLabel("Specify a word to match:"),
            BorderLayout.WEST);
    panel.add(jtfFilter, BorderLayout.CENTER);

    setLayout(new BorderLayout());
    add(panel, BorderLayout.SOUTH);
    add(new JScrollPane(jTable), BorderLayout.CENTER);

    jtfFilter.getDocument().addDocumentListener(new DocumentListener(){

        @Override
        public void insertUpdate(DocumentEvent e) {
            String text = jtfFilter.getText();

            if (text.trim().length() == 0) {
                rowSorter.setRowFilter(null);
            } else {
                rowSorter.setRowFilter(RowFilter.regexFilter("(?i)" + text));
            }
        }

        @Override
        public void removeUpdate(DocumentEvent e) {
            String text = jtfFilter.getText();

            if (text.trim().length() == 0) {
                rowSorter.setRowFilter(null);
            } else {
                rowSorter.setRowFilter(RowFilter.regexFilter("(?i)" + text));
            }
        }

        @Override
        public void changedUpdate(DocumentEvent e) {
            throw new UnsupportedOperationException("Not supported yet."); //To change body of generated methods, choose Tools | Templates.
        }

    });
}

public static void main(String[] args) {
    SwingUtilities.invokeLater(new Runnable(){
        public void run() {
           JFrame frame = new JFrame("Row Filter");
           frame.add(new TestTableSortFilter());
           frame.pack();
           frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
           frame.setLocationRelativeTo(null);
           frame.setVisible(true);
        }

    });
}
}

并且它工作正常，但问题是当我尝试通过点击此行获取值

((DefaultTableModel) jTableZ.getjTable1().getModel()).getValueAt(row, 36))

它没有返回此程序搜索的行，而是返回的行首先放在jTable中，没有搜索。让我尝试解释：当我在jTable中有多行并且搜索将其缩小为1行时，我会得到旧jTable中的第一行，而不是行搜索它的行。如何解决这个问题？

我忘了写我的数据不像这里那样

private Vector<Vector<String>> data

Answer 1

要搜索JTable，请尝试以下方法：

For(int row=0;row<C.ROWS; row++){
   For(int col=0;col<C.COLS; col++){
     if(table.getValueAt(row, col) == (Object) O){
         System.out.print(Hello); 
      }
   }
}

根据是否有所不同，您可以在此时使用JTable I对象的每个单元格中设置值。

Answer 2

我已经解决了它而不是getSelectedRow我必须将其转换为rowIndexModel以获取所选行

private int row = jTable1().getSelectedRow();
private int rowModel=jTable1().convertRowIndexToModel(row);

它运行良好：）

在jTable java中搜索元素

2 个答案: