Question

我有JTable，我希望表中的结果根据属性的特定值着色;这里comp为null：

private JNI18NTable _issueIncidentTable = new JNI18NTable(I18N_ID, COLUMNS, "issue.table.") {
        public Component prepareRenderer(TableCellRenderer renderer, int row, int col) {
            Component comp = super.prepareRenderer(renderer, row, col);

            int RangeIndex = Utilities.rowIndexToRangeIndex(_resultIter, row);
            Row theRow = _resultIter.getRowAtRangeIndex(RangeIndex);
            System.out.println("the row is" + theRow.getAttribute("Type"));
            boolean markRow = theRow != null && theRow.getAttribute("Type").toString().equals("c");
            boolean isSelected = isRowSelected(row);
            System.out.println("the comp" + comp);
            if (markRow) {
                comp.setForeground(Color.white);
                comp.setBackground(isRowSelected(row) ? Color.red.darker() : Color.red);
            } else {
                comp.setForeground(isSelected ? getSelectionForeground() : getForeground());
                comp.setBackground(isSelected ? getSelectionBackground() : getBackground());
            }
            return comp;
        }
    };

Answer 1

推断sscce，prepareRenderer()的实施似乎如下所示。您可能会在代码中的其他位置查找假定的问题。特别是，Utilities.rowIndexToRangeIndex()的实现应该使用表自己的 convert 方法。此外，始终使用isSelected的值。

import java.awt.Color;
import java.awt.Component;
import java.awt.EventQueue;
import javax.swing.JFrame;
import javax.swing.JScrollPane;
import javax.swing.JTable;
import javax.swing.table.DefaultTableModel;
import javax.swing.table.TableCellRenderer;

/**
 * @see http://stackoverflow.com/a/20684058/230513
 */
public class Test {

    private void display() {
        JFrame f = new JFrame("Test");
        JTable table = new JTable(new DefaultTableModel(10, 1) {

            @Override
            public Object getValueAt(int row, int col) {
                return "Row " + row;
            }
        }) {
            @Override
            public Component prepareRenderer(TableCellRenderer renderer, int row, int col) {
                Component comp = super.prepareRenderer(renderer, row, col);

                boolean isSelected = isRowSelected(row);
                if (row % 2 == 0) {
                    comp.setForeground(Color.white);
                    comp.setBackground(isSelected ? Color.red.darker() : Color.red);
                } else {
                    comp.setForeground(isSelected ? getSelectionForeground() : getForeground());
                    comp.setBackground(isSelected ? getSelectionBackground() : getBackground());
                }
                return comp;
            }
        };
        f.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
        f.add(new JScrollPane(table));
        f.pack();
        f.setLocationRelativeTo(null);
        f.setVisible(true);
    }

    public static void main(String[] args) {
        EventQueue.invokeLater(new Runnable() {

            @Override
            public void run() {
                new Test().display();
            }
        });
    }
}

如何根据属性的特定值为JTable的行着色

1 个答案: