java setSelectionBackGround（）不在jTable中设置颜色

Question

我认为是prepareRenderer(...)引起了麻烦。我创建jTable的代码是：-

jTable1 = new javax.swing.JTable(){
public Component prepareRenderer(TableCellRenderer renderer, int row, int column)
{
    Component c = super.prepareRenderer(renderer, row, column);
    if(column==0){
        c.setBackground(new java.awt.Color(223, 223, 223));
        c.setForeground(new Color(121, 63, 63));
        c.setFont(new java.awt.Font("Tahoma", java.awt.Font.BOLD, 12));
    }else{
        c.setBackground(new java.awt.Color(235, 235, 235));
        c.setFont(new java.awt.Font("Tahoma", java.awt.Font.PLAIN, 12));

    }
    return c;
}
};

现在，当我使用：-

    jTable1.setSelectionBackground(Color.BLACK);

它不起作用。知道怎么修吗

编辑：在完全删除prepareRenderer(...)方法后，我测试了该程序，可以进行黑色选择

Answer 1

终于找到了解决办法

public Component prepareRenderer(TableCellRenderer renderer, int row, int column)
{
Component c = super.prepareRenderer(renderer, row, column);
if(column==0){
    c.setBackground(new java.awt.Color(223, 223, 223));
    c.setForeground(new Color(121, 63, 63));
    c.setFont(new java.awt.Font("Tahoma", java.awt.Font.BOLD, 12));
}else{
    setBackground(new java.awt.Color(235, 235, 235));//c.setBackground(..) was the problem
    c.setFont(new java.awt.Font("Tahoma", java.awt.Font.PLAIN, 12));

}
return c;
}

Answer 2

这是DefaultTableCellRenderer的代码片段

    if (isSelected) {
        super.setForeground(fg == null ? table.getSelectionForeground()
                                       : fg);
        super.setBackground(bg == null ? table.getSelectionBackground()
                                       : bg);
    }

正如您所见，选择背景设置为渲染器（实际上是在调用super.prepareRenderer（渲染器、行、列））中）

之后，您只需根据列将背景重置为其中一种颜色。但无论哪一列编号，实际上都会替换选择背景

如果列=0呼叫

c.setBackground(getSelectionBackground());