对按钮事件进行了混淆

Question

I am trying to make a game in java, when a button is pressed I want it to change icon to indicate that button has been selected, then when it is clicked again I want to change it to its original icon to show it has been deselected.

public void actionPerformed(ActionEvent e)
{
    if(e.getSource() == b[7][4] && selected == false)
    {
        b[7][4].setIcon(selected);
        selected = true;
    }

    if(e.getSource() == b[7][4] && selected == true)
    {
        b[7][4].setIcon(king);
        selected = false;           
    }
}

This code currently does as I wish but it runs both at the same time thus changes the icon and then immediately changes it back to the original icon. How can I make it so one click changes it to selected and then the second click changes it back to a king icon?

Answer 1

Use else before the next if to chain them:

public void actionPerformed(ActionEvent e)
{
    if(e.getSource() == b[7][4] && selected == false)
    {
        b[7][4].setIcon(selected);
        selected = true;
    }
    else if(e.getSource() == b[7][4] && selected == true)
    {
        b[7][4].setIcon(king);
        selected = false;           
    }
}

This way the second if clause is only checked if the first condition evaluates to false .

Answer 2

一个更简单的解决方案：使用JToggleButton并通过setIcon(...)和setSelectedIcon(...)设置其Icon和selectedIcon。

Answer 3

Consider condensing your code into a single setIcon() call:

public void actionPerformed(ActionEvent e)
{
    if (e.getSource() == b[7][4]) {
        b[7][4].setIcon(selected ? king : defau1t);
        selected = !selected;
    }
}