禁用 JAVA 上的按钮，但保留唯一按钮

Question

我需要在单击它们时出现的唯一按钮会消失问题是当单击“唯一”按钮（见图）时，它们也会消失。我的代码：

private String namesArr[] = {"Yakir","Yarden","Igor","Maoz","Moshe","Israel","Tal","Haim","Nati","Mor","Daniel","Idan"};

private Button buttonArr[] = new Button[namesArr.length];

private Font font;

public StudentsGUI(String caption) {

    super(caption);

    addWindowListener(new WindowAdapter(){

            public void windowClosing(WindowEvent e){

                    dispose();

                        System.exit(0);

                        }

                    }); 

    this.setLayout(new GridLayout(3,3));

    font = new Font("Ariel",Font.BOLD,35);

    for(int i=0;i<namesArr.length;i++) {

        buttonArr[i] = new Button(" "+namesArr[(int)(Math.random()*namesArr.length)]);

        buttonArr[i].setFont(font);

        buttonArr[i].addActionListener(this);

        this.add(buttonArr[i]);

    }

    setLocation(800,500);

    setVisible(true);

    pack();

}

public void actionPerformed(ActionEvent e) {

    if (e.getSource() instanceof Button) {

        String btnText = ((Button)e.getSource()).getLabel();

        for(int i=0; i<buttonArr.length; i++) {

            if (buttonArr[i].getLabel().equals(btnText)) {

                this.remove(buttonArr[i]);

                pack();

            }

        }

    }

}

这张图帮助你理解：

因此，如果单击“Idan”，witch 是一个唯一名称，则不会发生任何事情，因为它只有一个实例，但是如果单击“Maoz”，则所有带有“Maoz”标题的按钮都会消失（这已经发生了）

Answer 1

按照@Freddy的回答使用集合应该会更好。但是，如果您要坚持使用数组，则应该执行以下操作（尽管尚未测试）public void actionPerformed(ActionEvent e) {&nbsp; &nbsp; &nbsp; &nbsp; if (e.getSource() instanceof Button) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; String btnText = ((Button)e.getSource()).getLabel();&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; int counter = 0;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; for(int i=0; i<buttonArr.length; i++) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (buttonArr[i].getLabel().equals(btnText)) counter++;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (count > 1) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;for(int j=0; j<buttonArr.length; j++) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (buttonArr[j].getLabel().equals(btnText))&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;this.remove(buttonArr[j]);&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;}&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; pack();&nbsp; &nbsp; &nbsp; &nbsp; }}

Answer 2

你的意思是这样的（代码可能有语法错误）？public void actionPerformed(ActionEvent e) {&nbsp; &nbsp; if (e.getSource() instanceof Button) {&nbsp; &nbsp; &nbsp; &nbsp; String btnText = ((Button)e.getSource()).getLabel();&nbsp; &nbsp; &nbsp; &nbsp; List<Button> btnList = new ArrayList<Button>();&nbsp; &nbsp; &nbsp; &nbsp; for(int i=0; i<buttonArr.length; i++) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (buttonArr[i].getLabel().equals(btnText)) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; btnList.add(buttonArr[i]);&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; //this.remove(buttonArr[i]);&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; //pack();&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; if (btnList.size() > 1) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; for (Iterator<Button> it = btnList.iterator(); it.hasNext()) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; this.remove(it.next());&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; pack();&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }}