java - android中如何判断点击了哪个动态按钮？

Question

我正在 oncreate 中创建一个动态按钮，我没有在 XML 中放置任何按钮。

LinearLayout table = (LinearLayout)findViewById(R.id.linearlayout);
Button btnAttendance,btnSched;
btnAttendance = new Button(this);
btnSched = new Button(this);
btnAttendance.setOnClickListener(this);
btnSched.setOnClickListener(this);
table.addView(btnAttendance);
        table.addView(btnSched);

//这是我在实现监听器时创建的 OnClick 方法:

@Override
    public void onClick(View arg0) {
        // TODO Auto-generated method stub
        BackgroundWorker fetchRiddles = new BackgroundWorker(DashActivity.this,"retrieveRiddles");
        switch (arg0.getId()) {
        case 1:  // in this case i usually put "R.id.Button1:" but my problem is that my buttons are dynamic
            fetchRiddles.execute("0,"+username);
            break;
        case 2:
            fetchRiddles.execute("1,"+username);
            break;
    }

如何使用 switch 确定点击了哪个按钮？我的按钮是动态创建的，无需在 xml 中放置任何按钮。

Answer 1

您可以通过编程方式设置 ID:

btnAttendance = new Button(this);
btnAttendance.setId(100);
btnSched = new Button(this);
btnSched.setId(200);

然后比较:

@Override
    public void onClick(View arg0) {
        // TODO Auto-generated method stub
        BackgroundWorker fetchRiddles = new BackgroundWorker(DashActivity.this,"retrieveRiddles");
        switch (arg0.getId()) {
        case 100:  // in this case i usually put "R.id.Button1:" but my problem is that my buttons are dynamic
            fetchRiddles.execute("0,"+username);
            break;
        case 200:
            fetchRiddles.execute("1,"+username);
            break;
    }