ImageButton错误onClick

时间:2012-08-31 09:16:24

标签: android switch-statement imagebutton

我在点击某些图像按钮并执行某些操作时调用函数click()。

public void onClick(View v)
{
    switch (v.getId()) {
    case R.id.btn_first_person:

        Log.i("utc","1st person");
        this.flip(R.id.btn_i);
        this.flip(R.id.btn_we);
        break;

    case R.id.btn_i:
        Log.i("utc","I clicked");
        this.change("I","play");
        break;

    case R.id.btn_we:
        Log.i("utc","We clicked");
        this.change("We","plays");
        break;

    default: Log.i("utc","default");
            Log.i("utc","asa" + v.getId());





    }
    }

“btn_i”按钮工作正常但是当我点击“btn_we”按钮时,它会直接进入默认状态。我不知道为什么?

public void change(String person, String verb)
{
    txtUsageFirst=(TextView)findViewById(R.id.utc_usage_one);
    txtUsageSecond=(TextView)findViewById(R.id.utc_usage_two);
    txtUsageFirst.setText(person);
    txtUsageSecond.setText(verb);

}

当我在第一种情况下调用此翻转函数(btn_first_person)时发生错误,在btnTemp =(Button)中查找findviewbyid(r); line [logcat不显示“check”,“1”]

public void flip(Integer r)
{
    Log.d("check","enetred flip");
    Log.d("check",r.toString());
    btnTemp=(Button) findViewById(r);
    Log.d("check","1");
    if(btnTemp.getVisibility()== View.VISIBLE)
    {
        Log.d("check","invisible now");
        btnTemp.setVisibility(View.GONE);
    }
    else
    {
        Log.d("check","visible now");
        btnTemp.setVisibility(View.VISIBLE);
    }



}

1 个答案:

答案 0 :(得分:0)

对于翻转功能,您将ImageButton转换为Button,这可能会产生错误。

相关问题