Question

我写了以下代码：

public class FragmentFirstPage extends Fragment implements OnClickListener{


View root;


@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle bundle){

    root = inflater.inflate(R.layout.fragment_page1, container, false); 

    Button bt = (Button) root.findViewById(R.id.button1);
    bt.setOnClickListener(this);

    Button mbt = (Button) root.findViewById(R.id.imageButton1);
    mbt.setOnClickListener(new View.OnClickListener() { 
    final ViewPager viewPager = (ViewPager) getActivity().findViewById (R.id.activity_main_viewpager);

         @Override
          public void onClick (View v){
          viewPager.setCurrentItem (viewPager.getCurrentItem() + 1); //or -1 to go previous

          } 
    });             
    return root;

}

@Override
public void onClick(View v) {
    // TODO Auto-generated method stub
    Intent myIntent = new Intent(getActivity(), SplashActivity.class);
    getActivity().startActivity(myIntent);

}

当我运行它时，它不显示任何错误但也不起作用。它显示一条错误消息，说不幸的是app已停止工作。我该怎么做才能使我的xml页面有2个按钮button1打开一个活动，imagebutton1显示下一个片段。这就是我想要实现的目标。

Answer 1

您需要做的就是检查onClick内部通过检查主叫视图的ID来检查哪个按钮。从那里你可以决定每个人会做什么。

public class beforemain extends Activity implements OnClickListener {

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.beforemain);

    Button mBtn1 = (Button) findViewById(R.id.button1);
    Button mBtn2 = (Button) findViewById(R.id.button2); //Just like #1
    mBtn1.setOnClickListener(this);
    mBtn2.setOnClickListener(this); //Also like #1

}

@Override
public void onClick(View v) {
    switch(v.getId()) { //Get the id of the button that was clicked
    case R.id.button1:
        Intent i = new Intent(beforemain.this, Splash1.class);
        startActivity(i);
        break;
    case R.id.button2:
        Intent i = new Intent(beforemain.this, Splash2.class);
        startActivity(i);
        break;
    }
}

Answer 2

按如下方式更改onClick方法的实现。

 @Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.beforemain);

    Button mBtn1 = (Button) findViewById(R.id.button1);
    mBtn1.setOnClickListener(this);
    Button mBtn2 = (Button) findViewById(R.id.button1);
    mBtn2.setOnClickListener(this);

}

@Override
public void onClick(View v) {
    if(mBtn1.getId()==v.getID())
   {
    Intent i1=new Intent(beforemain.this, Splash1.class);
    startActivity(i1);
   }
   else
   {
        Intent i=new Intent(beforemain.this, Splash2.class);
        startActivity(i);
   }
}

Answer 3

 Button mBtn1 = (Button) findViewById(R.id.button1);
Button mBtn2 = (Button) findViewById(R.id.button2); 
mBtn1.setOnClickListener(this);
mBtn2.setOnClickListener(this); 




 @Override
        public void onClick(View v) {
            // TODO Auto-generated method stub
            int id = v.getId();
            switch(id) {
            case R.id.button1:

       Intent i=new Intent(beforemain.this, Splash1.class);
        startActivity(i);

                break;
            case R.id.button2:

    Intent i=new Intent(beforemain.this, Splash2.class);
        startActivity(i);
                break;

        }

        }

Answer 4

您只需为所有按钮设置onClickListener，然后在if else或switch案例中比较其ID，并执行您想做的任何事情

例如。

@Override
public void onClick(View v) 
{
  if(v.getId()==R.id.button1)
  {
    Intent i=new Intent(beforemain.this, Splash1.class);
    startActivity(i);
  }
  else if(v.getId()==R.id.button2)
  {
    Intent i=new Intent(beforemain.this, Splash2.class);
    startActivity(i);}
  }
}

尝试使用2个不同的按钮打开2个不同的xml页面

4 个答案: