Java 如何使用弹出窗口中的另一个按钮创建按钮

我正在尝试创建一个应用程序，当点击弹出窗口中的按钮时，将在同一活动的另一个布局中创建一个新按钮

以下是popupwindow.xml的代码

<?xml version="1.0" encoding="utf-8"?>
<RelativeLayout
xmlns:android="http://schemas.android.com/apk/res/android" 
android:layout_width="match_parent"
android:layout_height="match_parent">

<Button
    android:id="@+id/start_goal"
    android:layout_width="wrap_content"
    android:layout_height="wrap_content"
    android:text="@string/start_goal" />


</RelativeLayout>

下面是MainActivity.java的java代码

public class MainActivity extends AppCompatActivity {

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

    Button setupgoal = (Button) findViewById(R.id.setupgoal);

    setupgoal.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {
            startActivity(new Intent(MainActivity.this,creategoal.class));
        }
    });
  }
}

---

您是否尝试过在XML上使用visibility:gone并在onClick方法上设置它？

没有，但这可能会产生什么影响？您可以在XML文件中使用visibility=gone创建按钮，然后在onClick侦听器上将其可见性设置为visible

public class creategoal extends Activity {
@Override
protected void onCreate(Bundle savedInstanceState){
    super.onCreate(savedInstanceState);

    setContentView(R.layout.popupwindow);

    DisplayMetrics dm = new DisplayMetrics();
    getWindowManager().getDefaultDisplay().getMetrics(dm);

    int width = dm.widthPixels;
    int height = dm.heightPixels;

    getWindow().setLayout((int)(width*.8),(int)(height*.6));
  }
}

public class MainActivity extends AppCompatActivity {

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

    Button setupgoal = (Button) findViewById(R.id.setupgoal);

    setupgoal.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {
            startActivity(new Intent(MainActivity.this,creategoal.class));
        }
    });
  }
}