Question

请有人帮忙 -

我有一个弹出窗口，我在一个名为pop的函数中创建了一个弹出窗口。我还有另一个功能，当用户点击弹出窗口中的按钮时，它会关闭popupWindow。我为解雇创建了一个单独的功能。我想将popupwindow传递给dismiss函数 - 这是最好的方法。我想重用dismiss函数。

public class MainActivity extends AppCompatActivity {

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

}

public void popup(View v) {
    LayoutInflater layoutInflater = (LayoutInflater) getBaseContext().getSystemService(LAYOUT_INFLATER_SERVICE);
    View layout = layoutInflater.inflate(R.layout.activity_alert_dialog,null);
    final PopupWindow popup= new PopupWindow(layout, LayoutParams.WRAP_CONTENT, LayoutParams.WRAP_CONTENT);

    View vi=(View) findViewByID(R.id.Layout);
    popup.showAtLocation(vi, 0, 20, -50);

}

public void dismiss(View v) {
     popup.dismiss();

} 
}

我的解雇方法是在资源布局文件中执行的：

<Button
    android:text="cancel"
    android:onClick="dismiss"/>

Answer 1

尝试以下代码：每次拨打popup()时都会创建弹出窗口。

public class MainActivity extends AppCompatActivity {

    private PopupWindow mPopup = null;
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
    }

    public void popup(View v) {
        LayoutInflater layoutInflater = (LayoutInflater) getBaseContext().getSystemService(LAYOUT_INFLATER_SERVICE);
        View layout = layoutInflater.inflate(R.layout.activity_alert_dialog,null);

        // Use global variable instead of local.
        //final PopupWindow popup= new PopupWindow(layout, LayoutParams.WRAP_CONTENT, LayoutParams.WRAP_CONTENT);
        mPopup= new PopupWindow(layout, LayoutParams.WRAP_CONTENT, LayoutParams.WRAP_CONTENT);

        View vi=(View) findViewByID(R.id.Layout);
        mPopup.showAtLocation(vi, 0, 20, -50);
    }

    public void dismiss(View v) {
        if (mPopup != null && mPopup.isShowing()) {
            mPopup.dismiss();
        }    
    } 
}

而且，我对mPopup.showAtLocation(vi, 0, 20, -50);的位置值有疑问。它是y = -50的正确位置吗？

我认为这个弹出窗口会显示出来。

Answer 2

为PopupWindow创建全局变量。

    private PopupWindow popup;

    public void popup(View v) {
        LayoutInflater layoutInflater = (LayoutInflater) getBaseContext().getSystemService(LAYOUT_INFLATER_SERVICE);
        View layout = layoutInflater.inflate(R.layout.activity_alert_dialog,null);
         popup= new PopupWindow(layout, LayoutParams.WRAP_CONTENT, LayoutParams.WRAP_CONTENT);

        View vi=(View) findViewByID(R.id.Layout);
        popup.showAtLocation(vi, 0, 20, -50);

    }

public void dismiss(View v) {
     popup.dismiss();

}

函数调用之外的弹出窗口

2 个答案: