Question

我在ListView中打开了一个OnItemClickListener，打开一个弹出窗口。但是，当我点击列表中的另一个项目时 - 另一个弹出窗口打开，我不知道如何获取已打开的窗口，以便我可以关闭它。任何帮助将非常感谢！

这是我打开弹出窗口的代码：

list.setOnItemClickListener(new AdapterView.OnItemClickListener() {

            @Override
            public void onItemClick(AdapterView<?> parent, View view,
                                    int position, long id) {

                LayoutInflater layoutInflater
                        = (LayoutInflater)getBaseContext()
                        .getSystemService(LAYOUT_INFLATER_SERVICE);
                View popupView =layoutInflater.inflate(R.layout.popup,null);


                Object checkExistence = parent.getItemAtPosition(position);
                int imgID = ctx.getResources().getIdentifier(checkExistence.toString().toLowerCase().replace(" ","_"), "drawable", ctx.getPackageName());
                ImageView imageView = (ImageView) popupView.findViewById(R.id.img1);

                imageView.setImageResource(imgID);

                final PopupWindow popupWindow = new PopupWindow(
                        popupView,
                        Toolbar.LayoutParams.WRAP_CONTENT,
                        Toolbar.LayoutParams.WRAP_CONTENT);

                Button btnDismiss = (Button)popupView.findViewById(R.id.dismiss);
                btnDismiss.setOnClickListener(new Button.OnClickListener(){

                    @Override
                    public void onClick(View v) {
                        // TODO Auto-generated method stub
                        popupWindow.dismiss();
                    }});

                popupWindow.showAsDropDown(list, 300, -1250);

            }
        });

Answer 1

你可以试试这个，

popupWindow.setBackgroundDrawable(new BitmapDrawable());
popupWindow.setOutsideTouchable(true);

或者

popupWindow.setFocusable(true);

Answer 2

如注释中所述，您可以使用范围较大的变量来保留对弹出窗口的引用。然后，您可以使用dismiss()关闭它。

// class member variable
PopupWindow popupWindow;

// ...

list.setOnItemClickListener(new AdapterView.OnItemClickListener() {

    @Override
    public void onItemClick(AdapterView<?> parent, View view, int position, long id) {

        // ...

        popupWindow = new PopupWindow(popupView, Toolbar.LayoutParams.WRAP_CONTENT, Toolbar.LayoutParams.WRAP_CONTENT);

        Button btnDismiss = (Button)popupView.findViewById(R.id.dismiss);
        btnDismiss.setOnClickListener(new Button.OnClickListener(){

            @Override
            public void onClick(View v) {

                // dismiss the popup window
                popupWindow.dismiss();
            }});

        // ...

    }
});

如何在android中关闭已打开的以编程方式弹出的窗口？

2 个答案: