android - 如何从弹出窗口获取输入

Question

//create inflater
final LayoutInflater inflater = (LayoutInflater) this
                .getSystemService(Context.LAYOUT_INFLATER_SERVICE);
//create popupwindow
    PopupWindow pw=new PopupWindow(inflater.inflate(R.layout.menu, (ViewGroup)findViewById(R.layout.dictionarylist)));

        Button Menu = (Button) findViewById(R.id.Menu);
        Menu.setOnClickListener(new Button.OnClickListener() {
            public void onClick(View v) {
                pw.showAtLocation(v, Gravity.CENTER, 0, 0);
                pw.update(0, 0, 200, 250);
                pw.setOutsideTouchable(false);
            }
        });

我想要的是在单击父 Activity 中的按钮时显示弹出窗口。弹出窗口有按钮，当点击按钮时它会执行一些功能。

Answer 1

您必须找到按钮的 View ，然后像这样为其分配监听器:

View pview=inflater.inflate(R.layout.menu, (ViewGroup)findViewById(R.layout.dictionarylist));

Button Menu = (Button) pview.findViewById(R.id.Menu);

Menu.setOnClickListener(new Button.OnClickListener() {
            public void onClick(View v) {
                pw.showAtLocation(v, Gravity.CENTER, 0, 0);
                pw.update(0, 0, 200, 250);
                pw.setOutsideTouchable(false);
            }

如果你还没有像这样初始化你的充气机:

Inflator inflator = LayoutInflater.from(this);