如何从弹出窗口获取输入

//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);
            }
        });

What i want is to show the popup window when i click the button in the parent activity. The popup window have buttons when onclick the button it do some functions.

我想要的是当我单击父活动中的按钮时显示弹出窗口。弹出窗口按下按钮时,按钮可以执行某些功能。

1 个解决方案

#1

You have to find the view of the button and then assign the listener to it like this:

你必须找到按钮的视图,然后像这样分配监听器:

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);
            }

Also initialize your inflator if you haven't already like this:

如果您还不喜欢这个,请初始化您的充气机:

Inflator inflator = LayoutInflater.from(this);