Question

我想在单击View时在弹出窗口中显示textView，但计算需要时间，所以我在AsyncTask中进行计算，但是如何在AsyncTask进程完成后立即显示弹出窗口？

public void onClick(View widget) {

        MyAsyncTask asyncTask = new MyAsyncTask(new AsyncResponse() {
            @Override
            public void processFinish(Object output) {
            meaning_result = (String) output;
            }
            });
         asyncTask.execute("xxxxx");
         showPopupWindow(widget);
}

这是我的第一个想法，但showPopupWindow(widget)首先执行，meaning_result尚未分配。如果分配了meaning_result，如何使showPopupWindow(widget)运行？

Answer 1

尝试这样的事情：

public void showPopUp(final View widget){
         runOnUiThread(new Runnable() {
            @Override
            public void run() {
                showPopupWindow(widget);
            }
        });
}

和：

public void onClick(final View widget) {

        MyAsyncTask asyncTask = new MyAsyncTask(new AsyncResponse() {
            @Override
            public void processFinish(Object output) {
                meaning_result = (String) output;
                showPopUp(widget);
            }
        });
        asyncTask.execute("xxxxx");

    }

希望这有帮助

Answer 2

您需要在AsynTask的postExecute中显示弹出窗口。

public void onClick(View widget) {

     AsyncTask asyncTask = new AsyncTask(){

        @Override
        protected long doInBackground(){
            //Do your cacultation
        }

        @Override
        protected void onPostExecute(Long result) {
             showPopupWindow(widget);
        }

        });
     asyncTask.execute("xxxxx");
}

如何在AsyncTask完成后showPopupWindow？

2 个答案: