Java/android 如何在延迟 3 秒后启动 AsyncTask？
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Java/android how to start an AsyncTask after 3 seconds of delay?
提问by lacas
How can an AsyncTask be started after a 3 second delay?
如何在延迟 3 秒后启动 AsyncTask？
采纳答案by Juhani
You can use Handler for that. Use postDelayed(Runnable, long) for that.
您可以为此使用 Handler。为此使用 postDelayed(Runnable, long) 。
回答by Zelimir
Use Handler class, and define Runnable handleMyAsyncTaskthat will contain code executed after 3000 msec delay:
使用 Handler 类，并定义 RunnablehandleMyAsyncTask将包含在 3000 毫秒延迟后执行的代码：
mHandler.postDelayed(handleMyAsyncTask, 1000*3);
回答by mani
You can use this piece of code to run after a 3 sec delay.
您可以使用这段代码在 3 秒延迟后运行。
new Timer().schedule(new TimerTask() {
@Override
public void run() {
// run AsyncTask here.
}
}, 3000);
回答by Facundo Olano
Using handlers as suggested in the other answers, the actual code is:
使用其他答案中建议的处理程序，实际代码是：
new Handler().postDelayed(new Runnable() {
@Override
public void run() {
new MyAsyncTask().execute();
}
}, 3000);
