int delay = 1000; // delay for 1 sec.
int period = 10000; // repeat every 10 sec.
Timer timer = new Timer();
timer.scheduleAtFixedRate(new TimerTask()
{
public void run()
{
displayData(); // display the data
}
}, delay, period);
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和别的:
while(needToDisplayData)
{
displayData(); // display the data
Thread.sleep(10000); // sleep for 10 seconds
}
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它们都不起作用(应用程序强制关闭).我还可以尝试其他什么选择?
dil*_*lix 27
您的代码失败,因为您在后台线程中执行睡眠,但显示数据必须在UI线程中执行.
您必须从runOnUiThread(Runnable)运行displayData或定义处理程序并向其发送消息.
例如:
(new Thread(new Runnable()
{
@Override
public void run()
{
while (!Thread.interrupted())
try
{
Thread.sleep(1000);
runOnUiThread(new Runnable() // start actions in UI thread
{
@Override
public void run()
{
displayData(); // this action have to be in UI thread
}
});
}
catch (InterruptedException e)
{
// ooops
}
}
})).start(); // the while thread will start in BG thread
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使用onPostDelayed()从任何View或Handler访问.您可以通过不创建Timer或新建来节省内存Thread.
private final Handler mHandler = new Handler();
private final Runnable mUpdateUI = new Runnable() {
public void run() {
displayData();
mHandler.postDelayed(mUpdateUI, 1000); // 1 second
}
}
};
mHandler.post(mUpdateUI);
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