等待 5 秒

Question

我想在启动另一个公共 void 方法之前等待 5 秒。线程睡眠对我不起作用。如果有一种不使用线程的wait() 方式，我很想知道这一点。

public void check(){
    //activity of changing background color of relative layout
}

我想在更改相对布局颜色之前等待 3 秒。

Answer 1

我更喜欢的是

(new Handler()).postDelayed(this::here is your method,2000);

Answer 2

只需使用 lambda 添加单行

(new Handler()).postDelayed(this::yourMethod, 5000);

编辑澄清：yourMethod 指的是您要在 5000 毫秒后执行的方法。

Answer 3

这对我有用：

    val handler = Handler()
    handler.postDelayed({
        // your code to run after 2 second
    }, 2000)

Answer 4

我对我的 android 应用程序使用以下代码（与您之前看到的相同），我需要等待一些线程才能开始我的新方法。它工作正常。

 Handler handler = new Handler();
        handler.postDelayed(new Runnable() {
            public void run() {
                // yourMethod();
            }
        }, 5000);   //5 seconds

Answer 5

导入使用：import android.os.Handler;

 new Handler().postDelayed(new Runnable() {
            public void run() {
                // yourMethod();
            }
        }, 5000);   //5 seconds

Answer 6

我在另一个问题上发布了this answer，但它也可能对您有所帮助。

类：

import android.os.Handler;
import android.os.Looper;

public class Waiter {

WaitListener waitListener;
int waitTime = 0;
Handler handler;
int waitStep = 1000;
int maxWaitTime = 5000;
boolean condition = false;

public Waiter(Looper looper, final int waitStep, final int maxWaitTime){

    handler = new Handler(looper);
    this.waitStep = waitStep;
    this.maxWaitTime = maxWaitTime;

}

public void start(){

    handler.post(new Runnable() {
        @Override
        public void run() {

            waitListener.checkCondition();

            if (condition) {

                waitListener.onConditionSuccess();

            } else {
                if (waitTime <= maxWaitTime) {

                    waitTime += waitStep;
                    handler.postDelayed(this, waitStep);

                } else {

                    waitListener.onWaitEnd();
                }
            }
        }
    });

}

public void setConditionState(boolean condition){
    this.condition = condition;
}

public void setWaitListener(WaitListener waitListener){
    this.waitListener = waitListener;
}

}

界面：

public interface WaitListener {

public void checkCondition();

public void onWaitEnd();

public void onConditionSuccess();

}

使用示例：

ConnectivityManager mConnMgr = (ConnectivityManager) getSystemService(Context.CONNECTIVITY_SERVICE);
final int result = mConnMgr.startUsingNetworkFeature(ConnectivityManager.TYPE_MOBILE, "enableMMS");

final Waiter waiter = new Waiter(getMainLooper(), 1000, 5000);
waiter.setWaitListener(new WaitListener() {

            @Override
            public void checkCondition() {
                Log.i("Connection", "Checking connection...");
                NetworkInfo networkInfo = mConnMgr.getNetworkInfo(ConnectivityManager.TYPE_MOBILE_MMS);
                waiter.setConditionState(networkInfo.isConnected());
            }

            @Override
            public void onWaitEnd() {
                Log.i("Connection", "No connection for sending");
                //DO
            }

            @Override
            public void onConditionSuccess() {
                Log.i("Connection", "Connection success, sending...");
                //DO
            }

});

waiter.start();

Answer 7

您可以使用 java 处理程序来完成您的任务：

Handler handler = new Handler();
handler.postDelayed(new Runnable() {
    public void run() {
     // Actions to do after 5 seconds
    }
}, 5000);

更多信息请阅读以下网址：

https://developer.android.com/reference/android/os/Handler.html

Answer 8

看看这是否适合你。一定要导入android.os.Handler

      Handler handler = new Handler();
            handler.postDelayed(new Runnable() {
                public void run() {
                    // yourMethod();
                }
            }, 5000);   //5 seconds