Java 线程睡眠使其他线程等待
我有一个任务,当为用户生成随机密码时,短信应该在4分钟后发出,但欢迎短信应该立即发出。由于密码我设置第一,需要发送4分钟后,我使线程睡眠(不能使用ExecutorServices),并欢迎短信线程开始 代码如下:Java 线程睡眠使其他线程等待,java,multithreading,java-threads,Java,Multithreading,Java Threads,我有一个任务,当为用户生成随机密码时,短信应该在4分钟后发出,但欢迎短信应该立即发出。由于密码我设置第一,需要发送4分钟后,我使线程睡眠(不能使用ExecutorServices),并欢迎短信线程开始 代码如下: String PasswordSMS="Dear User, Your password is "+'"'+"goody"+'"'+" Your FREE recharge service is LIVE now!"; String welcomeSMS="Dear goody, W
String PasswordSMS="Dear User, Your password is "+'"'+"goody"+'"'+" Your FREE
recharge service is LIVE now!";
String welcomeSMS="Dear goody, Welcome to XYZ";
try {
Thread q=new Thread(new GupShupSMSUtill(PasswordSMS,MOB_NUM));
Thread.sleep(4 * 60 * 1000);
q.start();
GupShupSMSUtill sendWelcomesms2=new GupShupSMSUtill(welcomeSMS, MOB_NUM);
Thread Bal3=new Thread(sendWelcomesms2);
Bal3.start();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
</code>
String PasswordSMS = "Dear User, Your password is " + "\"" + "goody" + "\"" + " Your FREE recharge service is LIVE now!";
String welcomeSMS = "Dear goody, Welcome to XYZ";
try
{
GupShupSMSUtill sendWelcomesms2 = new GupShupSMSUtill(welcomeSMS, MOB_NUM);
Thread Bal3 = new Thread(sendWelcomesms2);
Bal3.start();
Thread q = new Thread(new GupShupSMSUtill(PasswordSMS, MOB_NUM));
q.start();
latch.await(); // Wait
}
catch (InterruptedException e)
{
e.printStackTrace();
}
因此,如果我改变顺序,线程sendWelcomesms2立即启动。我必须发送欢迎短信,然后发送密码短信(4分钟后)如何实现
注意:两条短信都在4分钟后发出
Thread.sleep(4 * 60 * 1000);
延迟当前正在运行的线程的执行,直到等待时间结束,才会执行q.start()。此顺序没有意义。在为
q
发出start
命令之前,您正在休眠当前线程
您可能希望在
GupShupSMSUtill()中发出sleep命令(
(可能将其签名更改为类似于GupShupSMSUtill(PasswordSMS,MOB_NUM,sleeptime)
,以便能够控制它的睡眠时间)。只有在
Thread q=new Thread(new GupShupSMSUtill(PasswordSMS,MOB_NUM));
被执行。您的线程在
q.start();
被执行。因此,如果希望在主线程睡眠时运行q线程,则应按以下顺序编写代码行:
Thread q=new Thread(new GupShupSMSUtill(PasswordSMS,MOB_NUM)); // Create thread
q.start(); // start thread
Thread.sleep(4 * 60 * 1000); // suspend main thread for 4 sec
您可以使用join():
或闩锁:
private static java.util.concurrent.CountDownLatch latch = new java.util.concurrent.CountDownLatch(1);
以及守则:
String PasswordSMS="Dear User, Your password is "+'"'+"goody"+'"'+" Your FREE
recharge service is LIVE now!";
String welcomeSMS="Dear goody, Welcome to XYZ";
try {
Thread q=new Thread(new GupShupSMSUtill(PasswordSMS,MOB_NUM));
Thread.sleep(4 * 60 * 1000);
q.start();
GupShupSMSUtill sendWelcomesms2=new GupShupSMSUtill(welcomeSMS, MOB_NUM);
Thread Bal3=new Thread(sendWelcomesms2);
Bal3.start();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
</code>
String PasswordSMS = "Dear User, Your password is " + "\"" + "goody" + "\"" + " Your FREE recharge service is LIVE now!";
String welcomeSMS = "Dear goody, Welcome to XYZ";
try
{
GupShupSMSUtill sendWelcomesms2 = new GupShupSMSUtill(welcomeSMS, MOB_NUM);
Thread Bal3 = new Thread(sendWelcomesms2);
Bal3.start();
Thread q = new Thread(new GupShupSMSUtill(PasswordSMS, MOB_NUM));
q.start();
latch.await(); // Wait
}
catch (InterruptedException e)
{
e.printStackTrace();
}
在线程“q”的末尾:
提示-在这种情况下不要使用Thread.sleep(x)。因为您首先为主线程调用
sleep
,然后为第二个线程调用start
。请逐步运行该代码。从构造函数内部调用sleep()
?听起来像是糟糕的设计。在一个好的设计中,每种方法都有一个职责。构造函数的职责是初始化其类的实例<代码>睡眠()不初始化任何内容。