1.1消息顺序的场景
场景1:一个queue,多个consumer
一个queue,有多个consumer去消费,这样就会造成顺序的错误,consumer从MQ里面读取数据是有序的,但是每个consumer的执行时间是不固定的,无法保证先读到消息的consumer一定先完成操作,这样就会出现消息并没有按照顺序执行,造成数据顺序错误。,
人话就是:我确实保证了消息是按按顺序接的。但是由于每一个消息执行的是时间不一样。如果我前面的执行消息比较长,会导致我后面的操作比前面的操作更早执行。就出现了顺序错误
consumer多线程消费
一个queue对应一个consumer,但是consumer里面进行了多线程消费,这样也会造成消息消费顺序错误。
人话:就是线程的执行是强制式的,你永远不知道下一个执行的会是哪个指令。
2.解决方案:
单消费者、单线程消费就行。
如果你非要多线程,你可以参考下,多线程输出a,b,c
/*** @author XDarker* 2018-5-17*/
public class Main {public static void main(String[] args) throws InterruptedException {int num = 1;//当前正在执行线程的标记ABCPrint print = new ABCPrint(num); Thread threadA = new Thread(new RunnableA(print));Thread threadB = new Thread(new RunnableB(print));Thread threadC = new Thread(new RunnableC(print));threadA.start();Thread.sleep(500);threadB.start();Thread.sleep(500);threadC.start();}
}
class RunnableA implements Runnable{private ABCPrint print;public RunnableA(ABCPrint print) {super();this.print = print; }@Overridepublic void run() { print.PrintA();}
}
class RunnableB implements Runnable{private ABCPrint print;public RunnableB(ABCPrint print) {super();this.print = print;}@Overridepublic void run() { print.PrintB();}
}
class RunnableC implements Runnable{private ABCPrint print;public RunnableC(ABCPrint print) {super();this.print = print;}@Overridepublic void run() { print.PrintC();}
}
class ABCPrint {private int num;//当前正在执行线程的标记public ABCPrint(int num) {super();this.num = num;}public void PrintA(){for (int j = 0; j < 2; j++)//表示 循环打印2轮synchronized(this){while(num != 1){try {this.wait();} catch (InterruptedException e) {e.printStackTrace();}}for (int i = 0; i < 3; i++) {//表示 打印3次System.out.println("A"); }//打印A线程执行完 ,通知打印B线程num = 2; this.notifyAll(); }}public void PrintB(){for (int j = 0; j < 2; j++)//表示 循环打印2轮synchronized(this){while(num != 2){try {this.wait();} catch (InterruptedException e) {e.printStackTrace();}}for (int i = 0; i < 2; i++) {//表示 打印2次System.out.println("B"); }//打印B线程执行完 ,通知打印C线程num = 3; this.notifyAll(); }}public void PrintC(){for (int j = 0; j < 2; j++)//表示 循环打印2轮synchronized(this){while(num != 3){try {this.wait();} catch (InterruptedException e) {e.printStackTrace();}}System.out.println("C"); //打印C线程执行完 ,通知打印A线程num = 1; this.notifyAll(); }}
}