t;8"
3) "baidu.com"
4) "9"
5) "google.com"
6) "10"
查询元素的score值
redis> ZSCORE page_rank bing.com
"8"
移除单个元素
redis> ZREM page_rank google.com
(integer) 1
redis> ZRANGE page_rank 0 -1 WITHSCORES
1) "bing.com"
2) "8"
3) "baidu.com"
4) "9"
那么如何实现呢?我们将订单超时时间戳与订单号分别设置为 score 和 member,系统扫描第一个元素判断是否超时,具体如下图所示
实现一
package com.rjzheng.delay4;
import redis.clients.jedis.Jedis;
import redis.clients.jedis.JedisPool;
import redis.clients.jedis.Tuple;
import java.util.Calendar;
import java.util.Set;
public class AppTest {
private static final String ADDR = "127.0.0.1";
private static final int PORT = 6379;
private static JedisPool jedisPool = new JedisPool(ADDR, PORT);
public static Jedis getJedis() {
return jedisPool.getResource();
}
//生产者,生成5个订单放进去
public void productionDelayMessage() {
for (int i = 0; i < 5; i++) {
//延迟3秒
Calendar cal1 = Calendar.getInstance();
cal1.add(Calendar.SECOND, 3);
int second3later = (int) (cal1.getTimeInMillis() / 1000);
AppTest.getJedis().zadd("OrderId", second3later, "OID0000001" + i);
System.out.println(System.currentTimeMillis() + "ms:redis生成了一个订单任务:订单ID为" + "OID0000001" + i);
}
}
//消费者,取订单
public void consumerDelayMessage() {
Jedis jedis = AppTest.getJedis();
while (true) {
Set<Tuple> items = jedis.zrangeWithScores("OrderId", 0, 1);
if (items == null || items.isEmpty()) {
System.out.println("当前没有等待的任务");
try {
Thread.sleep(500);
} catch (InterruptedException e) {
e.printStackTrace();
}
continue;
}
int score = (int) ((Tuple) items.toArray()[0]).getScore();
Calendar cal = Calendar.getInstance();
int nowSecond = (int) (cal.getTimeInMillis() / 1000);
if (nowSecond >= score) {
String orderId = ((Tuple) items.toArray()[0]).getElement();
jedis.zrem("OrderId", orderId);
System.out.println(System.currentTimeMillis() + "ms:redis消费了一个任务:消费的订单OrderId为" + orderId);
}
}
}
public static void main(String[] args) {
AppTest appTest = new AppTest();
appTest.productionDelayMessage();
appTest.consumerDelayMessage();
}
}
此时对应输出如下
可以看到,几乎都是 3 秒之后,消费订单。
然而,这一版存在一个致命的硬伤,在高并发条件下,多消费者会取到同一个订单号,我们上测试代码 ThreadTest
package com.rjzheng.delay4;
import java.util.concurrent.CountDownLatch;
public class ThreadTest {
private static final int threadNum = 10;
private static CountDownLatch cdl = newCountDownLatch(threadNum);
static class DelayMessage implements Runnable {
public void run() {
try {
cdl.await();
} catch (InterruptedException e) {
e.printStackTrace();
}
AppTest appTest = new AppTest();
appTest.consumerDelayMessage();
}
}
public static void main(String[] args) {
AppTest appTest = new AppTest();
appTest.productionDelayMessage();
for (int i = 0; i < threadNum; i++) {
new Thread(new DelayMessage()).start();
cdl.countDown();
}
}
}
输出如下所示
显然,出现了多个线程消费同一个资源的情况。
解决方案
(1)用分布式锁,但是用分布式锁,性能下降了,该方案不细说。
(2)对 ZREM 的返回值进行判断,只有大于 0 的时候,才消费数据,于是将 consumerDelayMessage()方法里的
if(nowSecond >= score){
String orderId = ((Tuple)items.toArray()[0]).getElement();
jedis.zrem("OrderId", orderId);
System.out.println(System.currentTimeMillis()+"