? ? ? ?最近在做一個查詢優化時，考慮到一次查詢耗時較多，所以打算用多線程來做，之前是串行查詢。比如一個用戶查詢觸發50個線程，在只有一個用戶的情況下，性能比串行查詢確實提高了許多，但當多個用戶同時觸發查詢的時候，CPU卻飆高得很厲害，比串行查詢好不了多少。

? ? ? ?因為一次查詢是同步查詢，即要等待所有的線程完成后才返回結果，所以一開始想到的辦法是每次查詢都起一個線程池，這個線程池里面有50個線程，這個線程池阻塞等待所有的線程完成然后返回結果，從而造成50個用戶同時查詢時起了50個線程池，cpu資源消耗殆盡。能不能做到所用用戶查詢觸發的線程統一由一個線程池控制呢？百度后終于找到了解決辦法。

1 import java.util.concurrent.CountDownLatch; 2 3 public class ThreadA implements Runnable { 4 5 private CountDownLatch cDownLatch; 6 7 private String name; 8 9 public ThreadA(CountDownLatch cDownLatch, String threadName) { 10 this.cDownLatch = cDownLatch; 11 this.name = threadName; 12 } 13 14 @Override 15 public void run() { 16 System.out.println(name + " begin."); 17 try { 18 Thread.sleep(500); 19 } catch (InterruptedException e) { 20 e.printStackTrace(); 21 } 22 System.out.println(name + " end."); 23 cDownLatch.countDown(); 24 } 25 } 線程執行體 1 import java.util.concurrent.BlockingQueue; 2 import java.util.concurrent.LinkedBlockingQueue; 3 import java.util.concurrent.ThreadPoolExecutor; 4 import java.util.concurrent.TimeUnit; 5 6 public class MyThreadPool extends ThreadPoolExecutor { 7 8 private static MyThreadPool myPool = new MyThreadPool(10, 100, 1, 9 TimeUnit.DAYS, new LinkedBlockingQueue<Runnable>()); 10 11 private MyThreadPool(int corePoolSize, int maximumPoolSize, 12 long keepAliveTime, TimeUnit unit, BlockingQueue<Runnable> workQueue) { 13 super(corePoolSize, maximumPoolSize, keepAliveTime, unit, workQueue); 14 } 15 16 public static MyThreadPool getInstance() { 17 return myPool; 18 } 19 } 線程池 import java.util.concurrent.CountDownLatch;public class MainThread {public static void main(String[] args) {ThreadExecutor executor = new ThreadExecutor("A", 3);executor.execute();} }class ThreadExecutor {private String prefix;private int size;public ThreadExecutor(String prefix, int size) {this.prefix = prefix;this.size = size;}public void execute() {System.out.println(prefix + " begin.");ThreadA temp = null;CountDownLatch cDownLatch = new CountDownLatch(size);for (int i = 0; i < size; i++) {temp = new ThreadA(cDownLatch, prefix + i);MyThreadPool.getInstance().execute(temp);}try {cDownLatch.await();} catch (InterruptedException e) {e.printStackTrace();}System.out.println(prefix + " end.");} } 主線程

執行結果

A begin.
A0 begin.
A1 begin.
A2 begin.
A1 end.
A0 end.
A2 end.
A end.

?

轉載于:https://www.cnblogs.com/BensonHe/p/3150516.html

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