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JAVA异步调用再度改进
NEW A（MANY）-》SEND MESSAGE(UNIQUE)NEW B（MANY）-》SET MESSAGE(UNIQUE)
ONLY ONE C（IF A GET MESSAGE A SENT，C remove MESSAGE FROM HASHMAP）
public class AClient2C2 {
private final long timeout = 10000;
public void test(String taskId) {
long count = 0;
while (CData2C2.getInstance().isHasTask(taskId) && count < timeout) {
try {
System.out.println("wait+"+count);
Thread.sleep(1000);
count += 1000;
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
if(count>=timeout) {
System.out.println("超时");
}
}
public static void main(String[] args){
//B的方法会改变CData2C2的值
CData2C2.getInstance().regesitTaskId("aaaa");
AClient2C2 ac = new AClient2C2();
ac.test("aaaa");
//A的方法获取CData2C2的值会把对象从HASHMAP里删除
CData2C2.getInstance().removeTaskId("aaaa");
}
}
import java.util.HashMap;
import java.util.Map;
public class BService2C2 {
}
import java.util.HashMap;
import java.util.Map;
public class CData2C2 {
private static CData2C2 instance;
private static Map<String,String> map = new HashMap<String,String>();
private CData2C2() {
}
public static CData2C2 getInstance(){
if(instance==null){
instance = new CData2C2();
}
return instance;
}
public void regesitTaskId(String taskId){
map.put(taskId, taskId);
}
public void removeTaskId(String taskId){
map.remove(taskId);
}
public boolean isHasTask(String taskId){
if(null!=map.get(taskId)){
return Boolean.TRUE;
}
return Boolean.FALSE;
}
}
最后更新：2017-04-03 18:52:01