如何确保只启动一个线程

Question

情况是,我必须确保当我尝试getConnection时只创建一个RecoveryThread,如果它在PrimaryData Source上的getConnection失败,那么我的代码是:

public Connection getConnection() throws SQLException {
        if (isFailedOver()) {
            try {
                return failoverDataSource.getConnection();
            } catch (SQLException e) {
                throwBigError();
            }
        }
        Connection connection = null;
        try {
            connection = dataSource.getConnection();
            return connection;
        }
        catch (SQLException unexpected) {
            return requestFailover();
        }
    }

    private Connection requestFailover() throws SQLException {
        this.dbFailoverMutex.requestFailover();
        DBFailoverRecoveryService recoveryService = new DBFailoverRecoveryService(this.dbFailoverMutex,this.dataSource);
        Thread recoveryServiceThread = new Thread(recoveryService, "DBFailover Recovery Service");
        recoveryServiceThread.start();
        try {
            return failoverDataSource.getConnection();
        } catch (SQLException e) {
            throwBigError();
        }
        return null;
    }

如果有两个不同的线程尝试getConnection,这可能最终调用requestFailover()方法两次,当它被调用两次时,这将最终创建两个recoveryService线程,我该怎么做以确保永远不会发生？

在此先感谢您的帮助.

Answer 1

我该怎么做才能确保永远不会发生？

要考虑的一件事是切换到使用Executors.newSingleThreadExecutor()只会分叉一个线程来运行的东西.然后,您可以根据需要提交任意数量的任务,而无需担心它们重叠.

private final ExecutorService threadPool =
      Executors.newSingleThreadExecutor(/* pass in ThreadFactory to set name */);
...
DBFailoverRecoveryService recoveryService =
       new DBFailoverRecoveryService(this.dbFailoverMutex, this.dataSource);
threadPool.submit(recoveryService);

与往常一样ExecutorService,您需要在threadPool.shutdown()将最后一个任务提交到池中时调用,否则它将挂起您的应用程序.您可以添加一个Datasource.destroy();方法来执行此操作.