Question

我有以下代码：

  type Request = EitherT[IO, Throwable, KkProducerRecordMetadata]

  def create(producer: => KkProducerCreator)
  : IO[Producer[String, String]]
  = IO {
    try {
      new KafkaProducer[String, String](properties(producer))
    } catch {
      case e: InstanceAlreadyExistsException => ???
    }

  }

  def send(producer: => IO[Producer[String, String]])
          (record: => KkProducerRecord)
  : Request = EitherT(for {
    p <- producer
    m <- IO {
      try {
        //If MaxBlockMs is not set,
        //then after 60s it will throw an exception
        val pr = new ProducerRecord[String, String](record.topic, record.key, record.value)
        val meta = p.send(pr).get()
        p.flush()

        Right(KkProducerRecordMetadata(meta.hasOffset,
          meta.hasTimestamp,
          meta.offset,
          meta.partition,
          meta.timestamp,
          meta.topic))
      } catch {
        case e: Exception => Left(e)
      }
    }
  } yield m)

  def close(producer: => IO[Producer[String, String]])
  : IO[Unit]
  = producer.map { p =>
    p.flush()
    p.close()
  }

函数create创建了一个kafka生成器并包装到IO中，因为它可能会产生副作用。

函数send会将消息发送给kafka，正如您在第一个参数中看到的那样，它需要producer。

我正在使用以下功能：

  //Creates a producer
  private val pSignIn: IO[Producer[String, String]] =
    KkProducer.create(KkProducerCreator(sys.env.get("KAFKA_SERVER").get,
      "AUTH-SIGNIN-PRODUCER",
      List(MaxBlockMsConfig(4000))))

并发送消息给kafka：

KkProducer.send(pSignIn)(KkProducerRecord(AuthTopology.SignInReqTopic,
        AuthTopology.SignInKey, a))

正如您所看到的，每次调用send函数时，都会创建一个新的生产者实例，然后我得到一个例外：

javax.management.InstanceAlreadyExistsException: kafka.producer:type=app-info,id=AUTH-SIGNIN-PRODUCER

如何防止生产者以功能方式创建两次？我想创建一个生产者地图和身份证作为关键词？ StateT解决方案是什么？

Answer 1

确保您创建一次生产者。将生产者实例分配给val或lazy val，然后将val传递给函数。

将send和close函数的参数从名称调用更改为普通参数（按值调用）。

每次通过变量名称引用时，通过为每个send调用创建生产者来评估按名称参数调用参数。

声明你send如下

def send(producer: IO[Producer[String, String]])(record: => KkProducerRecord)

同样适用于close

示例：

def foo(code: => Int) = code + code + 1

def bar(code: Int) = code + code + 1

Scala REPL

scala> :paste
// Entering paste mode (ctrl-D to finish)

def foo(code: => Int) = code + code + 1

def bar(code: Int) = code + code + 1


// Exiting paste mode, now interpreting.

foo: (code: => Int)Int
bar: (code: Int)Int

scala> foo({println("evaluated"); 1})
evaluated
evaluated
res1: Int = 3

scala> bar({println("evaluated"); 1})
evaluated
res2: Int = 3

注意：如果foo println块被评估两次，不像bar。

防止两次创建对象

1 个答案:

Scala REPL