如何使用ValueMapper使用Scala更改Kafka Streams 10.2中的值类型

Question

我正在尝试将基于Scala的Kafka Streams应用程序从0.10.0.0升级到0.10.2.1,我无法弄清楚如何让应用程序进行编译.

我在文档中找到的示例使用mapValue但不更改值类型.我使用Scala的2.11与-Xexperimental编译器标志按这个.

码

class MyStream() {
  def startMyStream(): Unit = {
    val kStreamBuilder = new KStreamBuilder
    val kStream = kStreamBuilder.stream("myTopic")

    kStream.mapValues(new ValueMapper[AnyRef, Double]() {
      override def apply(value: Any) = 6.3
    })

    val kafkaStreams = new KafkaStreams(kStreamBuilder,  new Properties)
    kafkaStreams.start()
  }
}

编译错误

no type parameters for method mapValues: (x$1: org.apache.kafka.streams.kstream.ValueMapper[_, _ <: VR])org.apache.kafka.streams.kstream.KStream[Nothing,VR] exist so that it can be applied to arguments (org.apache.kafka.streams.kstream.ValueMapper[AnyRef,Double]{})
 --- because ---
argument expression's type is not compatible with formal parameter type;
  found   : org.apache.kafka.streams.kstream.ValueMapper[AnyRef,Double]
  required: org.apache.kafka.streams.kstream.ValueMapper[_, _ <: ?VR]
Note: AnyRef <: Any, but Java-defined trait ValueMapper is invariant in type V.
You may wish to investigate a wildcard type such as `_ <: Any`. (SLS 3.2.10)
    kStream.mapValues(new ValueMapper[AnyRef, Double]() {
            ^
type mismatch;
  found   : org.apache.kafka.streams.kstream.ValueMapper[AnyRef,Double]{}
  required: org.apache.kafka.streams.kstream.ValueMapper[_, _ <: VR]
    kStream.mapValues(new ValueMapper[AnyRef, Double]() {
                      ^
two errors found

表示为java类,编译很好.

public class MyStream {
    public void startMyStream() {
        KStreamBuilder kStreamBuilder = new KStreamBuilder();
        KStream kStream = kStreamBuilder.stream("myTopic");

        kStream.mapValues(new ValueMapper<Object, Double>() {
            @Override
            public Double apply(Object value) {
                return 6.3;
            }
        });

        KafkaStreams kafkaStreams = new KafkaStreams(kStreamBuilder, new Properties());
        kafkaStreams.start();
    }
}

如何编译Scala版本？

---

解决方案基于这个答案

三种工作方式,两种方式不工作.

class MyStream() {
  def startMyStream(): Unit = {
    val kStreamBuilder = new KStreamBuilder

    // Explicit tying here is not required to compile and run. 
    val kStream: KStream[Array[Byte], String] = kStreamBuilder.stream("myTopic")

    // Does not compile
    kStream.mapValues(new ValueMapper[AnyRef, Double]() {
      override def apply(value: AnyRef) = 6.3
    })

    // Does not compile
    kStream.mapValues(_ => 6.3)

    // Works
    kStream.mapValues[Double](new ValueMapper[AnyRef, Double]() {
      override def apply(value: AnyRef) = 6.3
    })

    // Works, requires compiler option -Xexperimental
    kStream.mapValues[Double](_ => 6.3)

    // Works, requires compiler option -Xexperimental
    kStream.mapValues[Double](convert)
    def convert(string: String): Double = 6.3

    val kafkaStreams = new KafkaStreams(kStreamBuilder, new Properties)
    kafkaStreams.start()
  }
}

---

更新:无效的解决方案

尝试1如建议的那样向kStream添加显式类型(val kStream: KStream[Array[Byte], String] = kStreamBuilder.stream("myTopic"))仍然无法编译并导致此错误.

no type parameters for method mapValues: (x$1: org.apache.kafka.streams.kstream.ValueMapper[_ >: String, _ <: VR])org.apache.kafka.streams.kstream.KStream[Array[Byte],VR] exist so that it can be applied to arguments (org.apache.kafka.streams.kstream.ValueMapper[AnyRef,Double]{})
--- because ---
argument expression's type is not compatible with formal parameter type;
  found   : org.apache.kafka.streams.kstream.ValueMapper[AnyRef,Double]
  required: org.apache.kafka.streams.kstream.ValueMapper[_ >: String, _ <: ?VR]
Note: AnyRef <: Any, but Java-defined trait ValueMapper is invariant in type V.
You may wish to investigate a wildcard type such as `_ <: Any`. (SLS 3.2.10)
kStream.mapValues(new ValueMapper[AnyRef, Double]() {
  ^
  type mismatch;
  found   : org.apache.kafka.streams.kstream.ValueMapper[AnyRef,Double]{}
  required: org.apache.kafka.streams.kstream.ValueMapper[_ >: String, _ <: VR]
  kStream.mapValues(new ValueMapper[AnyRef, Double]() {

尝试2添加上述内容并使用SAM转换"以避免显式写出匿名类的实例化"(kStream.mapValues(_ => 6.3))导致此编译器错误.

no type parameters for method mapValues: (x$1: org.apache.kafka.streams.kstream.ValueMapper[_ >: String, _ <: VR])org.apache.kafka.streams.kstream.KStream[Array[Byte],VR] exist so that it can be applied to arguments (org.apache.kafka.streams.kstream.ValueMapper[String,Double] with Serializable)
--- because ---
argument expression's type is not compatible with formal parameter type;
  found   : org.apache.kafka.streams.kstream.ValueMapper[String,Double] with Serializable
  required: org.apache.kafka.streams.kstream.ValueMapper[_ >: String, _ <: ?VR]
Note: String <: Any (and org.apache.kafka.streams.kstream.ValueMapper[String,Double] with Serializable <: org.apache.kafka.streams.kstream.ValueMapper[String,Double]), but Java-defined trait ValueMapper is invariant in type V.
You may wish to investigate a wildcard type such as `_ <: Any`. (SLS 3.2.10)
kStream.mapValues(_ => 6.3)
        ^
type mismatch;
found   : org.apache.kafka.streams.kstream.ValueMapper[String,Double] with Serializable
required: org.apache.kafka.streams.kstream.ValueMapper[_ >: String, _ <: VR]
kStream.mapValues(_ => 6.3)
                    ^

Answer 1

这里有两个问题.首先,您不指定流的类型:

val kStream: KStream[Array[Byte], String] = kStreamBuilder.stream("myTopic")

这实际上是您看到错误的原因 - 因为您没有指定类型,Scala会将它们推断为某些默认值,这几乎总是不是您想要的.

其次,由于您已启用-Xexperimental,因此您可以依赖SAM转换来避免显式写出匿名类的实例化:

kStream.mapValues(_ => 6.3)

更新:似乎由于某种原因,Scala编译器无法正确推断匿名函数/ SAM实例的输出类型.我能够通过以下小调整成功编译代码:

kStream.mapValues[Double](_ => 6.3)