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documentation中找到的示例使用mapValue，但它不会更改值类型。我按照this使用带有-Xexperimental编译器标志的Scala 2.11。

代码

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()
  }
}

编译器错误

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？

基于此answer

的解决方案

三种有效的方式，两种无效方式。

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 按照建议（val kStream: KStream[Array[Byte], String] = kStreamBuilder.stream("myTopic")）向kStream添加显式类型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 添加以上内容并将SAM转换用于＆＃34;以避免明确写出匿名类的实例化＆＃34; （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：似乎由于某些原因，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)

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

1 个答案: