這是可能的。您需要覆蓋org.apache.kafka.common.serialization
中定義的Deserializer<T>
接口,並且您需要通過包含Kafka參數的ConsumerStrategy[K, V]
類將key.deserializer
或value.deserializer
指向您的自定義類。例如:
import org.apache.kafka.common.serialization.Deserializer
class AvroDeserializer extends Deserializer[Array[Byte]] {
override def configure(map: util.Map[String, _], b: Boolean): Unit = ???
override def close(): Unit = ???
override def deserialize(s: String, bytes: Array[Byte]): Array[Byte] = ???
}
然後:
import org.apache.kafka.clients.consumer.ConsumerRecord
import org.apache.kafka.common.serialization.StringDeserializer
import org.apache.spark.streaming.kafka010._
import org.apache.spark.streaming.kafka010.LocationStrategies.PreferConsistent
import org.apache.spark.streaming.kafka010.ConsumerStrategies.Subscribe
import my.location.with.AvroDeserializer
val ssc: StreamingContext = ???
val kafkaParams = Map[String, Object](
"bootstrap.servers" -> "localhost:9092,anotherhost:9092",
"key.deserializer" -> classOf[StringDeserializer],
"value.deserializer" -> classOf[AvroDeserializer],
"group.id" -> "use_a_separate_group_id_for_each_stream",
"auto.offset.reset" -> "latest",
"enable.auto.commit" -> (false: java.lang.Boolean)
)
val topics = Array("sometopic")
val stream = KafkaUtils.createDirectStream[String, MyTypeWithAvroDeserializer](
ssc,
PreferConsistent,
Subscribe[String, String](topics, kafkaParams)
)