首页 > 解决方案 > 如何对 kafka 消费者应用程序进行功能测试?

问题描述

我想测试一个使用 Kafka 消息并将它们写入日志的应用程序。这是它在类似 scala 的伪代码中的近似表示:

import kafka.consumer.Consumer
import kafka.consumer.ConsumerConfig
import org.slf4j.LoggerFactory
import java.util.Properties
import java.util.HashMap

object ConsumerApp extends App {
  val topic = new HashMap[String, Integer]()
  topic.put("test", 1)
  val logger = LoggerFactory.getLogger(getClass().getName())
  val messageStream = Consumer
    .createJavaConsumerConnector(new ConsumerConfig(new Properties()))
    .createMessageStreams(topic)
    .get(topic).get(0)
  for (message <- messageStream) {
    val gotMessage = new String(message.message())
    logger.info(gotMessage)
  }
}

我想到的测试场景如下:

这是类似 Scala 的伪代码中的测试草稿:

import uk.org.lidalia.slf4jtest.TestLoggerFactory;
import uk.org.lidalia.slf4jtest.LoggingEvent.info;

abstract class UnitSpec extends FlatSpec with Matchers with EmbeddedKafka {

}

class ConsumerAppSpec extends UnitSpec {
  "ConsumerApp" should "consume and log messages from Kafka on specific topic" in {
    withRunningKafka {
      val consumer = ConsumerApp
      // interecept logger to be able to test that the kafka message is logged
      val logger = TestLoggerFactory.getTestLogger(consumer.getClass)
      // start the application, but beforehand do something to prevent it infinitely blocking
      ???
      consumer.main(Array())  

      // publish a test message    
      publishStringMessageToKafka("test", "TEST")
      // Confirm that the message has been properly logged
      ???
    }
    EmbeddedKafka.stop()
  }
}

我的问题是前三个问号处的测试代码。如果我执行 main() 方法,它不会终止,从而阻止执行其余的测试。

标签: scalatestingapache-kafkakafka-consumer-apiscalatest

解决方案


让它发挥作用...

object ConsumerApp extends App {

   def doStuff() {
      val topic = ...
      // more stuff ...
   }

   doStuff()
 }

 class ConsumerAppSpec {
   // ...
   consumer.doStuff()
   // ...
   publishMessage() 
 }

更新

我想,我误解了这个问题。而不是“不要调用main”,我应该说不要阻塞线程:)

 class ConsumerAppSpec {
   ...
   val futureResult = Future(consumer.doStuff)
   ...
   publishMessage()
 }

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