Question

Spring启动无法加载属性。以下是我通过yaml文件使用的属性。

 spring:
  kafka:
    bootstrap-servers: localhost:9092
    consumer:
      auto-commit-interval: 100
      enable-auto-commit: true
      group-id: ********************
      auto-offset-reset: earliest
      value-deserializer: org.springframework.kafka.support.serializer.JsonDeserializer
    producer:
      batch-size: 16384
      buffer-memory: 33554432
      retries: 0
      value-serializer: org.springframework.kafka.support.serializer.JsonSerializer
    listener:
      poll-timeout: 20000

我得到的例外是这个

引起：java.lang.IllegalAccessException：类org.apache.kafka.common.utils.Utils无法使用修饰符＆＃34访问org.springframework.kafka.support.serializer.JsonDeserializer类的成员被保护的＆＃34;

我认为构造函数受到保护。请提供一种实例化方法。

Answer 1

那是对的。参见：

protected JsonDeserializer() {
        this((Class<T>) null);
    }

    protected JsonDeserializer(ObjectMapper objectMapper) {
        this(null, objectMapper);
    }

    public JsonDeserializer(Class<T> targetType) {
        this(targetType, new ObjectMapper());
        this.objectMapper.configure(MapperFeature.DEFAULT_VIEW_INCLUSION, false);
        this.objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
    }

JsonDeserializer并非设计为由默认构造函数实例化，因为它需要知道要反序列化的targetType。

您可以将此类扩展为您的特定类型：

public class FooJsonDeserializer extends JsonDeserializer<Foo> { }

并将此作为该value-deserializer属性的类值使用。

或者您可以考虑自定义DefaultKafkaConsumerFactory：

@Bean
public ConsumerFactory<?, ?> kafkaConsumerFactory(KafkaProperties properties) {
    Map<String, Object> consumerProperties = properties.buildConsumerProperties();
    consumerProperties.put(CommonClientConfigs.METRIC_REPORTER_CLASSES_CONFIG,
            MyConsumerMetricsReporter.class);
    DefaultKafkaConsumerFactory<Object, Object> consumerFactory =
          new DefaultKafkaConsumerFactory<>(consumerProperties);
    consumerFactory.setValueDeserializer(new JsonDeserializer<>(Foo.class));
    return consumerFactory;
}

Spring Boot自动配置加载Spring Kafka属性失败

1 个答案: