Question

这是我的代表JSON的POJPO类

    @JsonProperty("employee_id")
    private String employeeId;
    private String name;

//  @JsonProperty("birth_date")
//  //@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd/MM/yyyy")
//  @JsonSerialize(using = LocalDateSerializer.class)
//    private LocalDate birthDate;


    public Employee(String employeeId, String name) {
        super();
        this.employeeId = employeeId;
        this.name = name;
    }

我的制作人课程。效果很好

@Autowired
private RabbitTemplate rabbitTemplate;

private ObjectMapper objectMapper = new ObjectMapper();

//@Scheduled(fixedRate=500)
public void sendHello (Employee emp ) throws JsonProcessingException
{

      String json = objectMapper.writeValueAsString(emp);
    rabbitTemplate.convertAndSend("course.employee", json );
}

我在RabbitMQ中的有效负载显示以下Json消息

headers:    
content_encoding:   UTF-8
content_type:   text/plain
Employee info ->{"name":"Employee 0","employee_id":"emp 0"}

我的消费者阶层。这是造成问题的原因。

私有ObjectMapper objectMapper = new ObjectMapper（）;

@RabbitListener(queues="course.employee" )
public void listen( String message ) throws InterruptedException, JsonParseException, JsonMappingException, IOException
{
    Employee emp ;

    emp= objectMapper.readValue(message, Employee.class);

//  Thread.sleep(ThreadLocalRandom.current().nextLong(2000));
}

下面是错误消息

org.springframework.amqp.rabbit.listener.exception.ListenerExecutionFailedException:
 Listener method 'public void
 com.example.hibernatemapping.rabbitmqProducer.HelloRabbitConsumer.listen(java.lang.String)
 throws java.lang.InterruptedException,com.fasterxml.jackson.core.JsonParseException,com.fasterxml.jackson.databind.JsonMappingException,java.io.IOException'
 threw exception    at
 Caused by: com.fasterxml.jackson.databind.exc.InvalidDefinitionException: Cannot construct instance of `com.example.hibernatemapping.rabbitmqProducer.Employee` (no Creators, like default construct, exist): cannot deserialize from Object value (no delegate- or property-based Creator)
 at [Source: (String)"{"name":"Employee 4","employee_id":"emp 4"}"; line: 1, column: 2]
    at com.fasterxml.jackson.databind.exc.InvalidDefinitionException.from(InvalidDefinitionException.java:67) ~[jackson-databind-2.9.8.jar:2.9.8]

Answer 1

编辑问题后，情况与下面有所不同。同样，错误消息告诉您所有您需要了解的信息：

没有像默认构造那样的创造者存在

添加这样的构造函数：

public Employee() {
}

日志告诉您，您输入的是

Employee info ->{"name":"Employee  0","employee_id":"emp 0"}

这不是有效的JSON。

不要这样做：

rabbitTemplate.convertAndSend("course.employee","Employee info ->" + json );

执行以下操作：

rabbitTemplate.convertAndSend("course.employee", json);

由以下原因引起的错误：com.fasterxml.jackson.core.JsonParseException：无法识别的令牌“ Employee”：正在期待（“ true”，“ false”或“ null”）

1 个答案: