spring - 如何调试 HttpMessageNotReadableException？

Question

我有一个这样的 DTO 对象:

data class ClientDto(
        var uuid: String,
        var ip: String,
        var lastSeen: Date? = Date()
) {
    internal fun toEntity() = Client(uuid=uuid, ip=ip, lastSeen=Date())
}

...和这样的 Controller :

@RestController
internal class ClientController {

    @Autowired
    private lateinit var service : ClientService

    @GetMapping("/clients")
    internal fun getClients() =  service.findAll()

    @PostMapping("/clients")
    internal fun postClient(@RequestBody client: ClientDto) = service.add(client)
}

现在我用 httpie 发布这样的东西:

http POST localhost:8080/clients uuid=my-uuid ip=192.123.31:8080

并得到:

{
    "error": "Bad Request",
    "exception": "org.springframework.http.converter.HttpMessageNotReadableException",
    "message": "JSON parse error: Can not construct instance of awesome.discovery.domain.dto.ClientDto: no suitable constructor found, can not deserialize from Object value (missing default constructor or creator, or perhaps need to add/enable type information?); nested exception is com.fasterxml.jackson.databind.JsonMappingException: Can not construct instance of awesome.discovery.domain.dto.ClientDto: no suitable constructor found, can not deserialize from Object value (missing default constructor or creator, or perhaps need to add/enable type information?)\n at [Source: java.io.PushbackInputStream@71c6ae97; line: 1, column: 2]",
    "path": "/clients",
    "status": 400,
    "timestamp": 1519587678605
}

1. 这篇文章有什么问题？
2. 更重要的是我该如何调试它，例如知道它尝试了什么以及如何了解它哪里出了问题？

Answer 1

1- 您的 DTO 需要提供一个默认构造函数(没有参数的空构造函数)。这就是错误所说的

2- Spring 根据RequestParam 的类型配置Jackson 使用反射自动反序列化JSON。您可以自定义此行为实现 JsonDeserializer。如果你想调试，你可以在用 PostMapping

注释的方法上放置一个断点