Json无法在Spring中反序列化实例错误

时间:2018-10-05 13:38:36

标签: java json spring http

我正在尝试发送json长列表并从db获取记录。

我的控制器是:

@Api(tags = Endpoint.RESOURCE_customer, description = "customer Resource")
@RestController
@RequestMapping(produces = MediaType.APPLICATION_JSON_VALUE,consumes = MediaType.APPLICATION_JSON_VALUE)
@ResponseStatus(HttpStatus.OK)
public class CustomerResourceController {

    private final customerService customerService;


    public CustomerResourceController(customerService customerService) {

        this.customerService = customerService;
    }

    @ApiOperation(
            value = "Return customer",
            response = customerDto.class, responseContainer="List"
    )
    @PostMapping(value = Endpoint.RRESOURCE_customer_ID)
    public List<customerDto> getCustomersByIds(@RequestBody List<Long> ids) {
        return customerService.findcustomerIds(ids);
    }

}

和客户端类是:

@Headers("Content-Type: " + MediaType.APPLICATION_JSON_VALUE)
public interface CustomerClient {

@RequestLine("POST /customer/customers/search")
    List<LocGrpDto> getCustomersByIds(@RequestBody @Validated List<Long> ids);
}

然后我使用JSON在邮递员中测试此服务:

  

{“ ids :: [1,7,8]}

但是我得到这个错误:

{
    "timestamp": "2018-10-05T13:29:57.645+0000",
    "status": 400,
    "error": "Bad Request",
    "message": "Could not read document: Can not deserialize instance of java.util.ArrayList out of START_OBJECT token\n at [Source: java.io.PushbackInputStream@3cb8b584; line: 1, column: 1]; nested exception is com.fasterxml.jackson.databind.JsonMappingException: Can not deserialize instance of java.util.ArrayList out of START_OBJECT token\n at [Source: java.io.PushbackInputStream@3cb8b584; line: 1, column: 1]",
    "path": "/api/v1/customer/customers/search",
    "errors": []
}

出什么问题了?您在这里看到任何问题,还是可能是由于我的服务类或dto类引起的?

1 个答案:

答案 0 :(得分:3)

尝试使用有效负载[1,7,8]而不是{"ids": [1,7,8]}进行请求。

您的JSON将转换为具有以下格式的请求正文。

class Body {

    private List<Long> ids;

    // constructor, getters and setters 

}

对于REST客户端,您可以查看RestTemplate

RestTemplate template;
List<Long> ids;
List<CustomerDto> = template.exchange(
            "/customer/customers/search",
            HttpMethod.POST,
            new HttpEntity<>(ids),
            new ParameterizedTypeReference<List<CustomerDto>>() {}).getBody()