如何在 spring Restapi 的 json 请求中的未知字段上抛出错误

我有一个 spring Rest api，它获取 json 数据并绑定到 pojo GetData。 每当我收到未知字段时，它都不会失败或抛出任何异常。我的要求是当它接收到 json 数据中的未知字段时应该抛出错误。

public ResponseEntity<Error> saveLocation(@Valid @RequestBody GetData getdata,BindingResult bindingResults) {

下面是我的 Pojo GetData

public class GetData{

@JsonProperty("deviceID")
@Pattern(regexp="^[\\p{Alnum}][-\\p{Alnum}\\p{L}]+[\\p{Alnum}]$",message = "Not a valid Device Id")
private String deviceID;


@JsonProperty("Coordinates")
@Pattern(regexp="^[\\p{Alnum}\\-][\\.\\,\\-\\_\\p{Alnum}\\p{L}\\s]+|",message = "Coordinates are not valid")
private String coordinates;}

下面是我的 json 请求。

{
"deviceID" : "01dbd619-843b-4197-b954",
"Coordinates" : "12.984012,80.246712",
}

现在，如果我发送带有额外字段的请求，例如国家/地区。它不会抛出任何错误。

{
    "deviceID" : "01dbd619-843b-4197-b954",
    "Coordinates" : "12.984012,80.246712",
    "country" : "dsasa"
}

请建议我如何在 json 请求中发送未知属性时出现错误

您可以尝试以下任何一种实现，它对我有用。您将必须重写另一种方法ResponseEntityExceptionHandler或通过使用ExceptionHandler.

1.通过重写ResponseEntityExceptionHandler的方法

import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
import org.springframework.http.HttpHeaders;
import org.springframework.http.HttpStatus;
import org.springframework.http.ResponseEntity;
import org.springframework.http.converter.HttpMessageNotReadableException;
import org.springframework.web.bind.annotation.ControllerAdvice;
import org.springframework.web.context.request.WebRequest;
import org.springframework.web.servlet.mvc.method.annotation.ResponseEntityExceptionHandler;


@ControllerAdvice
public class CustomExceptionHandler extends ResponseEntityExceptionHandler {

    private static final Logger log = LoggerFactory.getLogger(CustomExceptionHandler.class);

    //Other Handlers

    // Handle 400 Bad Request Exceptions

    @Override
    protected ResponseEntity<Object> handleHttpMessageNotReadable(HttpMessageNotReadableException ex, HttpHeaders headers, HttpStatus status, WebRequest request) {
        log.info(ex.getLocalizedMessage() + " ",ex);

        final CustomErrorMessage errorMessage = new CustomErrorMessage(ex.getLocalizedMessage(), InfoType.ERROR, HttpStatus.BAD_REQUEST, ex.fillInStackTrace().toString());
        return handleExceptionInternal(ex, errorMessage, headers, errorMessage.getStatus(), request);
    }

    //Other Handlers
}

除了上面的实现之外，如果您只想在请求负载中存在无法识别的属性或空属性并且存在空值时抛出错误（如下所示），您还可以尝试下面的实现JSON

   {
     "":""
   }

2. 使用异常处理程序

import com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException;
import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
import org.springframework.core.Ordered;
import org.springframework.core.annotation.Order;
import org.springframework.http.HttpStatus;
import org.springframework.http.ResponseEntity;
import org.springframework.web.bind.annotation.ControllerAdvice;
import org.springframework.web.bind.annotation.ExceptionHandler;
import org.springframework.web.bind.annotation.ResponseStatus;


@Order(Ordered.HIGHEST_PRECEDENCE)
@ControllerAdvice
public class GenericExceptionHandler {

    private static final Logger log = LoggerFactory.getLogger(GenericExceptionHandler.class);

    @ExceptionHandler(value = {UnrecognizedPropertyException.class})
    @ResponseStatus(HttpStatus.BAD_REQUEST)
    protected ResponseEntity<Object> handleUnrecognizedPropertyException(UnrecognizedPropertyException ex) {
        log.info(ex.getLocalizedMessage() + " ",ex);

        final String error = "JSON parse error: Unrecognized field " + "[ " + ex.getPropertyName() + " ]";

        final CustomErrorMessage errorMessage = new CustomErrorMessage(error, InfoType.ERROR, HttpStatus.BAD_REQUEST);
        return ResponseEntity.status(HttpStatus.BAD_REQUEST).body(errorMessage);
    }
}

Note :为了使上述两个实现正常工作，您需要在您的application.properties file.

spring.jackson.deserialization.fail-on-unknown-properties=true

希望对你有帮助 ：）