将另一个对象传递给当前的休息服务不会引发异常

Posted 2023-02-19

【中文标题】将另一个对象传递给当前的休息服务不会引发异常

【英文标题】：passing another object into current rest service does not throw an exception

【发布时间】：2019-12-06 00:45:25

【问题描述】：

这是与其他服务兼容的休息请求：

  "fromDate": 1562773101000,
  "toDate": 1563118701000,
  "turnOverType": 4,
  "fromAmount": 1,
  "toAmount": 10000000,
  "voucherDescription": null,
  "articleDescription": null,
  "referenceNumbers": [],
  "offset": 3,
  "pageSize": 20,
  "iban": "BLAHBLAHBLAHBLAH"

这是与请求不匹配的对应模型：

@XmlAccessorType(XmlAccessType.FIELD)
@XmlRootElement(name = "TransferRequestInquiryFilter")
public class TransferRequestInquiryFilter implements Serializable 
    @XmlElement(name = "sourceIbans")
    private List<String> sourceIbans;
    @XmlElement(name = "transferType")
    private TransferType transferType;
    @XmlElement(name = "fromTransferDate")
    private Timestamp fromTransferDate;
    @XmlElement(name = "toTransferDate")
    private Timestamp toTransferDate;
    @XmlElement(name = "fromRegistrationDate")
    private Timestamp fromRegistrationDate;
    @XmlElement(name = "toRegistrationDate")
    private Timestamp toRegistrationDate;
    @XmlElement(name = "trackingNumbers")
    private List<String> trackingNumbers;
    @XmlElement(name = "referenceNumbers")
    private List<String> referenceNumbers;
    @XmlElement(name = "transactionIds")
    private List<String> transactionIds;
    @XmlElement(name = "status")
    private TransactionStatus status;
    @XmlElement(name = "fromAmount")
    private Long fromAmount;
    @XmlElement(name = "toAmount")
    private Long toAmount;
    @XmlElement(name = "destinationIbans")
    private List<String> destinationIbans;

这是我的控制器..

@RequestMapping(value = "/inquiry", method = RequestMethod.POST)
public @ResponseBody
ResponseEntity<List<ExtendedTransferRequest>> transferInquiry(@RequestBody @Valid TransferRequestInquiryFilter transferRequestInquiryFilter
        , BindingResult bindingResult) 

    // when validation not works return bad request
    List<ErrorObject> errorObjects = requestInquiryValidator.validate(transferRequestInquiryFilter);
    if (errorObjects.size() > 0) 
        // just throw bad request and not detail of them
        return new ResponseEntity<>(HttpStatus.BAD_REQUEST);
    

    List<ExtendedTransferRequest> extendedTransferRequestList = new ArrayList<>();
    ExtendedTransferRequest extendedTransferRequest = new ExtendedTransferRequest();
    List<SettlementTransaction> settlementTransactionList = settlementSearch.findSettlement(transferRequestInquiryFilter);
    extendedTransferRequestList = TransferInquiryResponseMapper.INSTANCE.SettlementTransactionInquiryResponse(setlementTransactionList);
    return new ResponseEntity<>(extendedTransferRequestList, HttpStatus.OK);

只是 fromAmount 和 toAmount 填充。但我想在这种情况下得到一个例外，并向客户抛出一个错误的请求。我怎样才能做到这一点？如果我在 rest request 和 model 之间出现名称冲突或类型冲突，我需要处理它并向客户提出错误的请求。我正在使用 spring mvc 5 和 jackson-core 和 jackson-databind 2.9.4

【问题讨论】：

完全不清楚你的问题是什么。

服务参数与 rest request 不同，在这种情况下会出现 HttpMessageNotReadableException。但没有异常发生。

【参考方案1】：

使用validation-api，在控制器方法之前注释字段和@validated的正确验证，在RequestBody对象之前使用@valid会抛出正确的验证异常。