spring 中无法构造 `java.security.Timestamp` 错误的实例 Rest

Question

我有一个简单的 Rest 控制器，如下所示：

@PostMapping(value = "/DepositBalance")
public DepositBalanceResponse DepositBalance(@Valid @RequestBody DepositBalanceRequest requestDto) throws Exception {
    return depositService.GetDepositBalance(requestDto);
}

这是 DepositBalanceRequest 字段：

private String profileId;
private String userName;
@JsonFormat(pattern = "yyyy-MM-dd HH:mm:ss")
private Timestamp timestamp;

当我像这样通过 Postman 调用服务时：

{
    "profileId":"asdfasdf",
    "userName":"username",
    "timestamp":"2018-12-30 20:12:20",
}

我收到以下错误：

"Type definition error: [simple type, class java.security.Timestamp]; nested exception is com.fasterxml.jackson.databind.exc.InvalidDefinitionException: Cannot construct instance of `java.security.Timestamp` (no Creators, like default construct, exist): no String-argument constructor/factory method to deserialize from String value ('2018-12-30 20:12:20')\n at [Source: (PushbackInputStream); line: 4, column: 14] (through reference chain: com.api.v1.deposit.DepositBalanceRequest[\"timestamp\"])"

我想知道我是否错过了什么？

Answer 1

感谢M.Deinum，需要使用 LocalDateTime 而不是 Timestamp

private String profileId;
private String userName;
@JsonFormat(pattern = "yyyy-MM-dd HH:mm:ss")
private LocalDateTime timestamp;

spring 中无法构造 `java.security.Timestamp` 错误的实例 Rest

Cannot construct instance of `java.security.Timestamp` Error in spring Rest

spring

json

timestamp

jackson

spring-rest