如何使用Jackson解析LocalDate

Question

我在通过邮递员传递json数据时遇到问题。如果Mandetory日期带有@NotNull注释，没问题。另一个日期接受null。但是在更新时，该日期会产生问题。我使用Java1.8与Spring启动

以下链接我已经访问过，但不适合这个。杰克逊JSON解析错误的LocalDateTime解析：无法构造java.time.LocalDate的实例：没有String参数构造函数/工厂方法从String值反序列化https://stackoverflow.com/a/29959842/3415090JSONJavaSpring Boot中的8 LocalDateTime格式

我也用过

@JsonSerialize(using = LocalDateSerializer.class)
@JsonDeserialize(using = LocalDateDeserializer.class)

但是问题仍然存在。

我的用户体验

public class LeaveApplUx {

    @JsonIgnore
    @Size(max = 5, message = "Employee Code Must Be Within 4 To 5 Character Long Or Blank")
    private final String employeeCode;

    @NotNull(message = "Start Date Empty")
    @JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd-MM-yyyy")
    private final LocalDate startDate;

    @JsonIgnore
    @JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd-MM-yyyy")
    private final LocalDate rejoinDate;

    public LeaveApplUx(
            @Size(min = 4, max = 5, message = "Employee Code Must Be Within 4 To 5 Character Long Or Blank") 
        @JsonProperty("employeeCode") String employeeCode,
            @NotNull(message = "Start Date Empty") @JsonProperty("startDate") LocalDate startDate,
            @JsonProperty("rejoinDate") LocalDate rejoinDate) {
        this.employeeCode = employeeCode;
        this.startDate = startDate;
        this.rejoinDate = rejoinDate;
    }
//  GETTERS
}

在创作时，它工作得很好。

{
    "employeeCode": "B426",
    "startDate": "01-03-2023"
}

输入参数：{“员工代码”：“B426”、“开始日期”：{“年份”：2023、“月份”：“三月”、“月值”：3、“日月”：1、“跨年”：假、“日周”：“周三”、“日年”：60、“时代”：“CE”、“年表”：{“id”：“ISO”、“calendarType”：“iso8601”}}、“rejoinDate”：null}记录在DB中正确保存

但是在更新时，它会产生错误。

{
    "employeeCode": "B426",
    "startDate": "01-03-2023",
    "rejoinDate": "06-03-2023"
}

JSON parse error: 

Cannot deserialize value of type `java.time.LocalDate` from String "06-03-2023": Failed to deserialize java.time.LocalDate: (java.time.format.DateTimeParseException) Text '06-03-2023' could not be parsed at index 0; nested exception is com.fasterxml.jackson.databind.exc.InvalidFormatException: Cannot deserialize value of type `java.time.LocalDate` from String "06-03-2023": Failed to deserialize java.time.LocalDate: (java.time.format.DateTimeParseException) Text '06-03-2023' could not be parsed at index 0
     at [Source: (PushbackInputStream); line: 14, column: 19] (through reference chain: org.myapp.ux.hr.leave.LeaveApplUx["rejoinDate"])

Answer 1

由于您通过构造函数而不是setter设置值，因此您应该将@JsonFormat（...）放在构造函数参数上，而不是字段上。这应该可以解决它：

public class LeaveApplUx {

    @JsonIgnore
    private final String employeeCode;

    private final LocalDate startDate;

    @JsonIgnore
    private final LocalDate rejoinDate;

    public LeaveApplUx(@JsonProperty("employeeCode") String employeeCode,
                       @JsonProperty("startDate") @JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd-MM-yyyy") LocalDate startDate,
                       @JsonProperty("rejoinDate") @JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd-MM-yyyy") LocalDate rejoinDate) {
        this.employeeCode = employeeCode;
        this.startDate = startDate;
        this.rejoinDate = rejoinDate;
    }

    //getters
}