Jackson Cannot deserialize value of type `xxx` from String “xxx“ : Failed to deserialize xxx问题解决-蒲公英云

Jackson Cannot deserialize value of type `xxx` from String “xxx“ : Failed to deserialize xxx问题解决

问题描述：

com.fasterxml.jackson.databind.exc.InvalidFormatException: Cannot deserialize value of type `java.time.LocalDateTime` from String “2022-05-18 11:35:35”: Failed to deserialize java.time.LocalDateTime: (java.time.format.DateTimeParseException) Text ‘2022-05-18 11:35:35’ could not be parsed at index 10
at [Source: (String)”{“localDate”:”2022-05-18”,”localDateStr”:null,”localDateTime”:”2022-05-18 11:35:35”,”localDateTimeStr”:null}“; line: 1, column: 63] (through reference chain: com.xudongbase.common.model.TestModel[“localDateTime”])

问题分析：

1、由于yyyy-MM-dd HHss 格式的String数据不能转换LocalDateTime类型数据，导致报错。

/**
     * 测试单个实体反序列化和反序列化
     */
    @Test
    public void testModelSerializeAndDeserialize() throws Exception {
        TestModel testModel = new TestModel();
        testModel.setLocalDate(LocalDate.now());
        testModel.setLocalDateTime(LocalDateTime.now());
        String jsonStr = objectMapper.writeValueAsString(testModel);
        testModel = objectMapper.readValue(objectMapper.getFactory().createParser(jsonStr),TestModel.class);
    }

解决办法：LocalDateTime类型变量添加@JsonFormat注解指定日期格式即可。

@Data
public class TestModel {
    private LocalDate localDate;
    private String localDateStr;
    @JsonFormat(pattern = "yyyy-MM-dd HH:mm:ss",timezone="GMT+8")
    private LocalDateTime localDateTime;
    private String localDateTimeStr;
}