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

妖狐艹你老母 2023-09-29 14:58 14阅读 0赞

## 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 HH:mm:ss 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;
    }