@JsonFormat与@DateTimeFormat注解的使用和Timestamp取出来1970问题和@JSONField(name="Timestamp",deserializeUsing= FastJsonLocalDateTimeDeserializer.class)

总结： 

  注解@JsonFormat主要是后台到前台的时间格式的转换

  注解@DataFormAT主要是前后到后台的时间格式的转换

@JSONField 是解决

UserDto dto = JSONObject.parseObject(strJson, UserDto.class); 解决从Json里面取出来的时间戳 为1970的问题

public class FastJsonLocalDateTimeDeserializer implements ObjectDeserializer {

    private static List<DateTimeFormatter> dateTimeFormatters = new LinkedList<>();

    static {
        // Add your own formatter to there
        dateTimeFormatters.add(DateTimeFormatter.ofPattern("yyyy-MM-dd HH:mm:ss"));
        dateTimeFormatters.add(DateTimeFormatter.ofPattern("yyyy-MM-dd HH:mm:ss.SSS"));
        dateTimeFormatters.add(DateTimeFormatter.ofPattern("yyyy-MM-dd HH:mm:ss.SSSSSS"));
        dateTimeFormatters.add(DateTimeFormatter.ofPattern("yyyy-MM-dd HH:mm:ss.SSSSSSSSS"));
    }

    @SuppressWarnings("unchecked")
    @Override
    public LocalDateTime deserialze(DefaultJSONParser parser, Type type, Object fieldName) {
        final Long input = parser.lexer.longValue();
        LocalDateTime localDateTime = null;
        localDateTime=LocalDateTime.ofEpochSecond(input,0,ZoneOffset.of("+8"));
        Assert.notNull(localDateTime, "FastJson LocalDateTime use" +
                " FastJsonTimestampDeserializer format error: " + input);
        return localDateTime;
    }

    @Override
    public int getFastMatchToken() {
        return JSONToken.LITERAL_INT;
    }
}