Mysql數(shù)據(jù)庫中datetime、bigint、timestamp來表示時間選擇,誰來存儲時間效率最高

CREATE TABLE `users` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `time_date` datetime NOT NULL,
  `time_timestamp` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
  `time_long` bigint(20) NOT NULL,
  PRIMARY KEY (`id`),
  KEY `time_long` (`time_long`),
  KEY `time_timestamp` (`time_timestamp`),
  KEY `time_date` (`time_date`)
) ENGINE=InnoDB AUTO_INCREMENT=500003 DEFAULT CHARSET=latin1

/**
 * @author hetiantian
 * @date 2018/10/21
 * */
@Builder
@Data
public class Users {
    /**
     * 自增唯一id
     * */
    private Long id;

    /**
     * date類型的時間
     * */
    private Date timeDate;

    /**
     * timestamp類型的時間
     * */
    private Timestamp timeTimest

    /**
     * long類型的時間
     * */
    private long timeLong;
}

/**
 * @author hetiantian
 * @date 2018/10/21
 * */
@Mapper
public interface UsersMapper {
    @Insert("insert into users(time_date, time_timestamp, time_long) value(#{timeDate}, #{timeTimestamp}, #{timeLong})")
    @Options(useGeneratedKeys = true,keyProperty = "id",keyColumn = "id")
    int saveUsers(Users users);
}

public class UsersMapperTest extends BaseTest {
    @Resource
    private UsersMapper usersMapper;

    @Test
    public void test() {
        for (int i = 0; i  500000; i++) {
            long time = System.currentTimeMillis();
            usersMapper.saveUsers(Users.builder().timeDate(new Date(time)).timeLong(time).timeTimestamp(new Timestamp(time)).build());
        }
    }
}

select count(*) from users where time_date >="2018-10-21 23:32:44" and time_date ="2018-10-21 23:41:22"

select count(*) from users where time_timestamp >= "2018-10-21 23:32:44" and time_timestamp ="2018-10-21 23:41:22"

select count(*) from users where time_long >=1540135964091 and time_long =1540136482372

select time_date, count(*) from users group by time_date

select time_timestamp, count(*) from users group by time_timestamp

select * from users order by time_date

select * from users order by time_timestamp

select * from users order by time_long