在mysql，数据如下：
#查询某一用户该日抽奖时间select draw_time from user_draw_log where user_id = 1 and draw_date=‘2016-03-09‘ order by id;+---------------------+| draw_time           |+---------------------+| 2016-03-09 13:52:46 || 2016-03-09 13:52:53 || 2016-03-09 13:53:01 || 2016-03-09 13:53:13 || 2016-03-09 13:53:25 |...

想计算每次抽奖时间之间的间隔 以便判断是否是并发插入 我的方法如下使用一个临时变量记录前一次的抽奖时间

select draw_time, timediff(draw_time,@prev_time) diff,(@prev_time:=draw_time) from user_draw_log where user_id = 1 and draw_date=‘2016-03-09‘ order by id;+---------------------+------------------+-------------------------+| draw_time           | diff             | (@prev_time:=draw_time) |+---------------------+------------------+-------------------------+| 2016-03-09 13:52:46 | -00:08:28.000000 | 2016-03-09 13:52:46     || 2016-03-09 13:52:53 | 00:00:07.000000  | 2016-03-09 13:52:53     || 2016-03-09 13:53:01 | 00:00:08.000000  | 2016-03-09 13:53:01     || 2016-03-09 13:53:13 | 00:00:12.000000  | 2016-03-09 13:53:13     || 2016-03-09 13:53:25 | 00:00:12.000000  | 2016-03-09 13:53:25     || 2016-03-09 13:53:32 | 00:00:07.000000  | 2016-03-09 13:53:32     || 2016-03-09 13:53:38 | 00:00:06.000000  | 2016-03-09 13:53:38     |...

有没更方便的方法实现这一功能呢？对所有用户都求相邻记录时间差该如何操作？

hive做法如下：

1.Hive row_number() 函数的高级用法 row_num 按照某个字段分区显示第几条数据

select imei,ts,fuel_instant,gps_longitude,gps_latitude,row_number() over (PARTITION BY imei ORDER BY ts ASC) as row_num from sample_data_2

2.row_num 是相互连续的，join 自身，然后时间相减可求差
create table obd_20140101 as

　　select a.imei,a.row_num,a.ts,COALESCE(unix_timestamp(a.ts, ‘yyyy-MM-dd HH:mm:ss.S‘), 0) - unix_timestamp(b.ts, ‘yyyy-MM-dd HH:mm:ss.S‘) as intervel ,a.fuel_instant,a.gps_speed as obd_speed,a.gps_status,a.gps_longitude,a.gps_latitude,a.direct_angle,a.obdspeed from obddata_20140101 a join obddata_20140101 b on a.imei = b.imei and a.row_num = b.row_num +1

事实上该方法有更加简便的方法，那就是hive的分析窗口函数：

create table obd_20140101 as

select imei,ts as ts1,fuel_instant,gps_longitude,gps_latitude,lead(ts,1,ts) over  (PARTITION BY imei ORDER BY ts ASC)  as ts2 from sample_data_2;

这样，数据会按imei分组，并按时间排序。接下来的时间相减就简单了。

select a.imei,a.row_num,a.ts,COALESCE(unix_timestamp(a.ts1, ‘yyyy-MM-dd HH:mm:ss.S‘), 0) - unix_timestamp(a.ts2, ‘yyyy-MM-dd HH:mm:ss.S‘) as intervel ,a.fuel_instant,a.gps_speed as obd_speed,a.gps_status,a.gps_longitude,a.gps_latitude,a.direct_angle,a.obdspeed from obddata_20140101 a;

hive实现根据用户分组，按用户记录求上下两条记录的时间差