Date 蜂巢中的日期差异，差异应以hh:mm:ss为单位

我试图找出连续几行中两个日期之间的差异。我正在使用hive中的窗口功能，即，

lag

但不同之处在于，输出的格式应为

hh:mm:ss

例如：

• 日期1是2017-08-15 02:00:32
• 日期2是2017-08-15 02:00:20

输出应为：

00:00:12

我尝试使用的查询：

select from_unixtime(column_name),
(lag(unix_timestamp(from_unixtime(column_name)),1,0)
over(partition by column_name)-
unix_timestamp(from_unixtime(column_name))) as Duration from table_name;

但这会将输出返回为

12

（在上面的示例中）

更新 我已经使用bigint数据类型将该列存储在表中。时间是纪元格式的。我们在查询中使用from_unixtime将其转换为可读日期。时间戳中的样本值

1502802618

---

1502786788

只要时差小于24小时，答案就会相关

hive> with t as (select timestamp '2017-08-15 02:00:32' as ts1,timestamp '2017-08-15 02:00:20' as ts2)
    > select  ts1 - ts2   as diff
    > from    t
    > ;
OK
diff
0 00:00:12.000000000

给定时间戳

hive> with t as (select timestamp '2017-08-15 02:00:32' as ts1,timestamp '2017-08-15 02:00:20' as ts2)
    > select  split(ts1 - ts2,'[ .]')[1]  as diff
    > from    t
    > ;
OK
diff
00:00:12

给定字符串

hive> with t as (select '2017-08-15 02:00:32' as ts1,'2017-08-15 02:00:20' as ts2)
    > select  split(cast(ts1 as timestamp) - cast(ts2 as timestamp),'[ .]')[1]  as diff
    > from    t
    > ;
OK
diff
00:00:12

---

只要时差小于24小时，答案将是相关的

hive> with t as (select timestamp '2017-08-15 02:00:32' as ts1,timestamp '2017-08-15 02:00:20' as ts2)
    > select  ts1 - ts2   as diff
    > from    t
    > ;
OK
diff
0 00:00:12.000000000

给定时间戳

hive> with t as (select timestamp '2017-08-15 02:00:32' as ts1,timestamp '2017-08-15 02:00:20' as ts2)
    > select  split(ts1 - ts2,'[ .]')[1]  as diff
    > from    t
    > ;
OK
diff
00:00:12

给定字符串

hive> with t as (select '2017-08-15 02:00:32' as ts1,'2017-08-15 02:00:20' as ts2)
    > select  split(cast(ts1 as timestamp) - cast(ts2 as timestamp),'[ .]')[1]  as diff
    > from    t
    > ;
OK
diff
00:00:12

---

只要时差小于24小时，答案将是相关的

hive> with t as (select 1502802618 as ts1,1502786788 as ts2)
    > select  from_unixtime(to_unix_timestamp('0001-01-01 00:00:00')+(ts1 - ts2))  as diff
    > from    t
    > ;
OK
diff
0001-01-01 04:23:50

---

---

只要时差小于24小时，答案将是相关的

hive> with t as (select 1502802618 as ts1,1502786788 as ts2)
    > select  from_unixtime(to_unix_timestamp('0001-01-01 00:00:00')+(ts1 - ts2))  as diff
    > from    t
    > ;
OK
diff
0001-01-01 04:23:50

---

---

我试图执行您的查询，但它抛出错误“没有与（timestamp，timestamp）匹配的org.apache.hadoop.hive.ql.udf.generic.GenericUDFOPMinus类的方法”。为了提供更多关于表模式的信息，我在bigint中存储了一列，它在epoch timestamp中。这是一个经过测试的代码。显然，您的Hive版本不支持时间戳减法。我们使用的是Cloudera-5.3.1/Hive-0.13.1-cdh5.3.1。此版本不支持3年前（2014年6月6日）的版本？@Shash？。。。No@Shash，（1）在正确的位置进行注释（2）您谈论的是基本SQL，只需将

ts1

和

ts2

替换为相关的表达式（列和滞后（…）超过（…））我试图执行您的查询，但它抛出错误“没有与org.apache.hadoop.hive.ql.udf.generic.GenericUDFOPMinus类匹配的方法”（timestamp，timestamp）”。为了提供更多关于表架构的信息，我在bigint中存储了列，它在epoch timestamp中。这是一个经过测试的代码。显然您的配置单元版本不支持时间戳减法。我们使用的是Cloudera-5.3.1/Hive-0.13.1-cdh5.3.1。这个版本不支持？@Shash，三年前的版本（2014年6月6日）？。。。No@Shash，（1）在正确的位置进行注释（2）您正在谈论的是基本SQL，只需将

ts1

和

ts2

替换为相关的表达式（列和滞后（…）超过（…）