我无法从示例记录中获得平均下载分钟,如ANSI-92 SQL或Impala SQL标准中所示。
以粗体显示的时间分享starttimestamp和stoptimestamp的时间空间(以分钟为单位)。如何获得平均下载
等等。
有什么建议吗?非常感谢你提前!
此致
Pozy
答案 0 :(得分:1)
select
(unix_timestamp(stoptimestamp)-unix_timestamp(starttimestamp)) / 60.0 diff_minutes
from your_table
使用unix_timestamp()计算以秒为单位的差值,然后除以60或60.0,具体取决于结果中所需的精度。
要计算多行的平均下载量,您需要使用SUM()来聚合字节,并使用时间单位。 您可能希望使用秒进行初始计算,然后除以60.0
以下示例是针对SQL Sever编写的,因为我没有使用Impala
declare @mytable table
([login_id] varchar(11), [starttimestamp_] datetime, [stoptimestamp_] datetime, [download_bytes] decimal(12,1))
;
INSERT INTO @mytable
([login_id], [starttimestamp_], [stoptimestamp_], [download_bytes])
VALUES
('abc@fcc.com', '2015-12-31 23:59:50', '2016-01-01 00:00:20', 438.0),
('abc@fcc.com', '2016-01-01 00:00:28', '2016-01-01 00:01:13', 2190.0),
('abc@fcc.com', '2016-01-01 00:01:21', '2016-01-01 00:01:54', 876.0),
('abc@fcc.com', '2016-01-01 00:01:59', '2016-01-01 00:02:34', 1168.0),
('abc@fcc.com', '2016-01-01 00:02:43', '2016-01-01 00:03:34', 1179.0)
;
select
sum(download_bytes) sum_bytes
, sum(datediff(second,starttimestamp_,stoptimestamp_)) sum_time_unit
, sum(download_bytes)/sum(datediff(second,starttimestamp_,stoptimestamp_)) avg_bytes_sec
, (sum(download_bytes)/sum(datediff(second,starttimestamp_,stoptimestamp_)))/60.0 avg_bytes_min
from @mytable
-- WHERE ...
-- GROUP BY ...
+===========+===============+===============+===============+
| sum_bytes | sum_time_unit | avg_bytes_sec | avg_bytes_min |
+===========+===============+===============+===============+
| 5851 | 194 | 30.159793 | 0.502663 |
+-----------+---------------+---------------+---------------+