在 Google BigQuery 中使用 TIME_DIFF 和多个条件

Posted 2023-03-24

【中文标题】在 Google BigQuery 中使用 TIME_DIFF 和多个条件

【英文标题】：Using TIME_DIFF with multiple conditions in Google BigQuery

【发布时间】：2021-04-08 03:01:04

【问题描述】：

我正在尝试在 Google BigQuery (SQL) 中计算特定日期的工作时间。

白天工作的工资是 10 美元，而夜间工作的工资是 15 美元。 白天时间定义为早上 6 点到晚上 10 点，而晚上时间定义为晚上 10 点到早上 6 点。

员工可以灵活地工作，因为他们是豪华轿车司机。

以下是我的表格示例：

id	start_at	end_at	date
abc123	04:00:00	07:00:00	2020-01-05
abc123	09:00:00	15:32:00	2020-01-05
abc123	23:00:00	23:35:00	2020-01-05
abc123	23:40:00	23:59:00	2020-01-05
abc123	23:59:00	01:35:00	2020-01-05
abc123	02:02:00	04:35:00	2020-01-06
abc123	05:40:00	06:59:00	2020-01-06

因此，实际工作时间是通过计算 start_at 和 end_at 之间的差异来计算的，但是白天和晚上的时间条件在我的查询中变得很麻烦..

*日期列基于 start_at。即使您从晚上 11:59 开始并在第二天凌晨 12:05 结束，日期也会跟随 start_at 而不是 end_at 的日期。

有什么想法吗？提前致谢！

【问题讨论】：

付款的最短时间单位是多少？也就是说，如果你工作 72 分钟，你会得到 1 小时的报酬吗？或超过1小时？你被这个数据模型困住了吗？即您可以添加到数据模型中吗？

【参考方案1】：

考虑以下解决方案

create temp function night_day_split(start_at time, end_at time, date date) as (array(
  select as struct 
    extract(date from time_point) day,
    if(extract(hour from time_point) between 6 and 22, 'day', 'night') day_night,
    count(1) minutes
  from unnest(generate_timestamp_array(
      timestamp(datetime(date, start_at)), 
      timestamp(datetime(if(start_at < end_at, date, date + 1), end_at)), 
      interval 1 minute
    )) time_point
  group by 1, 2
));
select id, day, 
  sum(if(day_night = 'day', minutes, null)) day_minutes,
  sum(if(day_night = 'night', minutes, null)) night_minutes
from yourtable, 
unnest(night_day_split(start_at, end_at, date)) v
group by id, day     

如果应用于您问题中的样本数据 - 输出是

【参考方案2】：

你可以试试下面的代码：-

with mytable as (
  select 'abc123' id, cast( '04:00:00' as time) start_dt, cast( '07:00:00' as time) end_dt, date('2020-01-05' ) date union all
  select 'abc123', cast(    '09:00:00' as time), cast(  '15:32:00' as time), date('2020-01-05') union all
  select 'abc123', cast(    '23:00:00' as time), cast(  '23:35:00' as time), date('2020-01-05' )  union all
select 'abc123', cast('23:40:00' as time), cast(    '23:59:00' as time), date('2020-01-05') union all
select 'abc123', cast ('23:59:00' as time), cast(   '01:35:00' as time), date('2020-01-05') union all
select 'abc123', cast('02:02:00' as time), cast(    '04:35:00' as time), date('2020-01-06') union all
select 'abc123', cast('05:40:00'  as time), cast(   '06:59:00' as time), date('2020-01-06')
)
select id, date, sum (value) as sal from(
select id, date, 
case when start_dt > cast(  '06:00:00' as time) and  end_dt < cast( '22:00:00' as time) and start_dt < end_dt  then (time_diff(end_dt, start_dt, Minute)/60) * 10
when start_dt < cast(   '06:00:00' as time) and  end_dt < cast( '06:00:00' as time) then (time_diff(end_dt, start_dt, Minute)/60) * 15
when start_dt < cast(   '06:00:00' as time) and  end_dt < cast( '22:00:00' as time) then (time_diff(cast(   '06:00:00' as time), start_dt, Minute)/60) * 15 + (time_diff( end_dt,cast(  '06:00:00' as time), Minute)/60) * 10
when start_dt > cast(   '22:00:00' as time) and  end_dt < cast( '06:00:00' as time) then (time_diff(cast(   '23:59:00' as time), start_dt, Minute)/60) * 15 + (time_diff( end_dt,cast(  '00:00:00' as time), Minute)/60) * 15
when start_dt > cast(   '22:00:00' as time) and  end_dt > cast( '22:00:00' as time) then (time_diff(end_dt, start_dt, Minute)/60) * 15
else 0
end as value
from mytable) group by id, date

输出：-

您可以进一步按月分组以获得月薪。