Sql 日期不同,但仅计算营业日
我有一个相当简单的问题;看起来是这样的:Sql 日期不同,但仅计算营业日,sql,google-bigquery,datediff,Sql,Google Bigquery,Datediff,我有一个相当简单的问题;看起来是这样的: SELECT order_date, pickup_date, DATE_DIFF(pickup_date,order_date, day) order_to_pickup FROM `orders.table` 唯一的问题是,我需要计算工作日的日期差异,而不是所有的日期 因此,与上面返回的查询不同: +------------+-------------+-----------------+ | order_date | pickup_date |
SELECT
order_date,
pickup_date,
DATE_DIFF(pickup_date,order_date, day) order_to_pickup
FROM
`orders.table`
唯一的问题是,我需要计算工作日的日期差异,而不是所有的日期
因此,与上面返回的查询不同:
+------------+-------------+-----------------+
| order_date | pickup_date | order_to_pickup |
+------------+-------------+-----------------+
| 3/29/19 | 4/3/19 | 5 |
| 3/29/19 | 4/2/19 | 4 |
+------------+-------------+-----------------+
我希望它能返回:
+------------+-------------+-----------------+
| order_date | pickup_date | order_to_pickup |
+------------+-------------+-----------------+
| 3/29/19 | 4/3/19 | 2 |
| 3/29/19 | 4/2/19 | 3 |
+------------+-------------+-----------------+
我认为如果你考虑两个日期之间的周数,就会有一个聪明的解决办法,但同时也有一种蛮力的方法:
CREATE TEMP FUNCTION BusinessDateDiff(start_date DATE, end_date DATE) AS (
(SELECT COUNTIF(MOD(EXTRACT(DAYOFWEEK FROM date), 7) > 1)
FROM UNNEST(GENERATE_DATE_ARRAY(
start_date, DATE_SUB(end_date, INTERVAL 1 DAY))) AS date)
);
对于您的输入,我得到:
CREATE TEMP FUNCTION BusinessDateDiff(start_date DATE, end_date DATE) AS (
(SELECT COUNTIF(MOD(EXTRACT(DAYOFWEEK FROM date), 7) > 1)
FROM UNNEST(GENERATE_DATE_ARRAY(
start_date, DATE_SUB(end_date, INTERVAL 1 DAY))) AS date)
);
WITH OrdersTable AS (
SELECT
DATE '2019-03-29' AS order_date,
DATE '2019-04-03' AS pickup_date UNION ALL
SELECT
'2019-03-29',
'2019-04-02'
)
SELECT
order_date,
pickup_date,
BusinessDateDiff(order_date, pickup_date) AS order_to_pickup
FROM OrdersTable
ORDER BY pickup_date
+------------+-------------+-----------------+
| order_date | pickup_date | order_to_pickup |
+------------+-------------+-----------------+
| 2019-03-29 | 2019-04-02 | 2 |
| 2019-03-29 | 2019-04-03 | 3 |
+------------+-------------+-----------------+
以下是一种基于Looker Discussion社区中的工作计算日期间工作日的工作方法。最初的示例是用于红移的,因此我将其改编为下面的BigQuery
SELECT
CAST(-1*(DATE_DIFF(DATE '2019-01-01', DATE '2019-01-31', DAY) - ((FLOOR(DATE_DIFF(DATE '2019-01-01', DATE '2019-01-31', DAY) / 7) * 2) +
CASE
WHEN EXTRACT(DAYOFWEEK FROM DATE '2019-01-01') - EXTRACT(DAYOFWEEK FROM DATE '2019-01-31') IN (1, 2, 3, 4, 5) AND EXTRACT(DAYOFWEEK FROM DATE '2019-01-31') != 0 THEN 2
ELSE 0
END +
CASE
WHEN EXTRACT(DAYOFWEEK FROM DATE '2019-01-01') != 0 AND EXTRACT(DAYOFWEEK FROM DATE '2019-01-31') = 0 THEN 1
ELSE 0
END +
CASE
WHEN EXTRACT(DAYOFWEEK FROM DATE '2019-01-01') = 0 AND EXTRACT(DAYOFWEEK FROM DATE '2019-01-31') != 0 THEN 1
ELSE 0 END)) AS int64) AS weekdays
将此应用于您的数据集,我们得到:
SELECT
order_date,
pickup_date,
CAST(-1*(DATE_DIFF(order_date, pickup_date, DAY) - ((FLOOR(DATE_DIFF(order_date, pickup_date, DAY) / 7) * 2) +
CASE
WHEN EXTRACT(DAYOFWEEK FROM order_date) - EXTRACT(DAYOFWEEK FROM pickup_date) IN (1, 2, 3, 4, 5) AND EXTRACT(DAYOFWEEK FROM pickup_date) != 0 THEN 2
ELSE 0
END +
CASE
WHEN EXTRACT(DAYOFWEEK FROM order_date') != 0 AND EXTRACT(DAYOFWEEK FROM pickup_date) = 0 THEN 1
ELSE 0
END +
CASE
WHEN EXTRACT(DAYOFWEEK FROM order_date) = 0 AND EXTRACT(DAYOFWEEK FROM pickup_date) != 0 THEN 1
ELSE 0 END)) AS int64) AS weekdays
FROM
`orders.table`
这应该是@Elliott Brossard提到的简化的非暴力解决方案:
select
order_date,
pickup_date,
case
when date_diff(pickup_date, order_date, week) > 0
then date_diff(pickup_date, order_date, day) - (date_diff(pickup_date, order_date, week) * 2)
else
date_diff(pickup_date, order_date, day)
end
from `orders.table`
另一种解决办法。删除周末天数:
SELECT
order_date,
pickup_date,
date_diff(pickup_date, order_date, day)
- date_diff(pickup_date, order_date, WEEK(SUNDAY))
- date_diff(pickup_date, order_date, WEEK(SATURDAY)) as order_to_pickup
FROM `orders.table`
工作日的定义是什么?每个工作日都是工作日吗?或者你有没有不算工作日的假日日历?@ElliottBrossard为了这个原因,我们可以说工作日严格来说是周一到周五。这可能会帮助你:是的!我想这正是我所需要的!非常感谢。