DATE_DIFF 但只计算工作日

Question

我有一个相当简单的查询；它看起来是这样的：

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，而不是所有天数。

所以不是上面的查询返回：

+------------+-------------+-----------------+
| order_date | pickup_date | order_to_pickup |
+------------+-------------+-----------------+
| 3/29/19    | 4/3/19      |               5 |
| 3/29/19    | 4/2/19      |               4 |
+------------+-------------+-----------------+

我想要它回来：

+------------+-------------+-----------------+
| order_date | pickup_date | order_to_pickup |
+------------+-------------+-----------------+
| 3/29/19    | 4/3/19      |               2 |
| 3/29/19    | 4/2/19      |               3 |
+------------+-------------+-----------------+

Answer 1

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`

Answer 2

如果你考虑两个日期之间的周数，我认为有一个聪明的解决方案，但同时这里有一个蛮力方法：

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 |
+------------+-------------+-----------------+

Answer 3

这是一种工作方法，用于根据 Looker Discourse 社区中的工作计算日期之间的工作日 here。原始示例是针对 Redshift 的，因此我在下面针对 BigQuery 对其进行了调整。

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`

Answer 4

替代解决方案。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`

Answer 5

扩展@khan 对日期落在星期六/星期日的某些组合的答案：

CASE 
  WHEN Completed_Date is null THEN null 
  when date_diff(completed_date,create_date, week) > 0 
    then date_diff(completed_date,create_date, day) - (date_diff(completed_date,create_date,week) * 2)
  else
    date_diff(completed_date,create_date, day) 
  end +
  CASE WHEN EXTRACT(DAYOFWEEK FROM Create_Date) = 1 and EXTRACT(DAYOFWEEK FROM Completed_Date) = 7  THEN -1
  WHEN EXTRACT(DAYOFWEEK FROM Create_Date) != 1 and EXTRACT(DAYOFWEEK FROM Completed_Date) = 1 THEN 1
  WHEN EXTRACT(DAYOFWEEK FROM Create_Date) = 7 and EXTRACT(DAYOFWEEK FROM Completed_Date) != 7 THEN 1
  ELSE 0 END

Answer 6

这是我计算两个日期之间工作日的解决方案：

DECLARE in_date_from DATE DEFAULT "2023-03-06";
DECLARE in_date_to DATE DEFAULT "2023-03-19";

SELECT
  DATE_DIFF(in_date_to, in_date_from, DAY) + 1 -
    DATE_DIFF(in_date_to, in_date_from, WEEK(SATURDAY)) -
      DATE_DIFF(in_date_to, in_date_from, WEEK(SUNDAY)) -
        IF(EXTRACT(DAYOFWEEK FROM in_date_from) IN (1,7), 1, 0) AS `working_days`;