Question

我有一个相当简单的查询；看起来像这样：

SELECT
order_date,
pickup_date,
DATE_DIFF(pickup_date,order_date, day) order_to_pickup
FROM
`orders.table`

唯一的问题是，我需要计算营业日中的日期差，而不是整天。

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

我想它返回：

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

Answer 1

如果您考虑两个日期之间的星期数，我认为有一个聪明的解决方案，但是与此同时，这是一种蛮力的方法：

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)
);

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 2

这里是一种工作方法，用于根据Looker话语社区here中的工作来计算日期之间的工作日。原始示例适用于Redshift，因此我在下面将其改编为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

将其应用于您的数据集，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 3

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 4

另一种解决方案。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`

DATE_DIFF，但仅计算工作日

4 个答案: