如何添加 SQL 以检查是否仅在最近的工作日停电？

Question

我运行对企业进行审计，以查看它们是否缺少收入或客户，并生成一个数据集，该数据集由企业未通过任何 revenue/client 指标（=$0/ 0) 生成以下数据集：

Date      |Business|Revenue1|Revenue2|Num_Clients1|Num_Clients2
4-18-2022 |1111    |[=11=]      |[=11=]      |0           |0    
4-15-2022 |1111    |[=11=]      |[=11=]      |0           |0    
4-18-2022 |2222    |[=11=]      |[=11=]      |0           |0    
4-14-2022 |2222    |[=11=]      |[=11=]      |0           |0         
4-15-2022 |3333    |[=11=]      |[=11=]      |0           |0             
4-13-2022 |3333    |[=11=]      |[=11=]      |0           |0

今天 = 4-19-2022

如果实践在给定的工作日有一个非零度量值，那么它不会在上述数据集中有一行。

我想获取在今天之前的最后两个连续工作日（前天和昨天）指标全为零的企业列表。

在这种情况下，这将是企业 1111。企业 2222 不符合条件，因为 4-15-2022 没有通过审核。业务 3333 不符合条件，因为 4-15-2022 和 4-13-2022 不是相对于今天的最近两个连续工作日。

如何在 SQL 中完成此操作？我正在考虑检查是否运行ning total = 2 但我不知道相对于过去两个工作日如何做。

Answer 1

使用您提供的示例数据。考虑下面的查询，它计算当前日期的前一天和前 2 天。过滤具有前一天和前 2 天的数据并按 Business 计数。如果一条记录得到 count = 2 那么它有前 2 天的数据。

CREATE TEMPORARY FUNCTION working_days_diff(the_date DATE, num_of_days INT64) AS
  (
    DATE_SUB(CASE WHEN EXTRACT(DAYOFWEEK FROM the_date) IN (2,3,4,5,6,7)
      THEN
        CASE WHEN (EXTRACT(DAYOFWEEK FROM the_date) - MOD(num_of_days, 5)) > 1
        THEN DATE_SUB(the_date, INTERVAL MOD(num_of_days, 5) DAY)
        ELSE DATE_SUB(the_date, INTERVAL (MOD(num_of_days, 5) + 2) DAY)
        END
      ELSE
        DATE_SUB(the_date, INTERVAL (MOD(num_of_days, 5) +1) DAY)
    END, INTERVAL DIV(num_of_days, 5) WEEK) 
); -- I got this function here -> 

with data as (
select date('2022-04-18') as Date, 1111 as Business, 0 as Revenue1, 0 as Revenue2, 0 as Num_Clients1, 0 as Num_Clients2,
union all select date('2022-04-15') as Date, 1111 as Business, 0 as Revenue1, 0 as Revenue2, 0 as Num_Clients1, 0 as Num_Clients2,
union all select date('2022-04-18') as Date, 2222 as Business, 0 as Revenue1, 0 as Revenue2, 0 as Num_Clients1, 0 as Num_Clients2,
union all select date('2022-04-14') as Date, 2222 as Business, 0 as Revenue1, 0 as Revenue2, 0 as Num_Clients1, 0 as Num_Clients2,
union all select date('2022-04-15') as Date, 3333 as Business, 0 as Revenue1, 0 as Revenue2, 0 as Num_Clients1, 0 as Num_Clients2,
union all select date('2022-04-13') as Date, 3333 as Business, 0 as Revenue1, 0 as Revenue2, 0 as Num_Clients1, 0 as Num_Clients2,
),

add_prev_days as (
select *, 
    date('2022-04-19') as today, -- replace with current_date()
    working_days_diff(DATE('2022-04-19'), 1)  as prev_day, -- replace with current_date()
    working_days_diff(DATE('2022-04-19'), 2)  as prev_2_day, -- replace with current_date()
from data
),

add_count as (
select *, 
    count(Business) over (partition by Business) as count_bus
from add_prev_days 
    where (prev_day = Date or prev_2_day = Date) 
)

select Date,
    Business,
    Revenue1,
    Revenue2,
    Num_Clients1, 
    Num_Clients2 
from add_count
    where count_bus = 2

示例输出：

注意：我将“2022-04-19”硬编码为当前日期，以便查询正常运行。将此值替换为 current_date() 以适合您的用例。

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