Sql 如何在postgres中获得下一个业务日期？

工作日为周一至周五

如果我有一个日期时间字段计划_，如何找到下一个业务日期并在列别名中返回

我试过另一个答案，但效果不理想

EXTRACT(ISODOW FROM v.scheduled_for)::integer) % 7 as next_business_day,

错误：

Query 1 ERROR: ERROR:  syntax error at or near ")"
LINE 3:  EXTRACT(ISODOW FROM v.scheduled_for)::integer % 7)  as next...

                                                      ^

编辑：

感谢您的建议，我尝试了以下方法：

SELECT
    v.id AS visit_id,
    (IF extract(''dow'' from v.scheduled_for) = 0 THEN
        return v.scheduled_for + 1::integer;
    ELSIF extract(''dow'' from v.scheduled_for) = 6 THEN
        return v.scheduled_for - 1::integer;
    ELSE
        return v.scheduled_for;
    ) as next_business_day, 
    '' as invoice_ref_code,

我得到的错误是：

Query 1 ERROR: ERROR:  syntax error at or near ")"
LINE 1: ) as next_business_day, 
    ^

---

要进行概括，您需要创建一个函数来计算从给定日期开始的下一个工作日

create or replace function utl_next_business_day(date_in date default current_date)
returns date
language sql immutable leakproof strict
as $$
    with cd as (select  extract(isodow from date_in)::integer d)  
    select case when d between 1 and 4 
                then date_in + 1
                else date_in + 1 + (7-d) 
            end 
       from cd;       
$$; 

--- any single date 
select current_date, utl_next_business_day();

-- over time span (short)
select gdate::date for_date, utl_next_business_day(gdate::date) next_business_day
  from generate_series( current_date, current_date + 14, interval '1 day') gdate;

-- around year end over a time span
with test_date (dt) as
     ( values (date '2019-12-31')
            , (date '2020-12-31'), (date '2021-12-31'),(date '2022-12-31') 
            , (date '2021-01-01'), (date '2022-01-01'),(date '2023-01-01')
            )
select dt, utl_next_business_day(dt) from test_date
order by dt;

或者是@Eric的日历表建议

-- create and populate work table
create table bus_day_calendar ( bus_day date);
insert into bus_day_calendar (bus_day) 
  select utl_next_business_day(gdate::date)
    from generate_series( date '2018-12-31', date '2023-01-01', interval '1 day') gdate
   where extract(isodow from gdate)::integer not in (6,7) ;

--- Function to return next business day   
create or replace function utl_next_cal_business_day(date_in date default current_date)
returns date
language sql stable leakproof strict
as $$
    select min(bus_day)
      from bus_day_calendar 
     where bus_day > date_in;      
$$;    

--- any single date 
select current_date, utl_next_cal_business_day();

-- over time span (short)
select gdate::date for_date, utl_next_cal_business_day(gdate::date) next_business_day
  from generate_series( current_date, current_date + 14, interval '1 day') gdate;

-- around year end over a time span
with test_date (dt) as
     ( values (date '2019-12-31')
            , (date '2020-12-31'), (date '2021-12-31'),(date '2022-12-31') 
            , (date '2021-01-01'), (date '2022-01-01'),(date '2023-01-01')
            )
select dt, utl_next_cal_business_day(dt) from test_date
order by dt; 

---

这两种方法目前都不能处理周一至周五的非工作日，但都可以修改以实现这一目的。由于日历表只需要删除ROE，因此如果有必要，我认为这将成为更好的方法

计数

（

和

）

。相同的数字？本文可能会提供您需要的代码：“工作日”数据操作最好使用在您的环境中创建和维护的日历表来实现。搜索“SQLCalendarTable”应该为您提供许多潜在的起点，以构建一个适合您业务需要的起点。