Question

我正在尝试在C shell中编写一个脚本，用于在8个工作日内从Informix数据库中选择数据。到目前为止，我有sql代码计算过去+周日和周六的8天它看起来像这样：

select *
from ekzo 
where datzah = today-
(case
        when weekday(today) = 1 then 12
        when weekday(today) = 2 then 12
        when weekday(today) = 3 then 12
        when weekday(today) = 4 then 10
        when weekday(today) = 5 then 10
        when weekday(today) = 6 then 10
        when weekday(today) = 0 then 11
        end)

我创建了表格“prazkal”，假期如下：

datpra  01.01.2014
nazpra  Nova Godina
krapra  SRI

datpra  06.01.2014
nazpra  Bogojavljanje ili Sveta tri kralja
krapra  PON

datpra  20.04.2014
nazpra  Uskrs
krapra  NED

datpra  21.04.2014
nazpra  Uskršnji ponedjeljak
krapra  PON

...

考虑到周末和假日，我不知道如何扩展我的sql来计算过去8个工作日。

Answer 1

我会用2个函数来做。第一个函数检查日期是否为假日：

create function is_holiday(d datetime year to day)
returning boolean;
    -- define hcnt integer;

    if weekday(d) = 0 or weekday(d) = 6 then
        return 't';
    end if;

    -- code that check if 'd' is marked as holiday in calendar
    --select count(*) into hcnt from prazkal where datpra = d;
    --if hcnt > 0 then
    --  return 't';
    --end if;

    return 'f';
end function;

第二个功能将日期减少几天，省略假期：

create function move_date_back(start_d datetime year to day, count_days integer)
returning datetime year to day;
define new_d datetime year to day;
define i integer;
    let i = 0;
    let new_d = start_d;

    while i < count_days
        let new_d = new_d - interval(1) day to day;
        if not is_holiday(new_d) then
            let i = i + 1;
        end if;
    end while;

    return new_d;
end function;

在Informix中计算过去8个工作日

1 个答案: