是否有任何方法可以使用DB2 SQL在Table Workorder中找到重叠时间

Question

我想使用DB2 SQL进行报告，该报告将在5月15日至17日之间显示持续时间(小时(列表。当前列表看起来像这样：

单击此处查看列表

您可以看到，从WO-391296(第二行(中，它在ACTSTART和ACTFINISH中有重叠的时间。

我已经尝试了重叠功能。这是我的查询：

SELECT
    *
FROM
    workorder a,
    workorder b
WHERE
    a.wonum <> b.WONUM
    AND (a.ACTSTART, a.ACTFINISH) OVERLAPS (b.ACTSTART, b.ACTFINISH)
    AND DATE(a.ACTSTART) = '2019-05-15'
    AND DATE(a.ACTFINISH) = '2019-05-17'
    AND a.assetnum = 'A0000004'
    AND DATE(b.ACTSTART) = '2019-05-15'
    AND DATE(b.ACTFINISH) = '2019-05-17'
    AND b.assetnum = 'A0000004' ;

但它不起作用，只是给我错误消息。：

SQL错误[42601]：在" tstart，a.actfinish(之后发现了意外的令牌"重叠"。预期令牌可能包括：" in" .. sqlcode = -104，sqlstate = 42601，驱动程序= 3.64.114

这是我实际的完整查询：

SELECT assetnum, wonum, worktype, actstart, actfinish, 
    (DAY(ACTFINISH-ACTSTART)*24 + HOUR(ACTFINISH-ACTSTART)) + FLOAT(MINUTE(ACTFINISH-ACTSTART))/60 AS wotime 
FROM ( 
    SELECT assetnum, wonum, WORKTYPE, ACTSTART, 
        (CASE WHEN ACTFINISH IS NULL THEN '2019-05-17-24.00.00' ELSE ACTFINISH END) AS ACTFINISH 
    FROM workorder ast 
    WHERE istask=0 AND siteid = 'SBY' AND WORKTYPE = 'CM' AND
        (ast.assetnum = 'A0000004' or 
         assetnum in (select distinct assetnum from asset ast where ast.parent = 'A0000004') or 
         assetnum in (select distinct assetnum from asset ast where ast.parent 
                  in (select distinct assetnum from asset ast where ast.parent = 'A0000004'))) AND 
        ( 
            (status IN ('COMP', 'CONFIRM', 'CLOSE') AND 
            (date(ACTFINISH) BETWEEN '2019-05-15' and '2019-05-17')) OR 
            (status IN ('INPRG', 'COMP', 'CONFIRM', 'CLOSE') AND 
            (date(ACTSTART) <= '2019-05-17' AND (date(ACTFINISH) > '2019-05-17' OR ACTFINISH IS null))) 
        ) 
); 

此列表的实际总持续时间为6小时，错误，因为我不想 SUM重叠时间。

因此，最终结果必须为3.5小时(没有WO-391296(。

Answer 1

2间隔重叠，如果：
start of the 1-st is <= of end of the 2-nd
和
start of the 2-nd is <= of end of the 1-st

create or replace function overlaps_dates (a_from date, a_to date, b_from date, b_to date)
returns int
deterministic
no external action
contains sql
return case when min(a_from, a_to) <= max(b_from, b_to) and min(b_from, b_to) <= max(a_from, a_to) then 1 else 0 end;