Question

我想使用DB2 SQL进行报告，该报告将显示5月15日至17日之间的持续时间（小时）列表。当前列表如下：

如您所见，在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' ;

但是它不起作用，只是给我错误消息。

SQL错误[42601]：在“ TSTART，a.ACTFINISH）”之后发现了意外的标记“ OVERLAPS”。预期的令牌可能包括：“ IN” .. SQLCODE = -104，SQLSTATE = 42601，DRIVER = 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，这是错误，因为我不想SUM重叠的时间。

因此，最终结果必须为3.5小时（没有WO-391296）。

Answer 1

2个间隔在以下情况下重叠：
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;

有什么方法可以使用DB2 SQL在表Workorder中查找重叠时间？

1 个答案: