日期减法与SQL一些条件

-2

SELECT 
    tkoh.ticketid, 
    tkoh.ownergroup, 
    tkoh.owndate, 
    tck.ACTUALFINISH 
FROM 
    tkownerhistory tkoh 
LEFT JOIN 
    ticket tck on tck.ticketid = tkoh.ticketid 
WHERE 
    tkoh.ticketid IN ('О1253565', 'О1253578') 
    AND tkoh.ownergroup IS NOT NULL 
ORDER BY 
    tkoh.ticketid, tkoh.owndate;

它返回是这样的：

，我需要时间来计算，其中每一组拥有的记录（当前记录的遗产 - 下一记录的遗产），但是，如果有最终所有者，则时间必须从ACTUALFINISH中减去，而不是从下一个记录中减去，因为它可以在事前完成。我用简单的IF在Excel file中完成了它，但是它在SQL中真的可行吗？

来源

2016-09-30 Cove

您的预计产量将_greatly_在这里帮助。 –

此外，请将数据发布为格式文本，而不是图片 – Aleksej

您希望人们在某个网站上打开由陌生人发布的Excel文件？你住哪个星球？ – mathguy

SELECT t.*, 
     trim(to_char(floor(t.res_sec/60/60),'00'))||':'||trim(to_char(floor(mod(t.res_sec/60,60)),'00'))||':'||trim(to_char(mod(t.res_sec,60),'00')) as res_hhmiss 
FROM (SELECT 
     tkoh.ticketid, 
     tkoh.ownergroup, 
     tkoh.owndate, 
     tck.ACTUALFINISH, 
     abs(tkoh.owndate - nvl(lead(tkoh.owndate) over(order by tkoh.ticketid, tkoh.owndate), tck.ACTUALFINISH))*24*60*60 as res_sec 
     FROM tkownerhistory tkoh 
     LEFT JOIN ticket tck on tck.ticketid = tkoh.ticketid 
     WHERE tkoh.ticketid IN ('О1253565', 'О1253578') and tkoh.ownergroup is not null 
     ORDER BY tkoh.ticketid, tkoh.owndate);

来源

2016-09-30 14:20:13 SkyWalker

SELECT 
    tkoh.ticketid, 
    tkoh.ownergroup, 
    tkoh.owndate, 
    tck.ACTUALFINISH, 

    coalesce(datediff(day, tkoh.owndate, lead(tkoh.owndate) over(order by tkoh.ticketid, tkoh.owndate)), 
    datediff(day, tkoh.owndate, tck.ACTUALFINISH)) As Diff 

FROM tkownerhistory tkoh 
LEFT JOIN ticket tck on tck.ticketid = tkoh.ticketid 
WHERE tkoh.ticketid IN ('О1253565', 'О1253578') and tkoh.ownergroup is not null 
ORDER BY tkoh.ticketid, tkoh.owndate;

来源

2016-09-30 14:46:42

日期减法与SQL一些条件

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