结合2个复杂的MySQL查询到一个

当前 - 表

ID - Unique identifier for device (Int) 
Location -Unique identifier for name of area (VarChar) 
Status - The current status of the device (VarChar) 
Time - DateTime of when the last connection was made to the device (DateTime)

历史 - 表

CID (Sorta unused, just as an AI field to store multiple old bits of data uniquely) (Int) 
ID - Unique identifier for device (Int) 
Location --Unique identifier for name of area (VarChar) 
Status - The current status of the device (VarChar) 
Time -- DateTime of when the last connection was made to the device (DateTime)

所以这是数据库如何看起来现在我的查询看起来像这样......

查询1

SELECT c.*, 
    if(HOUR(TIMEDIFF(NOW(), c.TIME)) >=1, 1, 0) as LatestOlderThanAnHour, 
    min(h.time) as EarliestTime, 
    (abs(timestampdiff(HOUR,NOW(),min(TIME)))/count(*)*100) as percentage 
FROM Current c 
JOIN Historical h on c.ID = h.ID 
WHERE c.Location like "MyLocation" 
group by c.ID

查询2

SELECT MAX(h.TIME) AS LastDown 
FROM TABLENAME 
WHERE h.STATUS IN ('On-Login-Screen','IE-Window-Missing')

的最终目标是添加一个 “最后向下” 列由该查询返回的每一ID。我只是无法弄清楚如何去做。

来源

2013-05-07 A_Elric

'MAX（情况下h.status在（ '在登录屏幕'， 'IE-窗口 - 缺少'），然后小时。时间结束）' – 2013-05-07 20:20:59

^^将其作为答案，以便我可以接受它，并非常感谢你！ – 2013-05-07 20:34:07

max(case when h.status in ('On-Login-Screen','IE-Window-Missing') then h.time end)

或者我猜你也可以把它写

max(if(h.status in ('On-Login-Screen','IE-Window-Missing'), h.time, NULL))

来源

2013-05-07 20:36:49

结合2个复杂的MySQL查询到一个

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