[英]Returning the next-to-last entry using MySQL
一些信息:人們正在辦理登機手續,但他們沒有退房。 每次簽入都會在_checkins表中創建帶有時間戳,MemberID等的自動遞增的條目。
這是查詢需要返回的數據:
除了最后一部分,我都工作了。 我覺得LIMIT將成為解決方案的一部分,但我只是找不到正確實現它的方法。 到目前為止,這是我得到的:
SELECT m.ImageURI, m.ID, m.FirstName, m.LastName,
ROUND(time_to_sec(timediff(NOW(), MAX(ci.Created))) / 3600, 1) as
'HoursSinceCheckIn', CheckIns
FROM _checkins ci LEFT JOIN _members m ON ci.MemberID = m.ID
INNER JOIN(SELECT MemberID, COUNT(DISTINCT ID) as 'CheckIns'
FROM _checkins
WHERE(
Created BETWEEN NOW() - INTERVAL 30 DAY AND NOW()
)
GROUP BY MemberID
) lci ON ci.MemberID=lci.MemberID
WHERE(
ci.Created BETWEEN NOW() - INTERVAL 30 DAY AND NOW()
AND TIMESTAMPDIFF(HOUR, ci.Created, NOW()) < 2
AND ci.Reverted = 0
)
GROUP BY m.ID
ORDER BY CheckIns ASC
您可以大大簡化(也可以使代碼更安全):
SELECT _Members.ImageURI, _Members.ID, _Members.FirstName, _Members.LastName,
ROUND(TIME_TO_SEC(TIMEDIFF(NOW(), _FilteredCheckins.lastCheckin)) / 3600, 1) AS hoursSinceCheckIn, _FilteredCheckins.checkIns,
(SELECT MAX(_Checkins.created)
FROM _Checkins
WHERE _Checkins.memberId = _Members.ID
AND _Checkins.created < _FilteredCheckins.lastCheckin) AS previousCheckin
FROM _Members
JOIN (SELECT memberId, COUNT(*) AS checkIns, MAX(created) AS lastCheckin
FROM _Checkins
WHERE created >= NOW() - INTERVAL 30 DAY
GROUP BY memberId
HAVING lastCheckin >= NOW() - INTERVAL 2 HOURS) _FilteredCheckins
ON _FilteredCheckins.memberId = _Members.ID
ORDER BY _FilteredCheckins.checkIns ASC
_Checkins.id
是唯一的(應該是),所以COUNT(DISTINCT ID)
可以簡化為COUNT(*)
。 如果不是這種情況,則需要將其放回原處。 (旁注:請不要使用BETWEEN
, 尤其是日期/時間類型 )
(幽默的旁注:我一直在腦子里把它讀作“雞”。
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