Question

我有2个SQL表，一个用于subscribers (ID, Name)，第二个用于subscribers_Upgrades [Subscriber_ID (FK), Upgrade_Level, Upgrade_Date)

（有7个升级级别）

我需要一个查询来获取在最短时间段内升级的前10位订阅者Upgrade_Level和Upgrade Date针对特定Upgrade_Level（任何一种可用的7升级级别），并在特定的时间内

我们可以，例如提供像2B这样的2个条件 Where Upgrade_Level = 3 and Upgrade_Date between '01-April-2014' and '20-May-2016'

Answer 1

你的意思是最短的时期？谁在upgrade_date命令中最早升级了？

SELECT TOP 10 s.ID, 
    s.Name, 
    su.UpgradeLevel, 
    su.UpgradeDate 
    FROM Subscribers s 
    INNER JOIN SubscribersUpgrades su on s.ID = su.SubscriberId
    WHERE su.UpgradeLevel = 3 AND su.UpgradeDate BETWEEN '01-April-2014' AND '20-May-2016'
    ORDER BY su.UpgradeDate ASC

<强>更新

如果您只需要2个升级级别（级别2和级别3）的数据，这应该有效 -

SELECT TOP 10 s.Name, DATEDIFF(dd,t1.UpgradeDate,t2.UpgradeDate) UpgradePeriodInDays FROM

(SELECT su.SubscriberId, su.UpgradeDate FROM SubscribersUpgrades su
WHERE su.UpgradeLevel = 2 ) t1

INNER JOIN (SELECT su.SubscriberId, su.UpgradeDate FROM SubscribersUpgrades su
WHERE su.UpgradeLevel = 3) t2 on t1.SubscriberId = t2.SubscriberId

INNER JOIN Subscribers s on t1.SubscriberId = s.ID

WHERE t2.UpgradeDate BETWEEN '01-June-2016' AND '20-July-2016'
ORDER BY DATEDIFF(dd,t1.UpgradeDate,t2.UpgradeDate) ASC

加入＆amp;子查询计算最小日期

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