Question

我有一张访客表。访客具有以下列：

编号

StartTime （日期）

已购买（bool）

发货（bool）

对于过去7天内的每一天，我想选择3位当天访客 StartTime ：

访客总数
已购买 = true
已发货 = true

理想情况下，返回的结果是：

Day    Total    TotalPurchased    TotalShipped
1      100      67                42
2      82       61                27
etc...

我已经习惯了.NET Linq，所以这对我来说是一个很大的挑战。

到目前为止我提出的所有内容如下：

SELECT COUNT(*) AS Total
FROM [dbo].[Visitors]
WHERE DAY([StartTime]) = DAY(GETDATE())

它选择当天的总数就好了，不过我现在感觉很困难，所以如果有人能指出我正确的方向，它会很好。

Answer 1

过去7天使用Stanislav提出的查询，但使用此WHERE条款

   SELECT DAY([StartTime]) theDay,
           COUNT(*) AS Tot,
           SUM(CASE WHEN Purchased=true THEN 1 ELSE 0 END) as TotPurch,    
           SUM(CASE WHEN Shipped=true THEN 1 ELSE 0 END) as TotShip
    FROM [dbo].[Visitors]
    WHERE [StartTime] BETWEEN GETDATE()-7 AND GETDATE()
    GROUP BY DAY([StartTime])

Answer 2

SELECT COUNT(*) AS Total,
       SUM(CASE WHEN Purchased=true THEN 1 ELSE 0 END) as TotalPurchased,    
       SUM(CASE WHEN Shipped=true THEN 1 ELSE 0 END) as TotalShipped
FROM [dbo].[Visitors]
WHERE DAY([StartTime]) = DAY(GETDATE())

并按照jarlh提到的那样添加GROUP BY DAY（[StartTime]）

Answer 3

这是一个简单的选择，可以为您提供所需的数据集

SELECT DATEDIFF(day,StartTime, getdate())+1 as [Day], -- Add 1 to display 1 to 7 instead of 0 to 6
COUNT(*) as Total, 
SUM(CASE WHEN Purchased = 1 THEN 1 ELSE 0 END) as TotalPurchased, 
SUM(CASE WHEN Shipped = 1 THEN 1 ELSE 0 END) AS TotalShipped 
FROM Visitors
WHERE DATEDIFF(day,startTime,GETDATE()) < 6
GROUP BY DATEDIFF(day,startTime,GETDATE())
ORDER BY 1

此查询不会考虑日期的时间部分。

Answer 4

由于该表在[dbo]中，这可能与Sql Server数据库有关。

在这种情况下，[已购买]和[已发货]列很可能是BIT字段。

此类BIT字段可以使用SUM和IIF或CASE WHEN计算并且DATEDIFF可用于选择之前的7天。

SELECT 
DAY([StartTime]) as [Day],
SUM(IIF([Purchased] = 1, 1, 0)) AS TotalPurchased,
SUM(case when [Shipped] = 1 then 1 else 0 end) AS TotalShipped,
COUNT(*) AS Total
FROM [dbo].[Visitors]
WHERE datediff(day, [StartTime], GETDATE()) BETWEEN 1 AND 7
GROUP BY DAY([StartTime])
ORDER BY 1;

如果该表在日期字段上编入索引，则此WHERE子句应该更快地产生结果：

WHERE [StartTime] between dateadd(day,-7,cast(GETDATE() as date)) and dateadd(day,-1,cast(GETDATE() as date))

每天选择多个COUNT个

4 个答案: