Question

我正在SQL Server中使用以下查询来查找最近7天（不包括今天的日期）内进行的distinct次登录：

SELECT TOP (7) CONVERT(date, LoginTime) AS ActivityDate, COUNT(DISTINCT LoginID) AS UserCount
FROM Login
WHERE CONVERT(date, LoginTime) < CONVERT(date, GETDATE())
GROUP BY CONVERT(date, LoginTime)  
ORDER BY ActivityDate DESC;

它生成以下输出：

ActivityDate | UserCount
----------------------
2019-02-21   | 2
2019-02-20   | 3
2019-02-19   | 2
2019-02-15   | 2
2019-02-14   | 1
2019-02-13   | 2
2019-02-12   | 3

我的期望是连续七个工作日结束（不像当前输出，其中2019-02-16之后缺少日期2019-02-17，2019-02-18和2019-02-19）。我需要它，如果缺少日期，则必须以0计数显示。

我的预期输出如下：

ActivityDate | UserCount
----------------------
2019-02-21   | 2
2019-02-20   | 3
2019-02-19   | 2
2019-02-18   | 0
2019-02-17   | 0
2019-02-16   | 0
2019-02-15   | 2

Answer 1

要查看特定值，该值必须来自一行。因此，要查看登录表上不存在的日期，必须将其生成为 somewhere 处的行。

您可以使用简单的递归CTE在特定时间间隔之间每天生成1行，然后使用LEFT JOIN来加入在该特定日期匹配的登录名。由于我们正在使用LEFT JOIN，因此仍会显示不匹配的内容。

DECLARE @GeneratingDateFrom DATE = DATEADD(DAY, -7, GETDATE())
DECLARE @GeneratingDateTo DATE = GETDATE()

;WITH GeneratedDates AS
(
    SELECT
        GeneratedDate = @GeneratingDateFrom

    UNION ALL

    SELECT
        GeneratedDate = DATEADD(DAY, 1, G.GeneratedDate)
    FROM
        GeneratedDates AS G
    WHERE
        DATEADD(DAY, 1, G.GeneratedDate) < @GeneratingDateTo
)
SELECT
    G.GeneratedDate,
    count(distinct L.LoginID) as UserCount
FROM 
    GeneratedDates AS G
    LEFT JOIN [Login] AS L ON G.GeneratedDate = CONVERT(date, L.LoginTime)
GROUP BY
    G.GeneratedDate
ORDER BY 
    G.GeneratedDate desc

Answer 2

只有7天，所以只需键入这些日期即可：

SELECT ActivityDate, COUNT(DISTINCT LoginID) AS UserCount
FROM (VALUES
    (CAST(CURRENT_TIMESTAMP - 1 AS DATE)), -- build the list of dates
    (CAST(CURRENT_TIMESTAMP - 2 AS DATE)),
    (CAST(CURRENT_TIMESTAMP - 3 AS DATE)),
    (CAST(CURRENT_TIMESTAMP - 4 AS DATE)),
    (CAST(CURRENT_TIMESTAMP - 5 AS DATE)),
    (CAST(CURRENT_TIMESTAMP - 6 AS DATE)),
    (CAST(CURRENT_TIMESTAMP - 7 AS DATE))
) datelist(ActivityDate)
LEFT JOIN Login ON CAST(LoginTime AS DATE) = ActivityDate
GROUP BY ActivityDate
ORDER BY ActivityDate DESC

Answer 3

您可以尝试一下。在这里，您需要获取第一个最小日期和最大日期。之后，您需要生成这两天之间的所有日期。最后，您需要同时加入两个表。

declare @MinDate date
declare @MaxDate date

select * into #temp from(
select top (7) CONVERT(date,LoginTime) as ActivityDate,count(distinct LoginID) as UserCount
        from Login
        where CONVERT(date,LoginTime )< convert(date,getdate())
        group by CONVERT(date,LoginTime )  
        order by ActivityDate desc; 
)a        

Set @MinDate = (select min (ActivityDate) from #temp)
Set @MaxDate = (select max (ActivityDate) from #temp)  

Select a.Date, isnull(b.UserCount,0) as UserCount from(
SELECT  TOP (DATEDIFF(DAY, @MinDate, @MaxDate) + 1)
        Date = DATEADD(DAY, ROW_NUMBER() OVER(ORDER BY a.object_id) - 1, @MinDate)
FROM    sys.all_objects a
        CROSS JOIN sys.all_objects b;
)a left join #temp b on a.Date = b.ActivityDate

您可以找到实时演示Here。我已将您的查询输出插入到临时表中，但是逻辑是相同的。

Answer 4

生成日期的最佳方法是将表中没有行的日期加入到calendar table中。

这是一个非常简单的一年日历表，基于this answer:

CREATE TABLE [Calendar]
(
    [CalendarDate] DATETIME
)

DECLARE @StartDate DATETIME
DECLARE @EndDate DATETIME
SET @StartDate = GETDATE()
SET @EndDate = DATEADD(d, 365, @StartDate)

WHILE @StartDate <= @EndDate
      BEGIN
             INSERT INTO [Calendar]
             (
                   CalendarDate
             )
             SELECT
                   @StartDate

             SET @StartDate = DATEADD(dd, 1, @StartDate)
      END

（您可以修改此查询，以在将来添加更多日期，这样一来就无需维护它了。）

现在，您可以像这样在查询中加入日历表：

select top (7) c.CalendarDate as ActivityDate,count(distinct LoginID) as UserCount
from Calendar c
left join Login l
    ON c.CalendarDate = CONVERT(date, l.LoginTime)
    and CONVERT(date,LoginTime )< convert(date,getdate())
group by c.CalendarDate 
order by c.CalendarDate desc;

值得占用的空间，在许多其他情况下也将派上用场。

如何在SQL Server中获得0值的缺失日期？

4 个答案: