如何在 SQL 服务器中获取值为 0 的缺失日期？

Question

我在 SQL 服务器中使用以下查询来查找 distinct 最近 7 天（不包括今天的日期）的登录次数：

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

我的期望是按顺序排列所有最后 7 天（不像当前输出那样 2019-02-16、2019-02-17 和 2019-02-18 在 2019-02-19 之后丢失).我需要它，如果缺少日期，则必须以 0 计数显示它。

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

你可以试试这个。在这里您需要获得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。最后你需要同时加入 table.

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

您可以找到现场演示 Here。我已将您的查询输出插入临时 table 但逻辑相同。

Answer 2

要查看特定值，该值必须来自一行。因此，要查看登录 table 中不存在的日期，您必须在某处 .

中生成它们作为行

您可以使用简单的递归 CTE 在特定时间间隔内每天生成 1 行，然后使用 LEFT JOIN 加入在该特定日期匹配的登录。不匹配的仍然会显示，因为我们使用的是 LEFT JOIN.

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 3

只有 7 天，所以只需输入这些日期：

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 4

生成 table 中没有行的日期以加入 calendar table 的最佳方法。

这是一个非常简单的日历 table 一年，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，以便暂时不需要维护它。）

现在您可以像这样在查询中加入日历 table：

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 space 它占用，它也会在许多其他情况下派上用场。

如何在 SQL 服务器中获取值为 0 的缺失日期？

How to get missing dates with 0 value in SQL Server?

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