T-SQL 联合返回重复的行（日期）

Question

我试图解决在特定日期范围内为每个日期返回总和的问题，无论数据是否存在一天。

我发现最好的方法是使用预先填充了所有日期的表格，选择日期范围并将其与我的数据合并。

由于某种原因无法离开加入工作，但工会看起来工作几乎完美。唯一的问题是它会返回我的数据表中存在数据的日期的重复项。

SELECT NULL AS Visitors
,D.DATE AS Day
FROM Database.support.dates D
WHERE D.DATE BETWEEN @Start_date
    AND @End_date

UNION

SELECT count(V.id) AS Visitors
,DATEADD(day, 0, DATEDIFF(day, 0, V.CreateTime)) AS Day
 FROM Database.Clients.Clients V
 WHERE V.CreateTime BETWEEN @Start_date
    AND @End_date
AND V.WADID = @WADID
AND (
    V.WAPID = @WAPID
    OR @WAPID IS NULL
    )
GROUP BY DATEADD(day, 0, DATEDIFF(day, 0, V.CreateTime))
ORDER BY Day DESC

前一天一直在研究它，但仍然无法正常工作：/

Answer 1

我认为将日历表加入当前查询实际上是正确的做法。话虽如此，您可以通过简单地汇总当天来纠正您当前的情况，例如使用MAX。默认情况下，每天的 NULL 值将被忽略，只要存在非空访问者计数：

WITH cte AS (
    SELECT NULL AS Visitors, D.DATE AS Day
    FROM Database.support.dates D
    WHERE D.DATE BETWEEN @Start_date AND @End_date

    UNION

    SELECT COUNT(V.id), DATEADD(day, 0, DATEDIFF(day, 0, V.CreateTime))
    FROM Database.Clients.Clients V
    WHERE V.CreateTime BETWEEN @Start_date AND @End_date AND
        V.WADID = @WADID AND (V.WAPID = @WAPID OR @WAPID IS NULL)
    GROUP BY DATEADD(day, 0, DATEDIFF(day, 0, V.CreateTime))
)

SELECT Day, MAX(Visitors) AS Visitors     -- filter off unwanted NULL values
FROM cte
GROUP BY Day
ORDER BY Day DESC;