Question

理想情况下，我希望下面的内容更加紧凑，即每周删除一周的总计重复次数，并在几周内进行某种形式的迭代。如果它可以通过存储过程总计是完美的，使用“周回来”int，它将整整几周，然后（wishlist项目）总计当前这周的数据。鉴于以下情况，这可能吗？即exec spGetTotalsWeeksBack 8，将看起来8周后，总共只有整整一周，（如果开始日期参数不是星期日，则转到下一个星期日作为第1周输出）。我也遇到ISNULL的问题，即使是0，它仍然没有返回所有客户。任何帮助都表示赞赏！

DECLARE @reportStartDate DATE = '2015-06-28' /*A Sunday, report start date*/

SELECT a.ClientName
    ,ISNULL(COUNT(CASE 
                WHEN DATEDIFF(week, @reportStartDate, RecordCreateDate) = 0
                    THEN 1
                END), 0) AS 'Week 1'
    ,ISNULL(COUNT(CASE 
                WHEN DATEDIFF(week, @reportStartDate, RecordCreateDate) = 1
                    THEN 1
                END), 0) AS 'Week 2'
    ,ISNULL(COUNT(CASE 
                WHEN DATEDIFF(week, @reportStartDate, RecordCreateDate) = 2
                    THEN 1
                END), 0) AS 'Week 3'
    ,ISNULL(COUNT(CASE 
                WHEN DATEDIFF(week, @reportStartDate, RecordCreateDate) = 3
                    THEN 1
                END), 0) AS 'Week 4'
FROM Client a WITH (NOLOCK)
LEFT JOIN SecondaryClient b WITH (NOLOCK) ON b.MainClientID = a.ClientID
LEFT JOIN RecordLog c WITH (NOLOCK) ON c.ClientID = b.ClientID
    AND c.RecordCreateDate BETWEEN @reportStartDate /*Sunday, report start date*/
        AND DATEADD(week, 4, @reportStartDate) /* Saturday, report end date */
LEFT JOIN IncomingData d WITH (NOLOCK) ON d.DataID = c.DataID
    AND d.DataReceiveDate > @reportStartDate
    AND d.DataSourceID = 10
WHERE a.StatusID = 1
    AND a.ClientID IN (
        SELECT clientid
        FROM clientsettings
        WHERE active = 1
        )
GROUP BY a.ClientName
ORDER BY a.ClientName

Answer 1

这里有2个问题。

“我在ISNULL遇到问题，即使0”仍然没有返回所有客户端

您遇到此问题，因为您正在对LEFT JOIN-ed表进行过滤。

将重写查询视为UNION ALL的2个查询，每个查询都使用INNER JOINS：

DECLARE @reportStartDate DATE = '2015-06-28' /*A Sunday, report start date*/

SELECT a.ClientName
    ,COUNT(CASE 
                WHEN DATEDIFF(week, @reportStartDate, RecordCreateDate) = 0
                    THEN 1
                END), 0) AS 'Week 1'
    ,COUNT(CASE 
                WHEN DATEDIFF(week, @reportStartDate, RecordCreateDate) = 1
                    THEN 1
                END), 0) AS 'Week 2'
    ,COUNT(CASE 
                WHEN DATEDIFF(week, @reportStartDate, RecordCreateDate) = 2
                    THEN 1
                END), 0) AS 'Week 3'
    ,COUNT(CASE 
                WHEN DATEDIFF(week, @reportStartDate, RecordCreateDate) = 3
                    THEN 1
                END), 0) AS 'Week 4'
FROM Client a WITH (NOLOCK)
JOIN SecondaryClient b WITH (NOLOCK) ON b.MainClientID = a.ClientID
JOIN RecordLog c WITH (NOLOCK) ON c.ClientID = b.ClientID
    AND c.RecordCreateDate BETWEEN @reportStartDate /*Sunday, report start date*/
        AND DATEADD(week, 4, @reportStartDate) /* Saturday, report end date */
JOIN IncomingData d WITH (NOLOCK) ON d.DataID = c.DataID
    AND d.DataReceiveDate > @reportStartDate
    AND d.DataSourceID = 10
WHERE a.StatusID = 1
    AND a.ClientID IN (
        SELECT clientid
        FROM clientsettings
        WHERE active = 1
        )
GROUP BY a.ClientName
UNION ALL -- non existent values
SELECT DISTINCT a.client, Week1=0, Week2=0,Week3=0, Week4=0
FROM Client a WITH (NOLOCK)
WHERE NOT EXISTS (
    SELECT * FROM 
    SecondaryClient b WITH (NOLOCK) ON b.MainClientID = a.ClientID
    JOIN RecordLog c WITH (NOLOCK) ON c.ClientID = b.ClientID
        AND c.RecordCreateDate BETWEEN @reportStartDate /*Sunday, report start date*/
            AND DATEADD(week, 4, @reportStartDate) /* Saturday, report end date */
    JOIN IncomingData d WITH (NOLOCK) ON d.DataID = c.DataID
        AND d.DataReceiveDate > @reportStartDate
        AND d.DataSourceID = 10
    WHERE a.StatusID = 1
        AND a.ClientID IN (
            SELECT clientid
            FROM clientsettings
            WHERE active = 1
            )
)

“如果它可以通过存储过程总计，使用”周回来“int，那将是整整几周，然后（wishlist项目）总计当前这周的数据，这将是完美的。这是可能的鉴于以下情况？即exec spGetTotalsWeeksBack 8，将在8周前看，总计只有整整一周“

我想你可以做GROUP BY (DATEDIFF(week, @reportStartDate, RecordCreateDate之类的事情然后PIVOT-ing结果。

虽然，我担心你无法避免对于Week0-WeekN列的动态sql。

按整周分组的总计

1 个答案: