Question

我创建了一个SQL查询，该查询返回一个人一个月在一个月中的总天数，现在我想添加总数列和3种叶子列，即[休闲假]，[带薪假]，[免费休假]。

简而言之，我想在[休假]，[带薪休假]和[免费休假]中添加总计列。

我的查询如下：

;WITH ToPivot
AS ( SELECT 
            time_tracker.date, 
            Users.FirstName + ' ' + Users.LastName AS username, 
            (CASE 
                WHEN 
                    (
                        (datepart(hour, chk_in)) is null 
                        OR (datepart(hour, chk_out)) is null
                    ) 
                    then 
                    0 
                WHEN 
                    (
                        (datepart(hour, chk_in)) >= 12 
                        OR (datepart(hour, chk_out)) < 16 
                        AND ((datepart(hour, chk_out)) - (datepart(hour, chk_in))) < 6
                    ) 
                    THEN 
                    0.5  
                ELSE 
                1 
            END) AS late, TypeOfLeaves.leave_type, Userleavetyp.no_of_days 
            FROM Users 
            INNER JOIN time_tracker ON Users.ID = time_tracker.fk_userid 
            INNER JOIN Userleavetyp ON Users.ID = Userleavetyp.fk_user 
            INNER JOIN TypeOfLeaves ON Userleavetyp.fk_tol = TypeOfLeaves.ID 
            WHERE (Users.FK_Status = 1)
), 
LateTotals AS (
    SELECT T.username, Total = SUM(late),T.leave_type,T.no_of_days 
    FROM ToPivot AS T 
    WHERE T.date BETWEEN '2018-07-01' AND '2018-07-31' GROUP BY T.username,T.leave_type,T.no_of_days
)

SELECT distinct(pv2.username) as original ,PV2.*,L.Total 
FROM ToPivot AS P
PIVOT (SUM(late) FOR date IN ("2018-07-01", "2018-07-02", "2018-07-03", "2018-07-04", "2018-07-05", "2018-07-06", "2018-07-07", "2018-07-08", "2018-07-09", "2018-07-10", "2018-07-11", "2018-07-12", "2018-07-13", "2018-07-14", "2018-07-15", "2018-07-16", "2018-07-17", "2018-07-18", "2018-07-19", "2018-07-20", "2018-07-21", "2018-07-22", "2018-07-23", "2018-07-24", "2018-07-25", "2018-07-26", "2018-07-27", "2018-07-28", "2018-07-29", "2018-07-30", "2018-07-31")) AS pv1
PIVOT (SUM(no_of_days) FOR leave_type IN ([Casual Leave], [Paid Leave], [Complimentary Leave])) AS pv2
LEFT JOIN LateTotals AS L ON L.username = pv2.username

此代码的输出如下：

Answer 1

我相信您只需要按以下方式更改选择语句。从这个：

SELECT distinct(pv2.username) as original ,PV2.*,L.Total

对此：

SELECT distinct(pv2.username) as original ,PV2.*,L.Total,
    CASE WHEN PV2.[Casual Leave] IS NULL THEN 0 ELSE PV2.[Casual Leave] +
    CASE WHEN PV2.[Paid Leave] IS NULL THEN 0 ELSE PV2.[Paid Leave] +
    CASE WHEN PV2.[Complimentary Leave] IS NULL THEN 0 ELSE PV2.[Complimentary Leave]
    AS [Total Leave]

获取两列具有多个枢轴的总和

1 个答案: