Question

我试图让SQL Server动态数据透视表工作，允许我计算，然后总结列数。数据透视表的目的是创建个人在城市中停留的所有日期的报告以及总天数（一个月内）。因此，例如，A人每天都在6月入住 - 总数将是30.Person B仅在6月3日开始入住 - 总数将为27等。数据表仅包括Name，ArriveDate，DepartDate .. 。月份的日期是通过SQL查询创建的。

＆＃13;

+------+------------+------------+-------+-------+-------+-----+-------+-------+-------+
| Name | ArriveDate | DepartDate | 06-01 | 06-02 | 06-03 | ... | 06-29 | 06-30 | Total |
+------+------------+------------+-------+-------+-------+-----+-------+-------+-------+
| A    | 2014-06-01 | 2014-06-23 |     1 |     1 |     1 | ... |     1 |     1 |    30 |
| B    | 2014-06-02 | 2014-06-23 |     0 |     1 |     1 | ... |     1 |     1 |    27 |
| C    | 2014-06-02 | 2014-06-23 |     0 |     0 |     0 | ... |     1 |     1 |    16 |
+------+------------+------------+-------+-------+-------+-----+-------+-------+-------+

＆＃13;

这是我到目前为止的查询：

DROP TABLE #tempDates

DECLARE @cols AS NVARCHAR(MAX),
        @query  AS NVARCHAR(MAX)

;WITH cte (datelist, maxdate) AS
(
    SELECT MIN(ArriveDate) datelist,
           EOMONTH(GETDATE()) AS maxdate
    FROM Reservation
    UNION ALL
    SELECT dateadd(dd, 1, datelist), maxdate
    FROM cte
    WHERE datelist < maxdate
) 
SELECT c.datelist
INTO #tempDates
FROM cte c

SELECT @cols = STUFF(( SELECT distinct ',' + QUOTENAME(convert(CHAR(10), datelist, 120)) 
                       FROM #tempDates
                       FOR XML PATH(''), TYPE
                      ).value('.', 'NVARCHAR(MAX)') 
                     ,1,1,'')



SET @query = 'SELECT ID,
                     ArriveDate,
                     DepartDate,                     
                     ' + @cols + '
              FROM 
              (         
                SELECT r.ID,
                       r.ArriveDate,
                       r.DepartDate,
                       d.datelist, 
                       convert(CHAR(10), datelist, 120) PivotDate
                FROM 
                Reservation r       
                LEFT JOIN
                #tempDates d
                    ON d.datelist BETWEEN rg.ArriveDate AND GETDATE()
            ) x
            pivot 
            (
                COUNT(datelist)
                FOR PivotDate IN (' + @cols + ')
            ) p '

EXECUTE (@query)

Answer 1

以下是我使用动态交叉表的尝试：

SQL Fiddle

示例数据：

Name ArriveDate DepartDate
---- ---------- ----------
A    2015-07-01 2015-07-23
B    2015-07-02 2015-07-04
C    2015-07-03 2015-07-31

动态交叉表解决方案：

DECLARE @minDate AS DATE,
        @maxDate AS DATE

SELECT  @minDate = DATEADD(DAY, 1, EOMONTH(GETDATE(), -1)),
        @maxDate = EOMONTH(GETDATE())

CREATE TABLE #dates(dt DATE)

DECLARE @sql1 VARCHAR(MAX) = '',
        @sql2 VARCHAR(MAX) = '',
        @sql3 VARCHAR(MAX) = '';

WITH E1(N) AS(
    SELECT 1 FROM(VALUES (1),(1),(1),(1),(1),(1),(1),(1),(1),(1))t(N)
),
E2(N) AS(SELECT 1 FROM E1 a CROSS JOIN E1 b),
E4(N) AS(SELECT 1 FROM E2 a CROSS JOIN E2 b),
E8(N) AS(SELECT 1 FROM E4 a CROSS JOIN E4 b),
Tally(N) AS(
    SELECT TOP(DATEDIFF(DAY, @minDate, @maxDate) + 1)
        ROW_NUMBER() OVER(ORDER BY (SELECT NULL))
    FROM E8
)
INSERT INTO #dates
    SELECT DATEADD(DAY, N - 1, @minDate)
    FROM Tally

SELECT @sql1 =
'SELECT
    r.Name
    , r.ArriveDate
    , r.DepartDate' + CHAR(10)

SELECT @sql2 = @sql2 +
'   , SUM(CASE WHEN d.dt = CAST(''' + CONVERT(VARCHAR(8), dt, 112) + ''' AS DATE) THEN 1 ELSE 0 END) AS ' 
    + QUOTENAME(CONVERT(VARCHAR(10), dt, 120)) + CHAR(10)
FROM #dates
ORDER BY dt

SELECT @sql2 = @sql2 +
'   , COUNT(d.dt) AS [total]' + CHAR(10)

SELECT @sql3 =
'FROM Reservation r
LEFT JOIN #dates d
    ON d.dt BETWEEN r.ArriveDate AND r.DepartDate
GROUP BY
    r.Name, r.ArriveDate, r.DepartDate'

PRINT (@sql1 + @sql2 + @sql3)
EXEC (@sql1 + @sql2 + @sql3)

DROP TABLE #dates

结果：

Name ArriveDate DepartDate 2015-07-01  2015-07-02  2015-07-03  ..... 2015-07-29  2015-07-30  2015-07-31  total
---- ---------- ---------- ----------- ----------- ----------- ..... ----------- ----------- ----------- -----------
A    2015-07-01 2015-07-23 1           1           1           ..... 0           0           0           23
B    2015-07-02 2015-07-04 0           1           1           ..... 0           0           0           3
C    2015-07-03 2015-07-31 0           0           1           ..... 1           1           1           29

获取SQL动态（日历）数据透视表的总计

1 个答案: