Kembalikan kolom per tanggal dalam rentang

15

Katakanlah saya memiliki Tabel A: BookingsPerPerson

Person_Id    ArrivalDate    DepartureDate
123456       2012-01-01     2012-01-04
213415       2012-01-02     2012-01-07

Apa yang perlu saya capai dengan tampilan adalah sebagai berikut:

Person_Id    ArrivalDate    DepartureDate    Jan-01    Jan-02    Jan-03    Jan-04    Jan-05    Jan-06    Jan-07
123456       2012-01-01     2012-01-04       1         1         1         1
213415       2012-01-02     2012-01-07                 1         1         1         1         1         1

Sistem ini untuk acara, sehingga setiap pemesanan hotel bisa memakan waktu antara 1 hingga 15 hari tetapi tidak lebih dari itu. Setiap ide akan sangat dihargai.

Federico Giust
sumber

Jawaban:

27

Anda dapat menggunakan PIVOTfungsi untuk melakukan kueri ini. Jawaban saya akan mencakup versi statis dan dinamis karena terkadang lebih mudah untuk memahaminya menggunakan versi statis.

Static Pivot adalah saat Anda membuat hard-code semua nilai yang ingin Anda ubah menjadi kolom.

-- first into into a #temp table the list of dates that you want to turn to columns
;with cte (datelist, maxdate) as
(
    select min(arrivaldate) datelist, max(departuredate) maxdate
    from BookingsPerPerson
    union all
    select dateadd(dd, 1, datelist), maxdate
    from cte
    where datelist < maxdate
) 
select c.datelist
into #tempDates
from cte c

select *
from
(
    select b.person_id, b.arrivaldate, b.departuredate,
        d.datelist,
        convert(CHAR(10), datelist, 120) PivotDate
    from #tempDates d
    left join BookingsPerPerson b
        on d.datelist between b.arrivaldate and b.departuredate
) x
pivot
(
    count(datelist)
    for PivotDate in ([2012-01-01], [2012-01-02], [2012-01-03],
              [2012-01-04], [2012-01-05], [2012-01-06] , [2012-01-07])
) p;

Hasil (Lihat SQL Fiddle With Demo ):

PERSON_ID | ARRIVALDATE | DEPARTUREDATE | 2012-01-01 | 2012-01-02 | 2012-01-03 | 2012-01-04 | 2012-01-05 | 2012-01-06 | 2012-01-07
=====================================================================================================================================
123456    | 2012-01-01  | 2012-01-04    | 1          | 1          | 1          | 1          | 0          | 0          | 0
213415    | 2012-01-02  | 2012-01-07    | 0          | 1          | 1          | 1          | 1          | 1          | 1

Versi dinamis akan menghasilkan daftar nilai untuk diubah menjadi kolom:

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

;with cte (datelist, maxdate) as
(
    select min(arrivaldate) datelist, max(departuredate) maxdate
    from BookingsPerPerson
    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 person_id, arrivaldate, departuredate, ' + @cols + ' from 
             (
                select b.person_id, b.arrivaldate, b.departuredate,
                    d.datelist,
                    convert(CHAR(10), datelist, 120) PivotDate
                from #tempDates d
                left join BookingsPerPerson b
                    on d.datelist between b.arrivaldate and b.departuredate
            ) x
            pivot 
            (
                count(datelist)
                for PivotDate in (' + @cols + ')
            ) p '

execute(@query)

Hasilnya sama (lihat SQL Fiddle With Demo ):

PERSON_ID | ARRIVALDATE | DEPARTUREDATE | 2012-01-01 | 2012-01-02 | 2012-01-03 | 2012-01-04 | 2012-01-05 | 2012-01-06 | 2012-01-07
=====================================================================================================================================
123456    | 2012-01-01  | 2012-01-04    | 1          | 1          | 1          | 1          | 0          | 0          | 0
213415    | 2012-01-02  | 2012-01-07    | 0          | 1          | 1          | 1          | 1          | 1          | 1
Taryn
sumber
8

Saya sekolah tua, dan menemukan CASElebih mudah untuk berolahraga di kepala saya daripada PIVOT. Saya yakin bluefeet akan segera muncul dan membuat saya malu, tetapi sementara itu Anda dapat bermain dengan query SQL dinamis ini. Dengan asumsi meja Anda menyimpan DATEdan tidak DATETIME(atau lebih buruk lagi, VARCHAR):

USE tempdb;
GO

CREATE TABLE dbo.a
(
   Person_Id INT, 
   ArrivalDate DATE, 
   DepartureDate DATE
);

INSERT dbo.a SELECT 123456, '2012-01-01', '2012-01-04'
UNION ALL    SELECT 213415, '2012-01-02', '2012-01-07';

DECLARE @sql NVARCHAR(MAX) = N'SELECT Person_Id';

;WITH dr AS
(
  SELECT MinDate = MIN(ArrivalDate),
         MaxDate = MAX(DepartureDate)
  FROM dbo.a
),
n AS
(
  SELECT TOP (DATEDIFF(DAY, (SELECT MinDate FROM dr), (SELECT MaxDate FROM dr)) + 1)
   d = DATEADD(DAY, ROW_NUMBER() OVER (ORDER BY [object_id])-1, 
     (SELECT MinDate FROM dr))
 FROM sys.all_objects
)
SELECT @sql += ',
  ' + QUOTENAME(d) + ' = CASE WHEN ''' + CONVERT(CHAR(10), d, 120) 
  + ''' BETWEEN ArrivalDate AND DepartureDate THEN ''1'' ELSE '''' END' FROM n;

SELECT @sql += ' FROM dbo.a;'

EXEC sp_executesql @sql;
GO

DROP TABLE dbo.a;

Salah satu dari sedikit kasus, BTW, di mana saya bisa membenarkan menggunakan BETWEENuntuk pertanyaan rentang tanggal.

Aaron Bertrand
sumber
0

Bagaimana dengan ini untuk menghasilkan daftar tanggal

declare @Date01 as smalldatetime
declare @Date02 as smalldatetime
select @Date01 = min(periodstart), @Date02 = max(periodend)
    from BookingTable

declare @DateDiff as int
select @DateDiff = (select DATEDIFF(DAY, @Date01, @Date02))
;

WITH Tally (N) AS
(
    SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL))
    FROM sys.all_columns a CROSS JOIN sys.all_columns b
)
SELECT DATEADD(day, N, @Date01)
FROM Tally
where N <= @DateDiff
mko
sumber