Преобразовать эти строки в столбцы

Question

У меня довольно сложный запрос, и мне интересно, как я могу преобразовать их из строк в столбцы, чтобы они отображались в следующем формате. Среднее - это столбец, который следует использовать, общее количество просто для информации, но его не нужно свернуть.

отрывок из нескольких текущих результатов:

SELECT dayofweek                    AS Day, 
       Format(onhour, '00') + ':00' AS Hour, 
       Sum(totals)                  AS Total, 
       Avg(totals)                  AS Average 
FROM   (SELECT Datename(weekday, D.fordate) AS DayOfWeek, 
               H.onhour, 
               Isnull(T.totals, 0)          AS Totals 
        FROM   (SELECT DISTINCT T.fordate 
                FROM   #visit AS T) AS D 
               CROSS apply (VALUES(0), 
                                  (1), 
                                  (2), 
                                  (3), 
                                  (4), 
                                  (5), 
                                  (6), 
                                  (7), 
                                  (8), 
                                  (9), 
                                  (10), 
                                  (11), 
                                  (12), 
                                  (13), 
                                  (14), 
                                  (15), 
                                  (16), 
                                  (17), 
                                  (18), 
                                  (19), 
                                  (20), 
                                  (21), 
                                  (22), 
                                  (23)) AS H(onhour) 
               LEFT OUTER JOIN #visit AS T 
                            ON H.onhour = T.onhour 
                               AND D.fordate = T.fordate) a 
WHERE  onhour NOT IN ( 0, 1, 2, 3, 
                       4, 5, 6, 7, 
                       20, 21, 22, 23 ) 
GROUP  BY dayofweek, 
          onhour 
ORDER  BY CASE 
            WHEN dayofweek = 'Monday' THEN 1 
            WHEN dayofweek = 'Tuesday' THEN 2 
            WHEN dayofweek = 'Wednesday' THEN 3 
            WHEN dayofweek = 'Thursday' THEN 4 
            WHEN dayofweek = 'Friday' THEN 5 
            WHEN dayofweek = 'Saturday' THEN 6 
            WHEN dayofweek = 'Sunday' THEN 7 
          END ASC, 
          onhour 

Answer 1

Вы можете использовать запрос кросс-таблиц, чтобы получить свой результат.

SELECT CONVERT(varchar(8), DATEADD(hh, onhour, 0), 108) + ' - ' 
       + CONVERT(varchar(8), DATEADD(hh, onhour+1, 0), 108) AS Hour, 
       Sum(totals)                  AS Total, 
       Avg(CASE WHEN DayOfWeek = 'Monday'    THEN totals END) AS Monday  , 
       Avg(CASE WHEN dayofweek = 'Tuesday'   THEN totals END) AS Tuesday , 
       Avg(CASE WHEN dayofweek = 'Wednesday' THEN totals END) AS Wednesday, 
       Avg(CASE WHEN dayofweek = 'Thursday'  THEN totals END) AS Thursday, 
       Avg(CASE WHEN dayofweek = 'Friday'    THEN totals END) AS Friday  , 
       Avg(CASE WHEN dayofweek = 'Saturday'  THEN totals END) AS Saturday, 
       Avg(CASE WHEN dayofweek = 'Sunday'    THEN totals END) AS Sunday
FROM   (SELECT Datename(weekday, D.fordate) AS DayOfWeek, 
               H.onhour, 
               Isnull(T.totals, 0)          AS Totals 
        FROM   (SELECT DISTINCT T.fordate 
                FROM   #visit AS T) AS D 
               CROSS apply (VALUES(0), 
                                  (1), 
                                  (2), 
                                  (3), 
                                  (4), 
                                  (5), 
                                  (6), 
                                  (7), 
                                  (8), 
                                  (9), 
                                  (10), 
                                  (11), 
                                  (12), 
                                  (13), 
                                  (14), 
                                  (15), 
                                  (16), 
                                  (17), 
                                  (18), 
                                  (19), 
                                  (20), 
                                  (21), 
                                  (22), 
                                  (23)) AS H(onhour) 
               LEFT OUTER JOIN #visit AS T 
                            ON H.onhour = T.onhour 
                               AND D.fordate = T.fordate) a 
WHERE  onhour NOT IN ( 0, 1, 2, 3, 
                       4, 5, 6, 7, 
                       20, 21, 22, 23 ) 
GROUP  BY onhour 
ORDER  BY onhour;