6

次の結果を得るために、(SQL Reporting Services で使用するための) MS SQL クエリを探しています。

JobCode   0-10     11-20     21-30    31-40    41-50
A           1        2         0        0        2
B           0        2         2        0        0
C           3        0         1        1        0

現在のクエリ:

SELECT  RecordID
    , CAST(GETDATE() - JobStartDate AS Int) AS DaysSinceStart
    , Code
FROM    tblJobs

次の結果を返します。

RecordID | DaysSinceStart | Code
158987    11    A
158968     3    A
158972     4    C
158973    16    B
158974    23    C
158975    13    B
158976    45    A
158977    32    C
158985     9    C
158981    25    B
158982    47    A
158983    18    A
158978     5    C
158979    27    B

以下を使用して全体的な概要を返すことができますが、JobCode ごとに個々の結果を表示することはできません。

SELECT  Range AS [Day Range], COUNT(*) AS Jobs
FROM    (SELECT CASE WHEN DaysSinceStart BETWEEN 0 AND 9 THEN ' 0- 9'
                     WHEN DaysSinceStart BETWEEN 10 AND 19 THEN '10-19'
             WHEN DaysSinceStart BETWEEN 20 AND 29 THEN '20-29'
             WHEN DaysSinceStart BETWEEN 30 AND 39 THEN '30-39'
             WHEN DaysSinceStart BETWEEN 40 AND 49 THEN '40-49'
             WHEN DaysSinceStart BETWEEN 50 AND 59 THEN '50-59'
             WHEN DaysSinceStart BETWEEN 60 AND 69 THEN '60-69'
             WHEN DaysSinceStart BETWEEN 70 AND 79 THEN '70-79'
             WHEN DaysSinceStart BETWEEN 80 AND 89 THEN '80-89'
             WHEN DaysSinceStart BETWEEN 90 AND 99 THEN '90-99' 
             ELSE 'Over 100' END AS Range
        FROM    (SELECT DaysSinceStart
                 FROM tblJobs) AS derivedtbl_1) AS t
GROUP BY Range

コードフィールドごとの内訳を取得するのを手伝ってくれる人はいますか? よろしくお願いします。PS私はMS SQL Server 2012を使用しています

4

3 に答える 3

1

これを試してみてください -

クエリ:

SET NOCOUNT ON;

DECLARE @tblJobs TABLE
(
      RecordID INT
    , DaysSinceStart INT
    , Code CHAR(1)
)

INSERT INTO @tblJobs (RecordID, DaysSinceStart, Code)
VALUES 
    (158987,    11,    'A'), (158968,     3,    'A'),
    (158972,     4,    'C'), (158973,    16,    'B'),
    (158974,    23,    'C'), (158975,    13,    'B'),
    (158976,    45,    'A'), (158977,    32,    'C'),
    (158985,     9,    'C'), (158981,    25,    'B'),
    (158982,    47,    'A'), (158983,    18,    'A'),
    (158978,     5,    'C'), (158979,    27,    'B')

SELECT *
FROM (
    SELECT  
          t.Code
        , [Range] = 
            CASE 
                WHEN DaysSinceStart BETWEEN 0 AND 9   THEN '0-9'
                WHEN DaysSinceStart BETWEEN 10 AND 19 THEN '10-19'
                WHEN DaysSinceStart BETWEEN 20 AND 29 THEN '20-29'
                WHEN DaysSinceStart BETWEEN 30 AND 39 THEN '30-39'
                WHEN DaysSinceStart BETWEEN 40 AND 49 THEN '40-49'
                WHEN DaysSinceStart BETWEEN 50 AND 59 THEN '50-59'
                WHEN DaysSinceStart BETWEEN 60 AND 69 THEN '60-69'
                WHEN DaysSinceStart BETWEEN 70 AND 79 THEN '70-79'
                WHEN DaysSinceStart BETWEEN 80 AND 89 THEN '80-89'
                WHEN DaysSinceStart BETWEEN 90 AND 99 THEN '90-99' 
                ELSE 'Over 100' 
            END
    FROM @tblJobs t
) o
PIVOT
(
    COUNT(o.[Range])
    FOR [Range] IN (
        [0-9], [10-19], [20-29],
        [30-39], [40-49], [50-59],
        [60-69], [70-79], [80-89], 
        [90-99], [Over 100]
    )
) pt

出力：

Code 0-9         10-19       20-29       30-39       40-49       50-59       60-69       70-79       80-89       90-99       Over 100
---- ----------- ----------- ----------- ----------- ----------- ----------- ----------- ----------- ----------- ----------- -----------
A    1           2           0           0           2           0           0           0           0           0           0
B    0           2           2           0           0           0           0           0           0           0           0
C    3           0           1           1           0           0           0           0           0           0           0
于 2013-05-24T05:24:13.840 に答える
1

派生テーブル/CTE/テーブル変数/一時テーブル、または間隔を保持する通常のテーブルを作成します。そのテーブルを使用して、その間を結合し、datediff間隔JobStartDateをgetdate()取得します。次に、とpivotのそれぞれの値をカウントするために使用できます。Codeinterval

with C(low, high, interval) as
(
  select 0,  10, '0-10'  union all
  select 11, 20, '11-20' union all
  select 21, 30, '21-30' union all
  select 31, 40, '31-40' union all
  select 41, 50, '41-50' union all
  select 51, 2147483647, 'over 50'
)
select Code, [0-10], [11-20], [21-30], [31-40], [41-50], [over 50]
from  (
      select J.Code,
             J.RecordID,
             C.interval
      from tblJobs as J
        inner join C
          on datediff(day, J.JobStartDate, getdate()) between C.low and C.high
      ) as T
pivot (
      count(T.RecordID) 
      for T.Interval in ([0-10], [11-20], [21-30], [31-40], [41-50], [over 50])
      ) as P

SQL フィドル

于 2013-05-24T05:29:41.937 に答える