SQL 中的移位操作以获取计数答案

【问题标题】：Shift manipulation in SQL to get countsSQL 中的移位操作以获取计数
【发布时间】：2014-07-30 12:06:53
【问题描述】：

我参加了下表Attendance

EID 是员工 ID，在班次栏中，D 表示白班，N 表示夜班。

现在我正在尝试获取与每个员工有关的以下数据。

白班次数 - D 的计数，夜班次数 - N 的计数，工作天数 - 员工在任一班次或两个班次中工作的天数（即使是员工在同一天同时工作，也视为一天。）

我可以在三个不同的结果中获得所有三个信息，如下所示...

WITH CTE (EID, in_time, shift) AS  
(
SELECT EID, in_time, shift FROM Attendance 
WHERE (in_time BETWEEN CONVERT(DATETIME, '2014-01-07 00:00:00', 102) AND CONVERT(DATETIME, '2014-07-31 00:00:00', 102)) AND PID = 'A002'
)

SELECT     EID, COUNT(*) AS DayTotal 
FROM         CTE
WHERE     (shift = 'D')
GROUP BY EID

SELECT     EID, COUNT(*) AS NightTotal 
FROM         Attendance
WHERE     (shift = 'N')
GROUP BY EID
;

WITH CTE2 (EID, in_time, shift) AS  
(
SELECT EID, in_time, shift FROM Attendance 
WHERE (in_time BETWEEN CONVERT(DATETIME, '2014-01-07 00:00:00', 102) AND CONVERT(DATETIME, '2014-07-31 00:00:00', 102)) AND PID = 'A002'
)
SELECT EID, COUNT ( DISTINCT CONVERT (DATE, in_time)) AS [Days] 
FROM CTE2 
WHERE (shift = 'D' OR shift = 'N')
GROUP BY EID

但我想在单个结果（表）中使用它。所以我尝试了以下查询，但它没有给出预期的输出。

WITH CTE (EID, in_time, shift) AS  
(
SELECT EID, in_time, shift FROM Attendance 
WHERE (in_time BETWEEN CONVERT(DATETIME, '2014-01-07 00:00:00', 102) AND CONVERT(DATETIME, '2014-07-31 00:00:00', 102)) AND PID = 'A002'
)

SELECT EID, 
CASE WHEN Shift = 'D' THEN COUNT(Shift) END AS [Day],
CASE WHEN Shift = 'N' THEN COUNT(Shift) END AS [Night],   
COUNT ( DISTINCT CONVERT (DATE, in_time)) AS [Days]  
FROM         CTE 
GROUP BY EID, shift

您能告诉我一个方法吗？

预期结果

【问题讨论】：

标签： sql sql-server ado.net case common-table-expression

【解决方案1】：

我认为你可以使用条件聚合得到你想要的：

SELECT EID,
       sum(case when shift = 'd' then 1 else 0 end) as dayshifts,
       sum(case when shift = 'n' then 1 else 0 end) as nightshifts,
       count(*) as total
FROM Attendance a
WHERE (in_time BETWEEN CONVERT(DATETIME, '2014-01-07 00:00:00', 102) AND
                       CONVERT(DATETIME, '2014-07-31 00:00:00', 102)) AND
      PID = 'A002';

编辑：

如果您想计算不同日期的总数，请使用count(distinct)：

SELECT EID,
       sum(case when shift = 'd' then 1 else 0 end) as dayshifts,
       sum(case when shift = 'n' then 1 else 0 end) as nightshifts,
       count(distinct case when shift in ('d', 'n') then cast(in_time as date) end) as total
FROM Attendance a
WHERE (in_time BETWEEN CONVERT(DATETIME, '2014-01-07 00:00:00', 102) AND
                       CONVERT(DATETIME, '2014-07-31 00:00:00', 102)) AND
      PID = 'A002';

【讨论】：

如果有人同时上夜班，会不会在“总计”列中计算两次？
吉迪尔是正确的。它不应该计算两次。谢谢
第 4 行第一个 cast 应该是 case。现在它起作用了。谢谢。
@Chathuranga 。 . .谢谢你。固定。

【解决方案2】：

WITH cte (eid, in_time, shift) 
     AS (SELECT eid, 
                in_time, 
                shift 
         FROM   attendance 
         WHERE  ( in_time BETWEEN CONVERT(DATETIME, '2014-01-07 00:00:00', 102) 
                                  AND 
                                            CONVERT(DATETIME, 
                                            '2014-07-31 00:00:00', 
                                            102 
                                            ) ) 
                AND pid = 'A002') 

SELECT eid, 
       Sum(CASE 
               WHEN shift = 'D' THEN 1 
               ELSE 0 
             END)                               AS DayTotal, 
       Sum(CASE 
               WHEN shift = 'N' THEN 1 
               ELSE 0 
             END)                               AS NightTotal, 
       Count (DISTINCT CONVERT (DATE, in_time)) AS Days 
FROM   cte 
GROUP  BY eid

【讨论】：

它有效，但结果数字不匹配。嗯，价值观不现实，但条件似乎还可以？我会试一试。谢谢。
当 count 以这种方式使用时，它不关心条件并给出两个班次的计数。如果使用 sum 代替，它会给出正确的结果。但我不知道原因;)

【解决方案3】：

@Chathuranga，由于一天的日班和夜班应该算作一个，请让我知道以下解决方案是否适合您。

DECLARE @Attendance TABLE (EID INT,
                      PID CHAR(4),
                      In_Time DATETIME,
                      Out_Time DATETIME,
                      Shift CHAR(1))

INSERT INTO @Attendance
VALUES
('100', 'A001', '2014-07-01 07:00:00.000', '2014-07-01 19:30:00.000', 'D'),
('102', 'A001', '2014-07-01 19:30:00.000', '2014-07-02 07:00:00.000', 'N'),
('100', 'A001', '2014-07-01 19:30:00.000', '2014-07-02 07:00:00.000', 'N'),
('104', 'A001', '2014-07-02 07:00:00.000', '2014-07-02 19:30:00.000', 'D'),
('100', 'A001', '2014-07-03 19:30:00.000', '2014-07-04 07:00:00.000', 'N'),
('102', 'A001', '2014-07-03 19:30:00.000', '2014-07-04 07:00:00.000', 'N'),
('104', 'A001', '2014-07-03 07:00:00.000', '2014-07-03 19:30:15.000', 'D'),
('102', 'A001', '2014-07-04 07:00:00.000', '2014-07-04 19:30:00.000', 'D'),
('100', 'A001', '2014-07-04 07:00:00.000', '2014-07-04 19:30:10.000', 'D')

SELECT EID,
      SUM(CASE
              WHEN Shift = 'D' THEN 1
              ELSE 0
          END) AS DayShift,
      SUM(CASE
              WHEN Shift = 'N' THEN 1
              ELSE 0
          END) AS NightShift,
      COUNT(DISTINCT CAST(In_Time AS DATE)) AS DayTotal
FROM @Attendance
GROUP BY EID

【讨论】：