Question

这是我的表：

+-----+-----+-------------+-------------+-----+
|  ID | tFK |  schedDate  |  complDate  | res |
+=====+=====+=============+=============+=====+
|  1  |  2  | 2018-02-01  | 2018-02-01  |  2  |
+-----+-----+-------------+-------------+-----+
|  2  |  2  | 2018-02-02  | 2018-02-02  |  2  |
+-----+-----+-------------+-------------+-----+
|  3  |  2  | 2018-02-03  | 2018-02-03  |  2  |
+-----+-----+-------------+-------------+-----+
|  4  |  2  | 2018-02-04  | 2018-02-04  |  3  |
+-----+-----+-------------+-------------+-----+
|  5  |  2  | 2018-02-05  | 2018-02-05  |  2  |
+-----+-----+-------------+-------------+-----+
|  6  |  2  | 2018-02-06  | 2018-02-06  |  2  |
+-----+-----+-------------+-------------+-----+

我想只计算输入的tFK是否与之前相同的tFK输入的结果相同，无论日期是什么，这是我到目前为止的语法，但它不是我想要的输出。如何添加＆＃39; CONSECUTIVE＆＃39;我的查询的标准？

SELECT [taskFK],COUNT([result]) resultCount
FROM [cil].[cil].[schedule]
where scheduledDate between '2018-01-01' AND '2018-02-22'
AND completionDate IS NOT NULL
AND result=2
GROUP BY taskFK
HAVING COUNT([result]) > @variable
ORDER BY taskFK

编辑：

这将是预期的输出：如果@variable设置为2，则输出将为tFK = 2 如果@variable设置为3，则输出将为tFK = 1

由于

Answer 1

您可以使用LAG功能确定您的资源与该tfk的前一资产相同。由于我们不能在聚合函数中使用窗口函数，我们将使表变量或CTE计数并计算与之前的tfk结果匹配的记录。

DECLARE @schedule table (
id INT,
tfk INT,
scheddate DATE,
compldate DATE,
res INT
)

INSERT INTO @schedule VALUES
(1,2,'2018-02-01','2018-02-01',2),
(2,2,'2018-02-02','2018-02-02',2),
(3,2,'2018-02-03','2018-02-03',2),
(4,2,'2018-02-04','2018-02-04',3),
(5,2,'2018-02-05','2018-02-05',2),
(6,2,'2018-02-06','2018-02-06',2)

SELECT
    tfk,
    SUM(same_as_prev) AS sum_res
FROM (
    SELECT
        id,
        tfk,
        scheddate,
        compldate,
        res,
        CASE WHEN LAG(res) OVER (PARTITION BY tfk ORDER BY ID) = res THEN 1 ELSE 0 END AS same_as_prev
    FROM @schedule
    ) tbl
WHERE scheddate between '2018-01-01' AND '2018-02-22'
AND compldate IS NOT NULL
AND res = 2
GROUP BY tfk
HAVING SUM(same_as_prev) > 2
ORDER BY tfk

Answer 2

我假设您可以执行类似这样的操作来计算是否已存在行。

 SELECT 
 DISTINCT Count(tFK) AS [Count], tFK 
 FROM schedule
 Where res=3
 GROUP BY tFK
 HAVING Count(tFK)>0

如何计算具有相同连续值的行

2 个答案: