Question

我这里有一张桌子：table1

employee ID     Reporting Manager       EmployeeName    Week No.        Points
1                  Mr. A                    Bob         week1            -10
2                  Mr. A                    Deepak      week1             50
3                  Mr. B                    Brinda      week1             60
4                  Mr. B                    Chriss      week1            -10
1                  Mr. A                    Bob         week2            -10
2                  Mr. A                    Deepak      week2             40
3                  Mr. B                    Brinda      week2             20
4                  Mr. B                    Chriss      week2             90
1                  Mr. A                    Bob         week3             -10
2                  Mr. A                    Deepak      week3             30
3                  Mr. B                    Brinda      week3             40
4                  Mr. B                    Chriss      week3             50
1                  Mr. A                    Bob         week4            -10
2                  Mr. A                    Deepak      week4            -10
3                  Mr. B                    Brinda      week4            -10
4                  Mr. B                    Chriss      week4             30
1                  Mr. A                    Bob         week5            -10
2                  Mr. A                    Deepak      week5            -10
3                  Mr. B                    Brinda      week5             20
4                  Mr. B                    Chriss      week5             50

在“周无”栏中，我有连续5周的数据，每周我都要查看这些条件 -

1.每周我都会检查当前周的积分。如果本周的积分值为-10，而前一周的积分也是-10，那么在总积分字段的表2（下表）中添加-40那个特定的员工。（参见EmployeeName专栏）

2.我们检查当前周和前两周的分数。如果连续三周的分数为-10，则该人的奖励为-100，并添加到表2中的总分数字段。（下表）< / p>

3。类似地，对于连续四周，即当前周和前一个3周，如果点为-10，则将-200添加到table2的总点数字段。

employee ID     Employee Name    Total points
1                  Bob             -50
2                  Deepak          110
3                  Brinda          130
4                  Chriss          210

可能的解决方案（有问题）：

; WITH 
n as (
    select [EmployeeName], CAST(SUBSTRING([Week No.],5,10) AS INT) as wk, Points as pt  
    from YourTable --> Change this to your table
),
sc as (
    select w.*, case w.pt when -10 then (case w1.pt when -10 then (case w2.pt when -10 then (case w3.pt when -10 then -200 else -100 end) else -40 end) else 0 end) else 0 end x
    from n w
    left join n w1 on w.[EmployeeName] = w1.[EmployeeName] and w.wk = w1.wk+1
    left join n w2 on w.[EmployeeName] = w2.[EmployeeName] and w.wk = w2.wk+2
    left join n w3 on w.[EmployeeName] = w3.[EmployeeName] and w.wk = w3.wk+3
),
l as (
    select *, pt+x as total
    from sc
),
s as (
    select [EmployeeName], sum(total) total
    from l
    group by [EmployeeName]
)
select *
from s

任何人都可以帮我解决这个问题吗？我正在使用sql server 2012。

Answer 1

<强> SQL DEMO

WITH cte as (
    SELECT *, 
           LAG([Points], 1) OVER (PARTITION BY [employee ID] ORDER BY [Week No]) as prev1_points,
           LAG([Points], 2) OVER (PARTITION BY [employee ID] ORDER BY [Week No]) as prev2_points,
           LAG([Points], 3) OVER (PARTITION BY [employee ID] ORDER BY [Week No]) as prev3_points
    FROM employee
)
SELECT *,
       CASE WHEN [Points] = -10 AND prev1_points = -10  AND prev2_points = -10  AND prev3_points = -10 
            THEN -200
            WHEN [Points] = -10 AND prev1_points = -10  AND prev2_points = -10 
            THEN -100
            WHEN [Points] = -10 AND prev1_points = -10
            THEN -50
            ELSE 0
       END penalty       
FROM cte

最终输出

查找连续值并相应地给出积分

1 个答案: