Question

我需要更新此查询中的目标列

CREATE TABLE #test 
(
     id INT IDENTITY(1,1),  
     score int, 
     goal int
)

INSERT INTO #test(score) VALUES (80)
INSERT INTO #test(score) VALUES (86)
INSERT INTO #test(score) VALUES (90)
INSERT INTO #test(score) VALUES (90)
INSERT INTO #test(score) VALUES (90)
INSERT INTO #test(score) VALUES (80)
INSERT INTO #test(score) VALUES (95)
INSERT INTO #test(score) VALUES (95)
INSERT INTO #test(score) VALUES (95)
INSERT INTO #test(score) VALUES (95)
INSERT INTO #test(score) VALUES (95)
INSERT INTO #test(score) VALUES (89)

--- initial goal 5
-- if score is 90 or above for 3 continuous week then goal is 3
-- if score is 95 or above for 5 continuous week then goal is 1

Answer 1

快速笔记：

我修复了你的示例sql以添加日期。
我使用DATEADD功能轻松计算周数。
根据您的逻辑需求，您可能需要调整更新中的DATEADD函数。如果你想成为终极专属，那么你并不是100％明确，所以我认为它们最终是包容。

你应该可以做类似的事情：

UPDATE  #Test
SET     Goal = 3
WHERE   Id IN ( SELECT  Id
                FROM    #Test AS t
                OUTER APPLY ( SELECT    COUNT(1) AS 'Number'
                              FROM      #Test AS t2
                              WHERE     t2.Date BETWEEN DATEADD(WEEK, -2, t.Date) AND t.Date AND
                                        t2.Score >= 90
                            ) AS t2
                WHERE   t2.Number >= 3 );

UPDATE  #Test
SET     Goal = 1
WHERE   Id IN ( SELECT  Id
                FROM    #Test AS t
                OUTER APPLY ( SELECT    COUNT(1) AS 'Number'
                              FROM      #Test AS t2
                              WHERE     t2.Date BETWEEN DATEADD(WEEK, -4, t.Date) AND t.Date AND
                                        t2.Score >= 95
                            ) AS t2
                WHERE   t2.Number >= 5 );

最难的部分是计算“滚动”数字。我使用OUTER APPLY这样做，虽然可以通过几种不同的方式完成。几年前我问了一个类似的question，解决方法是使用CROSS APPLY。因此，您的里程可能会有所不同，而其他人可能会有不同的解决方案，但这样可以让您了解如何继续。

这是表结束时的结果集：

这是我用来生成表的sql。如果您计划在最终结果中使用临时表，请不要忘记放下临时表！

IF OBJECT_ID('tempdb..#Test') IS NOT NULL
    DROP TABLE #Test;

DECLARE @StartDate AS DATE = '2017-01-01';
CREATE TABLE #Test
    (
      Id INT IDENTITY(1, 1)
    , Score INT
    , Goal INT
    , Date DATE
    );
INSERT  INTO #Test
        ( Score, Date )
VALUES  ( 80, DATEADD(WEEK, 0, @StartDate) )  ,
        ( 86, DATEADD(WEEK, 1, @StartDate) )  ,
        ( 90, DATEADD(WEEK, 2, @StartDate) )  ,
        ( 90, DATEADD(WEEK, 3, @StartDate) )  ,
        ( 90, DATEADD(WEEK, 4, @StartDate) )  ,
        ( 80, DATEADD(WEEK, 5, @StartDate) )  ,
        ( 95, DATEADD(WEEK, 6, @StartDate) )  ,
        ( 95, DATEADD(WEEK, 7, @StartDate) )  ,
        ( 95, DATEADD(WEEK, 8, @StartDate) )  ,
        ( 95, DATEADD(WEEK, 9, @StartDate) )  ,
        ( 95, DATEADD(WEEK, 10, @StartDate) ) ,
        ( 89, DATEADD(WEEK, 11, @StartDate) );

DROP TABLE #Test;

根据tsql中其他列的先前值更新列

1 个答案: