Question

查找： max（abs（c（a = 2）-c（a = 1）））按b排序

给定表T

数学示例：

c(@2) - c(@1) in order of b
10    - 2     = 8
11    - 4     = 7
20    - 5     = 15  <-- max found

我可以获得b = 1差异的答案，但我想简化查询而不剪切并粘贴其他内容，其中b = 1，b = 2，b = 3等。不确定cte是否适用于此，还是使用递增b的递归查询？

最终工作查询：

SELECT
    coalesce(max(abs(rn.c - r0.c)), -999) AS rd --> RETURN nulls = -999.
FROM
    (SELECT b, c FROM t WHERE a = 1) AS r0
JOIN
    (SELECT b, c FROM t WHERE a = 2) AS rn
ON r0.b = rn.b;

Answer 1

create table stk_test(a int ,b int, c int)

insert into stk_test
select 1,1,2
union
select 1,2,4
union
select 1,3,5
union
select 2,1,10
union
select 2,2,11
union
select 2,3,20
union
select 3,1,40
union
select 3,2,40
union
select 3,3,40

select max(abs(a.a_c-b.b_c)) as Max_Diff_C from
(
    select c as a_c,b from stk_test ss
    where a=1
) a  join  
(
    select c  as b_c,b from stk_test ssr
    where a=2
)b on a.b=b.b

Answer 2

我还没明白你的意思。可能有帮助。

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经过https://kripken.github.io/sql.js/GUI/

的测试

Answer 3

这是使用Lead函数对数据进行排名的一种尝试

create table t(a int, b int , c int)

insert into t values(1,1,2)      
insert into t values(1,2,4)
insert into t values(1,3,5)
insert into t values(2,1,10)
insert into t values(2,2,11)
insert into t values(2,3,20)
insert into t values(3,1,40)
insert into t values(3,2,40)
insert into t values(3,3,40)

with data
  as (select * /*First rank the results on the basis of columns a and b*/    
            ,row_number() over(order by a,b) as rnk
        from t
     )
,temp_data
  as (select a      
            ,b
            ,c
            ,rnk
            ,abs(c-lead(c,3) over(order by rnk)) as max_rnk
       from data
      )
select a,max(max_rnk) /*Gets the max value of rank grouped by each a*/      
  from temp_data
group by a

使用dbfiddle进行演示

https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=b7db648a2512b623a7df4351326fa25d

SQL从同一列中找到最大增量值

3 个答案: