获取从 split_string 中提取的每个字符串的排名

Question

我有一个 TABLE 和 STR DNA DATA 以下 Table [DYS]

Id	DYS385
3	10-19
4	13-16
5	13-18
6	13-19
7	13-17
8	13-18
9	13-18
10	14-19
11	13-19
12	13-18

我正在使用以下脚本拆分 [DYS385]

的值

select top 10 id,[DYS385],t.Value 
from dys 
OUTER APPLY(select * from string_split([DYS385],'-')) t
where dys385 is not null

输出

Id	DYS385	VALUE
3	10-19	10
3	10-19	19
4	13-16	13
4	13-16	16
5	13-18	13
5	13-18	18
6	13-19	13
6	13-19	19
7	13-17	13
7	13-17	17

我想为每个值获取排名例子 10-19 => 10: 排名 1, 19: 排名 2

期望的结果：

Id	DYS385	VALUE	RANK
3	10-19	10	1
3	10-19	19	2
4	13-16	13	1
4	13-16	16	2
5	13-18	13	1
5	13-18	18	2
6	13-19	13	1
6	13-19	19	2
7	13-17	13	1
7	13-17	17	2

Answer 1

Split_string 有一个选项可以输出这个排名，称为序数。只需添加一个值为 1 的额外参数：

SELECT *FROM STRING_SPLIT('Lorem ipsum dolor sit amet.', ' ', 1);

那个returns:

value   ordinal
Lorem   1
ipsum   2
dolor   3
sit 4
amet.   5

在您的情况下，查询将是：

select top 10 id,[DYS385],t.* from dys OUTER APPLY(select * from string_split([DYS385],'-',1)) t where dys385 is not null

Answer 2

使用替代 string-split 方法，例如 XML 或 Json 可以 return 一个序数位置，例如：

create function dbo.SplitString(@string varchar(1000), @Delimiter varchar(10))
returns table
as
   return(  
      select j.[value], 1 + Convert(tinyint,j.[key]) Seq 
      from OpenJson(Concat('["',replace(@string,@delimiter, '","'),'"]')) j
   );


select value, Seq as [Rank]
from dbo.SplitString('10-19','-')
order by [Rank];

Example fiddle

Answer 3

鉴于此数据：

CREATE TABLE dbo.Something(Id int, DYS385 varchar(10));

INSERT dbo.Something(Id, DYS385) VALUES (3 , '10-19'), 
(4 , '13-16'), (5 , '13-18'), (6 , '13-19'),
(7 , '13-17'), (8 , '13-18'), (9 , '13-18'),
(10, '14-19'), (11, '13-19'), (12, '13-18');

确保按正确顺序排列 broken-up 字符串的另一种方法是使用 PARSENAME() 等技巧，尽管这可能对字符串长度以及点是否在其中有效很敏感数据：

SELECT s.Id, s.DYS385, value = v.v, [rank] = ROW_NUMBER() OVER 
  (PARTITION BY s.Id ORDER BY c.c DESC)
FROM dbo.Something AS s
CROSS APPLY (VALUES(1),(2)) AS c(c)
CROSS APPLY (VALUES(PARSENAME(REPLACE(s.DYS385,'-','.'),c.c))) AS v(v)
ORDER BY s.Id, [rank];

输出：

Id DYS385 value rank

3 10-19 10 1

3 10-19 19 2

4 13-16 13 1

4 13-16 16 2

5 13-18 13 1

5 13-18 18 2

6 13-19 13 1

6 13-19 19 2

7 13-17 13 1

7 13-17 17 2

8 13-18 13 1

8 13-18 18 2

9 13-18 13 1

9 13-18 18 2

10 14-19 14 1

10 14-19 19 2

11 13-19 13 1

11 13-19 19 2

12 13-18 13 1

12 13-18 18 2

示例db<>fiddle

Answer 4

我们还可以让recursivesplit_string函数returns对每一项的排序如下 Link dbfiddle

Create function Split_Recursive(@string nvarchar(max),@delimiter as varchar(1)) returns Table
as
return(
with cte as(
select 1 N,
case when charindex(@delimiter,@string,1) =0 then @string else
substring(@string,1,-1+charindex(@delimiter,@string,1)) end [Found],
case when charindex(@delimiter,@string,1)=0 then @string else substring(@string,1+charindex(@delimiter,@string,1) ,len(@string)) 
end [MYSTRING]
union all
select 1+n,
case when charindex(@delimiter,[MYSTRING],1)=0 then [MYSTRING] else
substring([MYSTRING],1,-1+charindex(@delimiter,[MYSTRING],1)) end,substring([MYSTRING],1+charindex(@delimiter,[MYSTRING]),len([MYSTRING])) from cte
where charindex(@delimiter,[MYSTRING],1)>0),
cte2 as (select N,Found from cte 
union select 1+t.N,Mystring from cte OUTER APPLY(select top 1 N from cte order by n desc)t where t.n=cte.n
)
select N Rank,Found from cte2
)

go

select * from dbo.Split_Recursive('12-16','-')

输出

Rank	Found
1	12
2	16

获取从 split_string 中提取的每个字符串的排名

Get the rank of each string extracted from split_string

sql-server

split

outer-join

sql-server-2017

row-number