使用SQL Server,我如何分割一个字符串,以便我可以访问项目x?

拿一根“你好,约翰·史密斯”的绳子。我如何通过空格分割字符串并访问索引1应该返回“约翰”的项目?


当前回答

修改@Aaron Bertrand的功能

CREATE FUNCTION [dbo].[SplitString]
(
    @List NVARCHAR(MAX),
    @Delim VARCHAR(255),
    @Idx int
)
RETURNS NVARCHAR(1000)
AS
BEGIN
    DECLARE @ValueTable TABLE(String NVARCHAR(50), Ind int)
    DECLARE @Value NVARCHAR(50)
    BEGIN
    INSERT INTO @ValueTable
    SELECT Value, idx FROM
        (SELECT [Value], idx = RANK() OVER (ORDER BY n) FROM 
              ( 
                SELECT n = Number, 
                [Value] = LTRIM(RTRIM(SUBSTRING(@List, [Number],
                CHARINDEX(@Delim, @List + @Delim, [Number]) - [Number])))
                FROM    
                        (SELECT Number = ROW_NUMBER() OVER (ORDER BY name)
                         FROM sys.all_objects) AS x
                WHERE Number <= LEN(@List)
                AND SUBSTRING(@Delim + @List, [Number], LEN(@Delim)) = @Delim
              ) AS y
          ) AS R WHERE idx = @Idx
    SET @Value = (SELECT String FROM @ValueTable)
    END
    RETURN @Value
END
GO

其他回答

可以利用Number表进行字符串解析。

创建一个物理数字表:

    create table dbo.Numbers (N int primary key);
    insert into dbo.Numbers
        select top 1000 row_number() over(order by number) from master..spt_values
    go

创建具有1000000行的测试表

    create table #yak (i int identity(1,1) primary key, array varchar(50))

    insert into #yak(array)
        select 'a,b,c' from dbo.Numbers n cross join dbo.Numbers nn
    go

创建函数

    create function [dbo].[ufn_ParseArray]
        (   @Input      nvarchar(4000), 
            @Delimiter  char(1) = ',',
            @BaseIdent  int
        )
    returns table as
    return  
        (   select  row_number() over (order by n asc) + (@BaseIdent - 1) [i],
                    substring(@Input, n, charindex(@Delimiter, @Input + @Delimiter, n) - n) s
            from    dbo.Numbers
            where   n <= convert(int, len(@Input)) and
                    substring(@Delimiter + @Input, n, 1) = @Delimiter
        )
    go

使用情况(在我的笔记本电脑上40秒内输出3mil行)

    select * 
    from #yak 
    cross apply dbo.ufn_ParseArray(array, ',', 1)

清理

    drop table dbo.Numbers;
    drop function  [dbo].[ufn_ParseArray]

这里的性能并不惊人,但在100万行表上调用函数并不是最好的主意。如果将字符串拆分到多行,我会避免使用该函数。

你可以在SQL用户定义函数解析带分隔符的字符串中找到有用的解决方案(来自代码项目)。

你可以使用这个简单的逻辑:

Declare @products varchar(200) = '1|20|3|343|44|6|8765'
Declare @individual varchar(20) = null

WHILE LEN(@products) > 0
BEGIN
    IF PATINDEX('%|%', @products) > 0
    BEGIN
        SET @individual = SUBSTRING(@products,
                                    0,
                                    PATINDEX('%|%', @products))
        SELECT @individual

        SET @products = SUBSTRING(@products,
                                  LEN(@individual + '|') + 1,
                                  LEN(@products))
    END
    ELSE
    BEGIN
        SET @individual = @products
        SET @products = NULL
        SELECT @individual
    END
END

递归CTE解决方案与服务器疼痛,测试它

MS SQL Server 2008模式设置:

create table Course( Courses varchar(100) );
insert into Course values ('Hello John Smith');

查询1:

with cte as
   ( select 
        left( Courses, charindex( ' ' , Courses) ) as a_l,
        cast( substring( Courses, 
                         charindex( ' ' , Courses) + 1 , 
                         len(Courses ) ) + ' ' 
              as varchar(100) )  as a_r,
        Courses as a,
        0 as n
     from Course t
    union all
      select 
        left(a_r, charindex( ' ' , a_r) ) as a_l,
        substring( a_r, charindex( ' ' , a_r) + 1 , len(a_R ) ) as a_r,
        cte.a,
        cte.n + 1 as n
    from Course t inner join cte 
         on t.Courses = cte.a and len( a_r ) > 0

   )
select a_l, n from cte
--where N = 1

结果:

|    A_L | N |
|--------|---|
| Hello  | 0 |
|  John  | 1 |
| Smith  | 2 |

I realize this is a really old question, but starting with SQL Server 2016 there are functions for parsing JSON data that can be used to specifically address the OP's question--and without splitting strings or resorting to a user-defined function. To access an item at a particular index of a delimited string, use the JSON_VALUE function. Properly formatted JSON data is required, however: strings must be enclosed in double quotes " and the delimiter must be a comma ,, with the entire string enclosed in square brackets [].

DECLARE @SampleString NVARCHAR(MAX) = '"Hello John Smith"';
--Format as JSON data.
SET @SampleString = '[' + REPLACE(@SampleString, ' ', '","') + ']';
SELECT 
    JSON_VALUE(@SampleString, '$[0]') AS Element1Value,
    JSON_VALUE(@SampleString, '$[1]') AS Element2Value,
    JSON_VALUE(@SampleString, '$[2]') AS Element3Value;

输出

Element1Value         Element2Value       Element3Value
--------------------- ------------------- ------------------------------
Hello                 John                Smith

(1 row affected)

使用STRING_SPLIT的现代方法需要SQL Server 2016及以上版本。

DECLARE @string varchar(100) = 'Hello John Smith'

SELECT
    ROW_NUMBER() OVER (ORDER BY value) AS RowNr,
    value
FROM string_split(@string, ' ')

结果:

RowNr   value
1       Hello
2       John
3       Smith

现在可以从行号中得到第n个元素。