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

在这里我发布了一个简单的解决方法

CREATE FUNCTION [dbo].[split](
          @delimited NVARCHAR(MAX),
          @delimiter NVARCHAR(100)
        ) RETURNS @t TABLE (id INT IDENTITY(1,1), val NVARCHAR(MAX))
        AS
        BEGIN
          DECLARE @xml XML
          SET @xml = N'<t>' + REPLACE(@delimited,@delimiter,'</t><t>') + '</t>'

          INSERT INTO @t(val)
          SELECT  r.value('.','varchar(MAX)') as item
          FROM  @xml.nodes('/t') as records(r)
          RETURN
        END

像这样执行函数

  select * from dbo.split('Hello John Smith',' ')

其他回答



    Alter Function dbo.fn_Split
    (
    @Expression nvarchar(max),
    @Delimiter  nvarchar(20) = ',',
    @Qualifier  char(1) = Null
    )
    RETURNS @Results TABLE (id int IDENTITY(1,1), value nvarchar(max))
    AS
    BEGIN
       /* USAGE
            Select * From dbo.fn_Split('apple pear grape banana orange honeydew cantalope 3 2 1 4', ' ', Null)
            Select * From dbo.fn_Split('1,abc,"Doe, John",4', ',', '"')
            Select * From dbo.fn_Split('Hello 0,"&""&&&&', ',', '"')
       */

       -- Declare Variables
       DECLARE
          @X     xml,
          @Temp  nvarchar(max),
          @Temp2 nvarchar(max),
          @Start int,
          @End   int

       -- HTML Encode @Expression
       Select @Expression = (Select @Expression For XML Path(''))

       -- Find all occurences of @Delimiter within @Qualifier and replace with |||***|||
       While PATINDEX('%' + @Qualifier + '%', @Expression) > 0 AND Len(IsNull(@Qualifier, '')) > 0
       BEGIN
          Select
             -- Starting character position of @Qualifier
             @Start = PATINDEX('%' + @Qualifier + '%', @Expression),
             -- @Expression starting at the @Start position
             @Temp = SubString(@Expression, @Start + 1, LEN(@Expression)-@Start+1),
             -- Next position of @Qualifier within @Expression
             @End = PATINDEX('%' + @Qualifier + '%', @Temp) - 1,
             -- The part of Expression found between the @Qualifiers
             @Temp2 = Case When @End < 0 Then @Temp Else Left(@Temp, @End) End,
             -- New @Expression
             @Expression = REPLACE(@Expression,
                                   @Qualifier + @Temp2 + Case When @End < 0 Then '' Else @Qualifier End,
                                   Replace(@Temp2, @Delimiter, '|||***|||')
                           )
       END

       -- Replace all occurences of @Delimiter within @Expression with '</fn_Split>&ltfn_Split>'
       -- And convert it to XML so we can select from it
       SET
          @X = Cast('&ltfn_Split>' +
                    Replace(@Expression, @Delimiter, '</fn_Split>&ltfn_Split>') +
                    '</fn_Split>' as xml)

       -- Insert into our returnable table replacing '|||***|||' back to @Delimiter
       INSERT @Results
       SELECT
          "Value" = LTRIM(RTrim(Replace(C.value('.', 'nvarchar(max)'), '|||***|||', @Delimiter)))
       FROM
          @X.nodes('fn_Split') as X(C)

       -- Return our temp table
       RETURN
    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 |

下面是一个函数,它将完成问题的目标,即分割字符串并访问项目X:

CREATE FUNCTION [dbo].[SplitString]
(
   @List       VARCHAR(MAX),
   @Delimiter  VARCHAR(255),
   @ElementNumber INT
)
RETURNS VARCHAR(MAX)
AS
BEGIN

       DECLARE @inp VARCHAR(MAX)
       SET @inp = (SELECT REPLACE(@List,@Delimiter,'_DELMTR_') FOR XML PATH(''))

       DECLARE @xml XML
       SET @xml = '<split><el>' + REPLACE(@inp,'_DELMTR_','</el><el>') + '</el></split>'

       DECLARE @ret VARCHAR(MAX)
       SET @ret = (SELECT
              el = split.el.value('.','varchar(max)')
       FROM  @xml.nodes('/split/el[string-length(.)>0][position() = sql:variable("@elementnumber")]') split(el))

       RETURN @ret

END

用法:

SELECT dbo.SplitString('Hello John Smith', ' ', 2)

结果:

John

Aaron Bertrand的回答很好,但也有缺陷。它不能准确地将空格作为分隔符处理(就像最初问题中的示例一样),因为长度函数将空格带在后面。

下面是他的代码,稍微调整了一下,允许使用空格分隔符:

CREATE FUNCTION [dbo].[SplitString]
(
    @List NVARCHAR(MAX),
    @Delim VARCHAR(255)
)
RETURNS TABLE
AS
    RETURN ( SELECT [Value] FROM 
      ( 
        SELECT 
          [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+'x')-1) = @Delim
      ) AS y
    );

几乎所有其他答案都是替换正在分割的字符串,这浪费了CPU周期并执行不必要的内存分配。

我在这里介绍了一种更好的进行字符串拆分的方法:http://www.digitalruby.com/split-string-sql-server/

代码如下:

SET NOCOUNT ON

-- You will want to change nvarchar(MAX) to nvarchar(50), varchar(50) or whatever matches exactly with the string column you will be searching against
DECLARE @SplitStringTable TABLE (Value nvarchar(MAX) NOT NULL)
DECLARE @StringToSplit nvarchar(MAX) = 'your|string|to|split|here'
DECLARE @SplitEndPos int
DECLARE @SplitValue nvarchar(MAX)
DECLARE @SplitDelim nvarchar(1) = '|'
DECLARE @SplitStartPos int = 1

SET @SplitEndPos = CHARINDEX(@SplitDelim, @StringToSplit, @SplitStartPos)

WHILE @SplitEndPos > 0
BEGIN
    SET @SplitValue = SUBSTRING(@StringToSplit, @SplitStartPos, (@SplitEndPos - @SplitStartPos))
    INSERT @SplitStringTable (Value) VALUES (@SplitValue)
    SET @SplitStartPos = @SplitEndPos + 1
    SET @SplitEndPos = CHARINDEX(@SplitDelim, @StringToSplit, @SplitStartPos)
END

SET @SplitValue = SUBSTRING(@StringToSplit, @SplitStartPos, 2147483647)
INSERT @SplitStringTable (Value) VALUES(@SplitValue)

SET NOCOUNT OFF

-- You can select or join with the values in @SplitStringTable at this point.