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)

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

我在网上寻找解决方案，下面的工作对我来说。 Ref。

然后像这样调用函数:

SELECT * FROM dbo.split('ram shyam hari gopal',' ')

SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO

CREATE FUNCTION [dbo].[Split](@String VARCHAR(8000), @Delimiter CHAR(1))       
RETURNS @temptable TABLE (items VARCHAR(8000))       
AS       
BEGIN       
    DECLARE @idx INT       
    DECLARE @slice VARCHAR(8000)        
    SELECT @idx = 1       
    IF len(@String)<1 OR @String IS NULL  RETURN       
    WHILE @idx!= 0       
    BEGIN       
        SET @idx = charindex(@Delimiter,@String)       
        IF @idx!=0       
            SET @slice = LEFT(@String,@idx - 1)       
        ELSE       
            SET @slice = @String       
        IF(len(@slice)>0)  
            INSERT INTO @temptable(Items) VALUES(@slice)       
        SET @String = RIGHT(@String,len(@String) - @idx)       
        IF len(@String) = 0 break       
    END   
    RETURN       
END

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

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)

解析姓和名的简单解决方案

DECLARE @Name varchar(10) = 'John Smith'

-- Get First Name
SELECT SUBSTRING(@Name, 0, (SELECT CHARINDEX(' ', @Name)))

-- Get Last Name
SELECT SUBSTRING(@Name, (SELECT CHARINDEX(' ', @Name)) + 1, LEN(@Name))

在我的例子中(在许多其他人中似乎也是如此……)，我有一个由一个空格隔开的姓和名列表。可以直接在选择语句中使用它来解析姓和名。

-- i.e. Get First and Last Name from a table of Full Names
SELECT SUBSTRING(FullName, 0, (SELECT CHARINDEX(' ', FullName))) as FirstName,
SUBSTRING(FullName, (SELECT CHARINDEX(' ', FullName)) + 1, LEN(FullName)) as LastName,
From FullNameTable