How to Split Strings in SQL Server

How to split a comma-separated value to columns

CREATE FUNCTION [dbo].[fn_split_string_to_column] (
    @string NVARCHAR(MAX),
    @delimiter CHAR(1)
    )
RETURNS @out_put TABLE (
    [column_id] INT IDENTITY(1, 1) NOT NULL,
    [value] NVARCHAR(MAX)
    )
AS
BEGIN
    DECLARE @value NVARCHAR(MAX),
        @pos INT = 0,
        @len INT = 0

    SET @string = CASE 
            WHEN RIGHT(@string, 1) != @delimiter
                THEN @string + @delimiter
            ELSE @string
            END

    WHILE CHARINDEX(@delimiter, @string, @pos + 1) > 0
    BEGIN
        SET @len = CHARINDEX(@delimiter, @string, @pos + 1) - @pos
        SET @value = SUBSTRING(@string, @pos, @len)

        INSERT INTO @out_put ([value])
        SELECT LTRIM(RTRIM(@value)) AS [column]

        SET @pos = CHARINDEX(@delimiter, @string, @pos + @len) + 1
    END

    RETURN
END

T-SQL split string

I've used this SQL before which may work for you:-

CREATE FUNCTION dbo.splitstring ( @stringToSplit VARCHAR(MAX) )
RETURNS
 @returnList TABLE ([Name] [nvarchar] (500))
AS
BEGIN

 DECLARE @name NVARCHAR(255)
 DECLARE @pos INT

 WHILE CHARINDEX(',', @stringToSplit) > 0
 BEGIN
  SELECT @pos  = CHARINDEX(',', @stringToSplit)  
  SELECT @name = SUBSTRING(@stringToSplit, 1, @pos-1)

  INSERT INTO @returnList 
  SELECT @name

  SELECT @stringToSplit = SUBSTRING(@stringToSplit, @pos+1, LEN(@stringToSplit)-@pos)
 END

 INSERT INTO @returnList
 SELECT @stringToSplit

 RETURN
END

and to use it:-

SELECT * FROM dbo.splitstring('91,12,65,78,56,789')

How do I split a delimited string so I can access individual items?

You may find the solution in SQL User Defined Function to Parse a Delimited String helpful (from The Code Project).

You can use this simple logic:

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

STRING_SPLIT in SQL Server 2012

Other approach is to use XML Method with CROSS APPLY to split your Comma Separated Data :

SELECT Split.a.value('.', 'NVARCHAR(MAX)') DATA
FROM
(
    SELECT CAST('<X>'+REPLACE(@ID, ',', '</X><X>')+'</X>' AS XML) AS String
) AS A
CROSS APPLY String.nodes('/X') AS Split(a);

Result :

DATA
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

Example :

DECLARE @ID NVARCHAR(300)= '1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20';
DECLARE @Marks NVARCHAR(300)= '0,1,2,5,8,9,4,6,7,3,5,2,7,1,9,4,0,2,5,0';
DECLARE @StudentsMark TABLE
(id    NVARCHAR(300),
 marks NVARCHAR(300)
); 
--insert into @StudentsMark 
;WITH CTE
     AS (
     SELECT Split.a.value('.', 'NVARCHAR(MAX)') id,
            ROW_NUMBER() OVER(ORDER BY
                             (
                                 SELECT NULL
                             )) RN
     FROM
     (
         SELECT CAST('<X>'+REPLACE(@ID, ',', '</X><X>')+'</X>' AS XML) AS String
     ) AS A
     CROSS APPLY String.nodes('/X') AS Split(a)),
     CTE1
     AS (
     SELECT Split.a.value('.', 'NVARCHAR(MAX)') marks,
            ROW_NUMBER() OVER(ORDER BY
                             (
                                 SELECT NULL
                             )) RN
     FROM
     (
         SELECT CAST('<X>'+REPLACE(@Marks, ',', '</X><X>')+'</X>' AS XML) AS String
     ) AS A
     CROSS APPLY String.nodes('/X') AS Split(a))
     INSERT INTO @StudentsMark
            SELECT C.id,
                   C1.marks
            FROM CTE C
                 LEFT JOIN CTE1 C1 ON C1.RN = C.RN;
SELECT *
FROM @StudentsMark;

How do I split a string into 2 parts based on a delimiter?

Here is a simple function that could be used to split string in DB:

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

it will return all the values by splitting with space.

Split String into given length but do not split word

Assuming you have a table of addresses, I'd use a recursive CTE.

On each iteration, find the last possible space to break on, then start the next iteration for the character after the space.

• take 31 characters
• reverse them
• find the position of the first space

Extra care to be taken for corner cases:

• The remaining string to be searched is less than 30 characters
• The current string being searched has no space in the first 31 characters

Using the following test data...

CREATE TABLE test (
  address   VARCHAR(MAX)
);

INSERT INTO
  test
VALUES
  ('216 Apartment123 AreaArea SampleWord1 Word2 MiddleTown Upper1Location Another5 NewYork'),
  ('216 Apartment123 AreaArea SampleWord1 Word2 MiddleTownxx Upper1LocationUpper1LocationUpper1Location Another5 NewYork'),
  ('216 Apartment123 AreaArea SampleWord1 Word2 MiddleTownxx Upper1LocationUpper1LocationUpper1Location Another5 NewYork x')

;

Using the following CTE...

DECLARE @chars BIGINT = 30;

WITH
  parts AS
(
  SELECT
    address,
    LEN(address)        AS length,
    CAST(0 AS BIGINT)   AS last_space,
    CAST(1 AS BIGINT)   AS next,
    address             AS fragment
  FROM
    test

  UNION ALL

  SELECT
    parts.address,
    parts.length,
    last_space.pos,
    parts.next + COALESCE(last_space.pos, @chars),
    SUBSTRING(parts.address, parts.next, COALESCE(last_space.pos - 1, @chars))
  FROM
    parts
  CROSS APPLY
  (
    SELECT
      @chars + 2
      -
      NULLIF(
        CHARINDEX(
          ' ',
          REVERSE(
            SUBSTRING(
              parts.address + ' ',
              parts.next,
              @chars + 1
            )
          )
        )
        , 0
      )
  )
    last_space(pos)
  WHERE
    parts.next <= parts.length
)

SELECT
  *, len(fragment) AS chars
FROM
  parts
WHERE
  next > 1
ORDER BY
  address,
  next

https://dbfiddle.uk/?rdbms=sqlserver_2019&fiddle=acd11f2bc73e5036bd82498ecf14b08f

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