T-SQL split string based on delimiter

问题:

I have some data that I would like to split based on a delimiter that may or may not exist.

Example data:

John/Smith
Jane/Doe
Steve
Bob/Johnson

I am using the following code to split this data into First and Last names:

SELECT SUBSTRING(myColumn, 1, CHARINDEX('/', myColumn)-1) AS FirstName,
       SUBSTRING(myColumn, CHARINDEX('/', myColumn) + 1, 1000) AS LastName
FROM   MyTable

The results I would like:

FirstName---LastName
John--------Smith
Jane--------Doe
Steve-------NULL
Bob---------Johnson

This code works just fine as long as all the rows have the anticipated delimiter, but errors out when a row does not:

"Invalid length parameter passed to the LEFT or SUBSTRING function."

How can a re-write this to work properly?

回答1:

May be this will help you.

SELECT SUBSTRING(myColumn, 1, CASE CHARINDEX('/', myColumn)
            WHEN 0
                THEN LEN(myColumn)
            ELSE CHARINDEX('/', myColumn) - 1
            END) AS FirstName
    ,SUBSTRING(myColumn, CASE CHARINDEX('/', myColumn)
            WHEN 0
                THEN LEN(myColumn) + 1
            ELSE CHARINDEX('/', myColumn) + 1
            END, 1000) AS LastName
FROM MyTable

回答2:

For those looking for answers for SQL Server 2016+. Use the built-in STRING_SPLIT function

Eg:

DECLARE @tags NVARCHAR(400) = 'clothing,road,,touring,bike'  

SELECT value  
FROM STRING_SPLIT(@tags, ',')  
WHERE RTRIM(value) <> '';  

Reference: https://msdn.microsoft.com/en-nz/library/mt684588.aspx

回答3:

SELECT CASE 
        WHEN CHARINDEX('/', myColumn, 0) = 0
            THEN myColumn
        ELSE LEFT(myColumn, CHARINDEX('/', myColumn, 0)-1)
        END AS FirstName
    ,CASE 
        WHEN CHARINDEX('/', myColumn, 0) = 0
            THEN ''
        ELSE RIGHT(myColumn, CHARINDEX('/', REVERSE(myColumn), 0)-1)
        END AS LastName
FROM MyTable

回答4:

Try filtering out the rows that contain strings with the delimiter and work on those only like:

SELECT SUBSTRING(myColumn, 1, CHARINDEX('/', myColumn)-1) AS FirstName,
       SUBSTRING(myColumn, CHARINDEX('/', myColumn) + 1, 1000) AS LastName
FROM   MyTable
WHERE CHARINDEX('/', myColumn) > 0

Or

SELECT SUBSTRING(myColumn, 1, CHARINDEX('/', myColumn)-1) AS FirstName,
       SUBSTRING(myColumn, CHARINDEX('/', myColumn) + 1, 1000) AS LastName
FROM   MyTable
WHERE myColumn LIKE '%/%'

回答5:

I just wanted to give an alternative way to split a string with multiple delimiters, in case you are using a SQL Server version under 2016.

The general idea is to split out all of the characters in the string, determine the position of the delimiters, then obtain substrings relative to the delimiters. Here is a sample:

-- Sample data
DECLARE @testTable TABLE (
    TestString      VARCHAR(50)
)
INSERT INTO @testTable VALUES 
    ('Teststring,1,2,3')
    ,('Test')

DECLARE @delimiter VARCHAR(1) = ','

-- Generate numbers with which we can enumerate
;WITH Numbers AS (
    SELECT 1 AS N

    UNION ALL 

    SELECT N + 1
    FROM Numbers 
    WHERE N < 255
), 
-- Enumerate letters in the string and select only the delimiters
Letters AS (
    SELECT  n.N
            , SUBSTRING(t.TestString, n.N, 1) AS Letter
            , t.TestString 
            , ROW_NUMBER() OVER (   PARTITION BY t.TestString
                                    ORDER BY n.N
                                ) AS Delimiter_Number 
    FROM Numbers n
        INNER JOIN @testTable t
            ON n <= LEN(t.TestString)
    WHERE SUBSTRING(t.TestString, n, 1) = @delimiter 

    UNION 

    -- Include 0th position to "delimit" the start of the string
    SELECT  0
            , NULL
            , t.TestString 
            , 0
    FROM @testTable t 
)
-- Obtain substrings based on delimiter positions
SELECT  t.TestString 
        , ds.Delimiter_Number + 1 AS Position
        , SUBSTRING(t.TestString, ds.N + 1, ISNULL(de.N, LEN(t.TestString) + 1) - ds.N - 1) AS Delimited_Substring 
FROM @testTable t
    LEFT JOIN Letters ds
        ON t.TestString = ds.TestString 
    LEFT JOIN Letters de
        ON t.TestString = de.TestString 
        AND ds.Delimiter_Number + 1 = de.Delimiter_Number  
OPTION (MAXRECURSION 0)

回答6:

The examples above work fine when there is only one delimiter, but it doesn't scale well for multiple delimiters. Note that this will only work for SQL Server 2016 and above.

/*Some Sample Data*/
DECLARE @mytable TABLE ([id] VARCHAR(10), [name] VARCHAR(1000));
INSERT INTO @mytable
VALUES ('1','John/Smith'),('2','Jane/Doe'), ('3','Steve'), ('4','Bob/Johnson')


/*Split based on delimeter*/
SELECT P.id, [1] 'FirstName', [2] 'LastName', [3] 'Col3', [4] 'Col4'
FROM(
    SELECT A.id, X1.VALUE, ROW_NUMBER() OVER (PARTITION BY A.id ORDER BY A.id) RN
    FROM @mytable A
    CROSS APPLY STRING_SPLIT(A.name, '/') X1
    ) A
PIVOT (MAX(A.[VALUE]) FOR A.RN IN ([1],[2],[3],[4],[5])) P

回答7:

ALTER FUNCTION [dbo].[split_string](
          @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