T-SQL基于分隔符拆分字符串

Question

我有一些数据，我想根据可能存在或不存在的分隔符进行拆分。

示例数据：

John/Smith
Jane/Doe
Steve
Bob/Johnson

我使用以下代码将此数据拆分为名字和姓氏：

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

结果我想：

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

只要所有行都具有预期的分隔符，此代码就可以正常工作，但是当行没有时出错：

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

如何重写这个才能正常工作？

Answer 1

可能这会对你有所帮助。

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

Answer 2

对于那些寻找SQL Server 2016+的答案的人。使用内置的STRING_SPLIT函数

例如：

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

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

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

Answer 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

Answer 4

尝试使用分隔符过滤掉包含字符串的行，并仅处理以下内容：

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

要么

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

Answer 5

我只想提供一种替代方法来分割带有多个分隔符的字符串，以防您在2016年使用SQL Server版本。

一般的想法是拆分字符串中的所有字符，确定分隔符的位置，然后获得相对于分隔符的子字符串。这是一个示例：

-- 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)

Answer 6

当只有一个分隔符时，上面的示例工作正常，但对于多个分隔符，它不能很好地扩展。请注意，这仅适用于SQL Server 2016及更高版本。

/*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

Answer 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