我有一些数据要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
我想要的结果：
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，错误就会出现：
"Invalid length parameter passed to the LEFT or SUBSTRING function."
如何重写这个才能正常工作？
答案 0 :(得分：57)
可能会对你有帮助。
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
答案 1 :(得分：7)
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) <> '';
答案 2 :(得分：6)
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
答案 3 :(得分：4)
尝试使用分隔符过滤掉包含字符串的行，并对其进行处理，例如：
SELECT SUBSTRING(myColumn, 1, CHARINDEX('/', myColumn)-1) AS FirstName,
SUBSTRING(myColumn, CHARINDEX('/', myColumn) + 1, 1000) AS LastName
FROM MyTable
WHERE CHARINDEX('/', myColumn) > 0
或者
SELECT SUBSTRING(myColumn, 1, CHARINDEX('/', myColumn)-1) AS FirstName,
SUBSTRING(myColumn, CHARINDEX('/', myColumn) + 1, 1000) AS LastName
FROM MyTable
WHERE myColumn LIKE '%/%'
答案 4 :(得分：1)
如果您使用的是2016年以下的SQL Server版本，我只是想提供一种使用多个定界符分割字符串的替代方法。
通常的想法是将字符串中的所有字符分开，确定分隔符的位置，然后获取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)
答案 5 :(得分：0)
上面的示例在只有一个定界符的情况下可以很好地工作，但是对于多个定界符却无法很好地扩展。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
答案 6 :(得分：0)
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
答案 7 :(得分：0)
这些都帮助我实现了这一目标。我仍然在2012年，但是现在有了一些快速操作，即使字符串具有不同数量的定界符，我也可以分割字符串，并从该字符串中获取第n个子字符串。它也很快。我知道这篇文章已经过时了，但是我花了很长时间才找到一些东西，因此希望对其他人有所帮助。
CREATE FUNCTION [dbo].[SplitsByIndex]
(@separator VARCHAR(20) = ' ',
@string VARCHAR(MAX),
@position INT
)
RETURNS VARCHAR(MAX)
AS
BEGIN
DECLARE @results TABLE
(id INT IDENTITY(1, 1),
chrs VARCHAR(8000)
);
DECLARE @outResult VARCHAR(8000);
WITH X(N)
AS (SELECT 'Table1'
FROM(VALUES(0), (0), (0), (0), (0), (0), (0), (0), (0), (0), (0), (0), (0), (0), (0), (0)) T(C)),
Y(N)
AS (SELECT 'Table2'
FROM X A1,
X A2,
X A3,
X A4,
X A5,
X A6,
X A7,
X A8), -- Up to 16^8 = 4 billion
T(N)
AS (SELECT TOP (ISNULL(LEN(@string), 0)) ROW_NUMBER() OVER(
ORDER BY
(
SELECT NULL
)) - 1 N
FROM Y),
Delim(Pos)
AS (SELECT t.N
FROM T
WHERE(SUBSTRING(@string, t.N, LEN(@separator + 'x') - 1) LIKE @separator
OR t.N = 0)),
Separated(value)
AS (SELECT SUBSTRING(@string, d.Pos + LEN(@separator + 'x') - 1, LEAD(d.Pos, 1, 2147483647) OVER(
ORDER BY
(
SELECT NULL
))-d.Pos - LEN(@separator))
FROM Delim d
WHERE @string IS NOT NULL)
INSERT INTO @results(chrs)
SELECT s.value
FROM Separated s
WHERE s.value <> @separator;
SELECT @outResult =
(
SELECT chrs
FROM @results
WHERE id = @position
);
RETURN @outResult;
END;
可以这样使用：
SELECT [dbo].[SplitsByIndex](' ',fieldname,2)
from tablename