使用SQL Server,我如何分割一个字符串,以便我可以访问项目x?
拿一根“你好,约翰·史密斯”的绳子。我如何通过空格分割字符串并访问索引1应该返回“约翰”的项目?
使用SQL Server,我如何分割一个字符串,以便我可以访问项目x?
拿一根“你好,约翰·史密斯”的绳子。我如何通过空格分割字符串并访问索引1应该返回“约翰”的项目?
当前回答
Alter Function dbo.fn_Split
(
@Expression nvarchar(max),
@Delimiter nvarchar(20) = ',',
@Qualifier char(1) = Null
)
RETURNS @Results TABLE (id int IDENTITY(1,1), value nvarchar(max))
AS
BEGIN
/* USAGE
Select * From dbo.fn_Split('apple pear grape banana orange honeydew cantalope 3 2 1 4', ' ', Null)
Select * From dbo.fn_Split('1,abc,"Doe, John",4', ',', '"')
Select * From dbo.fn_Split('Hello 0,"&""&&&&', ',', '"')
*/
-- Declare Variables
DECLARE
@X xml,
@Temp nvarchar(max),
@Temp2 nvarchar(max),
@Start int,
@End int
-- HTML Encode @Expression
Select @Expression = (Select @Expression For XML Path(''))
-- Find all occurences of @Delimiter within @Qualifier and replace with |||***|||
While PATINDEX('%' + @Qualifier + '%', @Expression) > 0 AND Len(IsNull(@Qualifier, '')) > 0
BEGIN
Select
-- Starting character position of @Qualifier
@Start = PATINDEX('%' + @Qualifier + '%', @Expression),
-- @Expression starting at the @Start position
@Temp = SubString(@Expression, @Start + 1, LEN(@Expression)-@Start+1),
-- Next position of @Qualifier within @Expression
@End = PATINDEX('%' + @Qualifier + '%', @Temp) - 1,
-- The part of Expression found between the @Qualifiers
@Temp2 = Case When @End < 0 Then @Temp Else Left(@Temp, @End) End,
-- New @Expression
@Expression = REPLACE(@Expression,
@Qualifier + @Temp2 + Case When @End < 0 Then '' Else @Qualifier End,
Replace(@Temp2, @Delimiter, '|||***|||')
)
END
-- Replace all occurences of @Delimiter within @Expression with '</fn_Split><fn_Split>'
-- And convert it to XML so we can select from it
SET
@X = Cast('<fn_Split>' +
Replace(@Expression, @Delimiter, '</fn_Split><fn_Split>') +
'</fn_Split>' as xml)
-- Insert into our returnable table replacing '|||***|||' back to @Delimiter
INSERT @Results
SELECT
"Value" = LTRIM(RTrim(Replace(C.value('.', 'nvarchar(max)'), '|||***|||', @Delimiter)))
FROM
@X.nodes('fn_Split') as X(C)
-- Return our temp table
RETURN
END
其他回答
如果数据库的兼容性级别为130或更高,则可以使用STRING_SPLIT函数和OFFSET FETCH子句按索引获取特定的项。
要获得索引N(从零开始)的项,可以使用以下代码
SELECT value
FROM STRING_SPLIT('Hello John Smith',' ')
ORDER BY (SELECT NULL)
OFFSET N ROWS
FETCH NEXT 1 ROWS ONLY
要检查数据库的兼容性级别,执行以下代码:
SELECT compatibility_level
FROM sys.databases WHERE name = 'YourDBName';
一个简单的优化算法:
ALTER FUNCTION [dbo].[Split]( @Text NVARCHAR(200),@Splitor CHAR(1) )
RETURNS @Result TABLE ( value NVARCHAR(50))
AS
BEGIN
DECLARE @PathInd INT
Set @Text+=@Splitor
WHILE LEN(@Text) > 0
BEGIN
SET @PathInd=PATINDEX('%'+@Splitor+'%',@Text)
INSERT INTO @Result VALUES(SUBSTRING(@Text, 0, @PathInd))
SET @Text= SUBSTRING(@Text, @PathInd+1, LEN(@Text))
END
RETURN
END
declare @strng varchar(max)='hello john smith'
select (
substring(
@strng,
charindex(' ', @strng) + 1,
(
(charindex(' ', @strng, charindex(' ', @strng) + 1))
- charindex(' ',@strng)
)
))
我使用弗雷德里克的答案,但这在SQL Server 2005中不起作用
我修改了它,我使用select with union all,它可以工作
DECLARE @str varchar(max)
SET @str = 'Hello John Smith how are you'
DECLARE @separator varchar(max)
SET @separator = ' '
DECLARE @Splited table(id int IDENTITY(1,1), item varchar(max))
SET @str = REPLACE(@str, @separator, ''' UNION ALL SELECT ''')
SET @str = ' SELECT ''' + @str + ''' '
INSERT INTO @Splited
EXEC(@str)
SELECT * FROM @Splited
结果集是:
id item
1 Hello
2 John
3 Smith
4 how
5 are
6 you
几乎所有其他答案都是替换正在分割的字符串,这浪费了CPU周期并执行不必要的内存分配。
我在这里介绍了一种更好的进行字符串拆分的方法:http://www.digitalruby.com/split-string-sql-server/
代码如下:
SET NOCOUNT ON
-- You will want to change nvarchar(MAX) to nvarchar(50), varchar(50) or whatever matches exactly with the string column you will be searching against
DECLARE @SplitStringTable TABLE (Value nvarchar(MAX) NOT NULL)
DECLARE @StringToSplit nvarchar(MAX) = 'your|string|to|split|here'
DECLARE @SplitEndPos int
DECLARE @SplitValue nvarchar(MAX)
DECLARE @SplitDelim nvarchar(1) = '|'
DECLARE @SplitStartPos int = 1
SET @SplitEndPos = CHARINDEX(@SplitDelim, @StringToSplit, @SplitStartPos)
WHILE @SplitEndPos > 0
BEGIN
SET @SplitValue = SUBSTRING(@StringToSplit, @SplitStartPos, (@SplitEndPos - @SplitStartPos))
INSERT @SplitStringTable (Value) VALUES (@SplitValue)
SET @SplitStartPos = @SplitEndPos + 1
SET @SplitEndPos = CHARINDEX(@SplitDelim, @StringToSplit, @SplitStartPos)
END
SET @SplitValue = SUBSTRING(@StringToSplit, @SplitStartPos, 2147483647)
INSERT @SplitStringTable (Value) VALUES(@SplitValue)
SET NOCOUNT OFF
-- You can select or join with the values in @SplitStringTable at this point.