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
首先,创建一个函数(使用CTE,公共表表达式不再需要临时表)
create function dbo.SplitString
(
@str nvarchar(4000),
@separator char(1)
)
returns table
AS
return (
with tokens(p, a, b) AS (
select
1,
1,
charindex(@separator, @str)
union all
select
p + 1,
b + 1,
charindex(@separator, @str, b + 1)
from tokens
where b > 0
)
select
p-1 zeroBasedOccurance,
substring(
@str,
a,
case when b > 0 then b-a ELSE 4000 end)
AS s
from tokens
)
GO
然后,像这样使用它作为任何表(或修改它以适应现有存储的proc)。
select s
from dbo.SplitString('Hello John Smith', ' ')
where zeroBasedOccurance=1
更新
以前的版本将失败的输入字符串长度超过4000个字符。这个版本考虑到了以下限制:
create function dbo.SplitString
(
@str nvarchar(max),
@separator char(1)
)
returns table
AS
return (
with tokens(p, a, b) AS (
select
cast(1 as bigint),
cast(1 as bigint),
charindex(@separator, @str)
union all
select
p + 1,
b + 1,
charindex(@separator, @str, b + 1)
from tokens
where b > 0
)
select
p-1 ItemIndex,
substring(
@str,
a,
case when b > 0 then b-a ELSE LEN(@str) end)
AS s
from tokens
);
GO
用法不变。
其他回答
I realize this is a really old question, but starting with SQL Server 2016 there are functions for parsing JSON data that can be used to specifically address the OP's question--and without splitting strings or resorting to a user-defined function. To access an item at a particular index of a delimited string, use the JSON_VALUE function. Properly formatted JSON data is required, however: strings must be enclosed in double quotes " and the delimiter must be a comma ,, with the entire string enclosed in square brackets [].
DECLARE @SampleString NVARCHAR(MAX) = '"Hello John Smith"';
--Format as JSON data.
SET @SampleString = '[' + REPLACE(@SampleString, ' ', '","') + ']';
SELECT
JSON_VALUE(@SampleString, '$[0]') AS Element1Value,
JSON_VALUE(@SampleString, '$[1]') AS Element2Value,
JSON_VALUE(@SampleString, '$[2]') AS Element3Value;
输出
Element1Value Element2Value Element3Value
--------------------- ------------------- ------------------------------
Hello John Smith
(1 row affected)
这里有一个UDF可以做到这一点。它将返回一个带分隔符的值的表,我还没有尝试所有的场景,但您的示例工作良好。
CREATE FUNCTION SplitString
(
-- Add the parameters for the function here
@myString varchar(500),
@deliminator varchar(10)
)
RETURNS
@ReturnTable TABLE
(
-- Add the column definitions for the TABLE variable here
[id] [int] IDENTITY(1,1) NOT NULL,
[part] [varchar](50) NULL
)
AS
BEGIN
Declare @iSpaces int
Declare @part varchar(50)
--initialize spaces
Select @iSpaces = charindex(@deliminator,@myString,0)
While @iSpaces > 0
Begin
Select @part = substring(@myString,0,charindex(@deliminator,@myString,0))
Insert Into @ReturnTable(part)
Select @part
Select @myString = substring(@mystring,charindex(@deliminator,@myString,0)+ len(@deliminator),len(@myString) - charindex(' ',@myString,0))
Select @iSpaces = charindex(@deliminator,@myString,0)
end
If len(@myString) > 0
Insert Into @ReturnTable
Select @myString
RETURN
END
GO
你可以这样称呼它:
Select * From SplitString('Hello John Smith',' ')
编辑:使用len>1处理分隔符的更新解决方案如下:
select * From SplitString('Hello**John**Smith','**')
如果任何人只想获得分离文本的一部分,可以使用这个
select * from from splitstringsep ('Word1 word2 word3',' ')
CREATE function [dbo].[SplitStringSep]
(
@str nvarchar(4000),
@separator char(1)
)
returns table
AS
return (
with tokens(p, a, b) AS (
select
1,
1,
charindex(@separator, @str)
union all
select
p + 1,
b + 1,
charindex(@separator, @str, b + 1)
from tokens
where b > 0
)
select
p-1 zeroBasedOccurance,
substring(
@str,
a,
case when b > 0 then b-a ELSE 4000 end)
AS s
from tokens
)
CREATE TABLE test(
id int,
adress varchar(100)
);
INSERT INTO test VALUES(1, 'Ludovic Aubert, 42 rue de la Victoire, 75009, Paris, France'),(2, 'Jose Garcia, 1 Calle de la Victoria, 56500 Barcelona, Espana');
SELECT id, value, COUNT(*) OVER (PARTITION BY id) AS n, ROW_NUMBER() OVER (PARTITION BY id ORDER BY (SELECT NULL)) AS rn, adress
FROM test
CROSS APPLY STRING_SPLIT(adress, ',')
如果你查看下面关于使用SQL分割字符串的SQL教程,你会发现许多函数可以用于在SQL Server上分割给定的字符串
例如,SplitAndReturnNth UDF函数可用于使用分隔符分割文本,并将第n块作为函数的输出返回
select dbo.SplitAndReturnNth('Hello John Smith',' ',2)