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
CREATE FUNCTION [dbo].[fnSplitString]
(
@string NVARCHAR(MAX),
@delimiter CHAR(1)
)
RETURNS @output TABLE(splitdata NVARCHAR(MAX)
)
BEGIN
DECLARE @start INT, @end INT
SELECT @start = 1, @end = CHARINDEX(@delimiter, @string)
WHILE @start < LEN(@string) + 1 BEGIN
IF @end = 0
SET @end = LEN(@string) + 1
INSERT INTO @output (splitdata)
VALUES(SUBSTRING(@string, @start, @end - @start))
SET @start = @end + 1
SET @end = CHARINDEX(@delimiter, @string, @start)
END
RETURN
END
并使用它
select *from dbo.fnSplitString('Querying SQL Server','')
其他回答
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
在这里我发布了一个简单的解决方法
CREATE FUNCTION [dbo].[split](
@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
像这样执行函数
select * from dbo.split('Hello John Smith',' ')
我在网上寻找解决方案,下面的工作对我来说。 Ref。
然后像这样调用函数:
SELECT * FROM dbo.split('ram shyam hari gopal',' ')
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE FUNCTION [dbo].[Split](@String VARCHAR(8000), @Delimiter CHAR(1))
RETURNS @temptable TABLE (items VARCHAR(8000))
AS
BEGIN
DECLARE @idx INT
DECLARE @slice VARCHAR(8000)
SELECT @idx = 1
IF len(@String)<1 OR @String IS NULL RETURN
WHILE @idx!= 0
BEGIN
SET @idx = charindex(@Delimiter,@String)
IF @idx!=0
SET @slice = LEFT(@String,@idx - 1)
ELSE
SET @slice = @String
IF(len(@slice)>0)
INSERT INTO @temptable(Items) VALUES(@slice)
SET @String = RIGHT(@String,len(@String) - @idx)
IF len(@String) = 0 break
END
RETURN
END
我一直在使用vzczc的答案使用递归cte的一段时间,但一直想更新它来处理可变长度分隔符,也处理字符串与前驱和滞后“分隔符”,如当你有一个csv文件的记录,如:
“鲍勃”,“史密斯”桑尼维尔”,“CA”
或者当你处理如下所示的六部分fqn时。我广泛地使用这些来记录subject_fqn的审计,错误处理等,parsename只处理四个部分:
[netbios_name].[machine_name].[instance].[database].[schema].[table].[column]
这是我的更新版本,感谢vzczc的原始帖子!
select * from [utility].[split_string](N'"this"."string"."gets"."split"."and"."removes"."leading"."and"."trailing"."quotes"', N'"."', N'"', N'"');
select * from [utility].[split_string](N'"this"."string"."gets"."split"."but"."leaves"."leading"."and"."trailing"."quotes"', N'"."', null, null);
select * from [utility].[split_string](N'[netbios_name].[machine_name].[instance].[database].[schema].[table].[column]', N'].[', N'[', N']');
create function [utility].[split_string] (
@input [nvarchar](max)
, @separator [sysname]
, @lead [sysname]
, @lag [sysname])
returns @node_list table (
[index] [int]
, [node] [nvarchar](max))
begin
declare @separator_length [int]= len(@separator)
, @lead_length [int] = isnull(len(@lead), 0)
, @lag_length [int] = isnull(len(@lag), 0);
--
set @input = right(@input, len(@input) - @lead_length);
set @input = left(@input, len(@input) - @lag_length);
--
with [splitter]([index], [starting_position], [start_location])
as (select cast(@separator_length as [bigint])
, cast(1 as [bigint])
, charindex(@separator, @input)
union all
select [index] + 1
, [start_location] + @separator_length
, charindex(@separator, @input, [start_location] + @separator_length)
from [splitter]
where [start_location] > 0)
--
insert into @node_list
([index],[node])
select [index] - @separator_length as [index]
, substring(@input, [starting_position], case
when [start_location] > 0
then
[start_location] - [starting_position]
else
len(@input)
end) as [node]
from [splitter];
--
return;
end;
go
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)