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
基于纯集的解决方案,使用TVF和递归CTE。您可以将此函数JOIN和APPLY到任何数据集。
create function [dbo].[SplitStringToResultSet] (@value varchar(max), @separator char(1))
returns table
as return
with r as (
select value, cast(null as varchar(max)) [x], -1 [no] from (select rtrim(cast(@value as varchar(max))) [value]) as j
union all
select right(value, len(value)-case charindex(@separator, value) when 0 then len(value) else charindex(@separator, value) end) [value]
, left(r.[value], case charindex(@separator, r.value) when 0 then len(r.value) else abs(charindex(@separator, r.[value])-1) end ) [x]
, [no] + 1 [no]
from r where value > '')
select ltrim(x) [value], [no] [index] from r where x is not null;
go
用法:
select *
from [dbo].[SplitStringToResultSet]('Hello John Smith', ' ')
where [index] = 1;
结果:
value index
-------------
John 1
其他回答
试试这个:
CREATE function [SplitWordList]
(
@list varchar(8000)
)
returns @t table
(
Word varchar(50) not null,
Position int identity(1,1) not null
)
as begin
declare
@pos int,
@lpos int,
@item varchar(100),
@ignore varchar(100),
@dl int,
@a1 int,
@a2 int,
@z1 int,
@z2 int,
@n1 int,
@n2 int,
@c varchar(1),
@a smallint
select
@a1 = ascii('a'),
@a2 = ascii('A'),
@z1 = ascii('z'),
@z2 = ascii('Z'),
@n1 = ascii('0'),
@n2 = ascii('9')
set @ignore = '''"'
set @pos = 1
set @dl = datalength(@list)
set @lpos = 1
set @item = ''
while (@pos <= @dl) begin
set @c = substring(@list, @pos, 1)
if (@ignore not like '%' + @c + '%') begin
set @a = ascii(@c)
if ((@a >= @a1) and (@a <= @z1))
or ((@a >= @a2) and (@a <= @z2))
or ((@a >= @n1) and (@a <= @n2))
begin
set @item = @item + @c
end else if (@item > '') begin
insert into @t values (@item)
set @item = ''
end
end
set @pos = @pos + 1
end
if (@item > '') begin
insert into @t values (@item)
end
return
end
像这样测试它:
select * from SplitWordList('Hello John Smith')
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','**')
这个模式工作得很好,可以进行推广
Convert(xml,'<n>'+Replace(FIELD,'.','</n><n>')+'</n>').value('(/n[INDEX])','TYPE')
^^^^^ ^^^^^ ^^^^
注意字段,索引和类型。
让一些表具有类似的标识符
sys.message.1234.warning.A45
sys.message.1235.error.O98
....
然后,你就可以写作了
SELECT Source = q.value('(/n[1])', 'varchar(10)'),
RecordType = q.value('(/n[2])', 'varchar(20)'),
RecordNumber = q.value('(/n[3])', 'int'),
Status = q.value('(/n[4])', 'varchar(5)')
FROM (
SELECT q = Convert(xml,'<n>'+Replace(fieldName,'.','</n><n>')+'</n>')
FROM some_TABLE
) Q
拆铸所有零件。
我知道已经晚了,但我最近有了这个需求,并提出了下面的代码。我没有选择使用用户定义的函数。希望这能有所帮助。
SELECT
SUBSTRING(
SUBSTRING('Hello John Smith' ,0,CHARINDEX(' ','Hello John Smith',CHARINDEX(' ','Hello John Smith')+1)
),CHARINDEX(' ','Hello John Smith'),LEN('Hello John Smith')
)
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