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
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
)
其他回答
可以利用Number表进行字符串解析。
创建一个物理数字表:
create table dbo.Numbers (N int primary key);
insert into dbo.Numbers
select top 1000 row_number() over(order by number) from master..spt_values
go
创建具有1000000行的测试表
create table #yak (i int identity(1,1) primary key, array varchar(50))
insert into #yak(array)
select 'a,b,c' from dbo.Numbers n cross join dbo.Numbers nn
go
创建函数
create function [dbo].[ufn_ParseArray]
( @Input nvarchar(4000),
@Delimiter char(1) = ',',
@BaseIdent int
)
returns table as
return
( select row_number() over (order by n asc) + (@BaseIdent - 1) [i],
substring(@Input, n, charindex(@Delimiter, @Input + @Delimiter, n) - n) s
from dbo.Numbers
where n <= convert(int, len(@Input)) and
substring(@Delimiter + @Input, n, 1) = @Delimiter
)
go
使用情况(在我的笔记本电脑上40秒内输出3mil行)
select *
from #yak
cross apply dbo.ufn_ParseArray(array, ',', 1)
清理
drop table dbo.Numbers;
drop function [dbo].[ufn_ParseArray]
这里的性能并不惊人,但在100万行表上调用函数并不是最好的主意。如果将字符串拆分到多行,我会避免使用该函数。
Aaron Bertrand的回答很好,但也有缺陷。它不能准确地将空格作为分隔符处理(就像最初问题中的示例一样),因为长度函数将空格带在后面。
下面是他的代码,稍微调整了一下,允许使用空格分隔符:
CREATE FUNCTION [dbo].[SplitString]
(
@List NVARCHAR(MAX),
@Delim VARCHAR(255)
)
RETURNS TABLE
AS
RETURN ( SELECT [Value] FROM
(
SELECT
[Value] = LTRIM(RTRIM(SUBSTRING(@List, [Number],
CHARINDEX(@Delim, @List + @Delim, [Number]) - [Number])))
FROM (SELECT Number = ROW_NUMBER() OVER (ORDER BY name)
FROM sys.all_objects) AS x
WHERE Number <= LEN(@List)
AND SUBSTRING(@Delim + @List, [Number], LEN(@Delim+'x')-1) = @Delim
) AS y
);
建立在@NothingsImpossible解决方案上,或者,更确切地说,评论投票最多的答案(就在接受的答案下面),我发现下面的快速和肮脏的解决方案满足了我自己的需求-它的好处是完全在SQL域内。
给定一个字符串"first;second;third;fourth;fifth",比如说,我想获取第三个标记。只有当我们知道字符串将有多少标记时,这才有效——在这种情况下,它是5。所以我的操作方式是将最后两个令牌切掉(内部查询),然后将前两个令牌切掉(外部查询)
我知道这是丑陋的,涵盖了我所处的具体情况,但我张贴它只是为了以防有人发现它有用。干杯
select
REVERSE(
SUBSTRING(
reverse_substring,
0,
CHARINDEX(';', reverse_substring)
)
)
from
(
select
msg,
SUBSTRING(
REVERSE(msg),
CHARINDEX(
';',
REVERSE(msg),
CHARINDEX(
';',
REVERSE(msg)
)+1
)+1,
1000
) reverse_substring
from
(
select 'first;second;third;fourth;fifth' msg
) a
) b
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
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')