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
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。
其他回答
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
修改@Aaron Bertrand的功能
CREATE FUNCTION [dbo].[SplitString]
(
@List NVARCHAR(MAX),
@Delim VARCHAR(255),
@Idx int
)
RETURNS NVARCHAR(1000)
AS
BEGIN
DECLARE @ValueTable TABLE(String NVARCHAR(50), Ind int)
DECLARE @Value NVARCHAR(50)
BEGIN
INSERT INTO @ValueTable
SELECT Value, idx FROM
(SELECT [Value], idx = RANK() OVER (ORDER BY n) FROM
(
SELECT n = Number,
[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)) = @Delim
) AS y
) AS R WHERE idx = @Idx
SET @Value = (SELECT String FROM @ValueTable)
END
RETURN @Value
END
GO
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
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。