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
递归CTE解决方案与服务器疼痛,测试它
MS SQL Server 2008模式设置:
create table Course( Courses varchar(100) );
insert into Course values ('Hello John Smith');
查询1:
with cte as
( select
left( Courses, charindex( ' ' , Courses) ) as a_l,
cast( substring( Courses,
charindex( ' ' , Courses) + 1 ,
len(Courses ) ) + ' '
as varchar(100) ) as a_r,
Courses as a,
0 as n
from Course t
union all
select
left(a_r, charindex( ' ' , a_r) ) as a_l,
substring( a_r, charindex( ' ' , a_r) + 1 , len(a_R ) ) as a_r,
cte.a,
cte.n + 1 as n
from Course t inner join cte
on t.Courses = cte.a and len( a_r ) > 0
)
select a_l, n from cte
--where N = 1
结果:
| A_L | N |
|--------|---|
| Hello | 0 |
| John | 1 |
| Smith | 2 |
其他回答
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
使用SQL Server 2016及以上版本。使用这段代码修剪字符串,忽略NULL值,并按正确的顺序应用行索引。它也适用于空格分隔符:
DECLARE @STRING_VALUE NVARCHAR(MAX) = 'one, two,,three, four, five'
SELECT ROW_NUMBER() OVER (ORDER BY R.[index]) [index], R.[value] FROM
(
SELECT
1 [index], NULLIF(TRIM([value]), '') [value] FROM STRING_SPLIT(@STRING_VALUE, ',') T
WHERE
NULLIF(TRIM([value]), '') IS NOT NULL
) R
以下是我的解决方案,可能会对某些人有所帮助。修改以上Jonesinator的回答。
如果我有一个带分隔符的INT值字符串,并希望返回一个INT表(然后我可以加入)。如。44岁的1,3343 6,8765年
创建一个UDF:
IF OBJECT_ID(N'dbo.ufn_GetIntTableFromDelimitedList', N'TF') IS NOT NULL
DROP FUNCTION dbo.[ufn_GetIntTableFromDelimitedList];
GO
CREATE FUNCTION dbo.[ufn_GetIntTableFromDelimitedList](@String NVARCHAR(MAX), @Delimiter CHAR(1))
RETURNS @table TABLE
(
Value INT NOT NULL
)
AS
BEGIN
DECLARE @Pattern NVARCHAR(3)
SET @Pattern = '%' + @Delimiter + '%'
DECLARE @Value NVARCHAR(MAX)
WHILE LEN(@String) > 0
BEGIN
IF PATINDEX(@Pattern, @String) > 0
BEGIN
SET @Value = SUBSTRING(@String, 0, PATINDEX(@Pattern, @String))
INSERT INTO @table (Value) VALUES (@Value)
SET @String = SUBSTRING(@String, LEN(@Value + @Delimiter) + 1, LEN(@String))
END
ELSE
BEGIN
-- Just the one value.
INSERT INTO @table (Value) VALUES (@String)
RETURN
END
END
RETURN
END
GO
然后得到表格结果:
SELECT * FROM dbo.[ufn_GetIntTableFromDelimitedList]('1,20,3,343,44,6,8765', ',')
1
20
3
343
44
6
8765
在join语句中:
SELECT [ID], [FirstName]
FROM [User] u
JOIN dbo.[ufn_GetIntTableFromDelimitedList]('1,20,3,343,44,6,8765', ',') t ON u.[ID] = t.[Value]
1 Elvis
20 Karen
3 David
343 Simon
44 Raj
6 Mike
8765 Richard
如果你想返回一个nvarchar列表而不是int,那么只需更改表定义:
RETURNS @table TABLE
(
Value NVARCHAR(MAX) NOT NULL
)
下面的示例使用递归CTE
更新18.09.2013
CREATE FUNCTION dbo.SplitStrings_CTE(@List nvarchar(max), @Delimiter nvarchar(1))
RETURNS @returns TABLE (val nvarchar(max), [level] int, PRIMARY KEY CLUSTERED([level]))
AS
BEGIN
;WITH cte AS
(
SELECT SUBSTRING(@List, 0, CHARINDEX(@Delimiter, @List + @Delimiter)) AS val,
CAST(STUFF(@List + @Delimiter, 1, CHARINDEX(@Delimiter, @List + @Delimiter), '') AS nvarchar(max)) AS stval,
1 AS [level]
UNION ALL
SELECT SUBSTRING(stval, 0, CHARINDEX(@Delimiter, stval)),
CAST(STUFF(stval, 1, CHARINDEX(@Delimiter, stval), '') AS nvarchar(max)),
[level] + 1
FROM cte
WHERE stval != ''
)
INSERT @returns
SELECT REPLACE(val, ' ','' ) AS val, [level]
FROM cte
WHERE val > ''
RETURN
END
演示SQLFiddle
递归CTE解决方案与服务器疼痛,测试它
MS SQL Server 2008模式设置:
create table Course( Courses varchar(100) );
insert into Course values ('Hello John Smith');
查询1:
with cte as
( select
left( Courses, charindex( ' ' , Courses) ) as a_l,
cast( substring( Courses,
charindex( ' ' , Courses) + 1 ,
len(Courses ) ) + ' '
as varchar(100) ) as a_r,
Courses as a,
0 as n
from Course t
union all
select
left(a_r, charindex( ' ' , a_r) ) as a_l,
substring( a_r, charindex( ' ' , a_r) + 1 , len(a_R ) ) as a_r,
cte.a,
cte.n + 1 as n
from Course t inner join cte
on t.Courses = cte.a and len( a_r ) > 0
)
select a_l, n from cte
--where N = 1
结果:
| A_L | N |
|--------|---|
| Hello | 0 |
| John | 1 |
| Smith | 2 |