我有一个表格列，其中包含abc_1_2_3_4.gif或zzz_12_3_3_45.gif等值。
我想在上面的值中找到每个下划线的索引 _。只有四个下划线，但鉴于它们可以在弦中的任何位置，我怎么能实现这一点呢？
答案 0 :(得分：34)
单程（2k8）;
select 'abc_1_2_3_4.gif ' as img into #T
insert #T values ('zzz_12_3_3_45.gif')
;with T as (
select 0 as row, charindex('_', img) pos, img from #T
union all
select pos + 1, charindex('_', img, pos + 1), img
from T
where pos > 0
)
select
img, pos
from T
where pos > 0
order by img, pos
>>>>
img pos
abc_1_2_3_4.gif 4
abc_1_2_3_4.gif 6
abc_1_2_3_4.gif 8
abc_1_2_3_4.gif 10
zzz_12_3_3_45.gif 4
zzz_12_3_3_45.gif 7
zzz_12_3_3_45.gif 9
zzz_12_3_3_45.gif 11
<强>更新
;with T(img, starts, pos) as (
select img, 1, charindex('_', img) from #t
union all
select img, pos + 1, charindex('_', img, pos + 1)
from t
where pos > 0
)
select
*, substring(img, starts, case when pos > 0 then pos - starts else len(img) end) token
from T
order by img, starts
>>>
img starts pos token
abc_1_2_3_4.gif 1 4 abc
abc_1_2_3_4.gif 5 6 1
abc_1_2_3_4.gif 7 8 2
abc_1_2_3_4.gif 9 10 3
abc_1_2_3_4.gif 11 0 4.gif
zzz_12_3_3_45.gif 1 4 zzz
zzz_12_3_3_45.gif 5 7 12
zzz_12_3_3_45.gif 8 9 3
zzz_12_3_3_45.gif 10 11 3
zzz_12_3_3_45.gif 12 0 45.gif
答案 1 :(得分：10)
您可以在位置+1
charindex('_', [TEXT], (charindex('_', [TEXT], 1))+1)
在+1是你想要找到的第n次。
答案 2 :(得分：7)
您可以function使用以下split the values至delimiter。它return a table并找到第n个匹配项，只需要select即可！或者将其更改为return您需要的内容而不是table。
CREATE FUNCTION dbo.Split
(
@RowData nvarchar(2000),
@SplitOn nvarchar(5)
)
RETURNS @RtnValue table
(
Id int identity(1,1),
Data nvarchar(100)
)
AS
BEGIN
Declare @Cnt int
Set @Cnt = 1
While (Charindex(@SplitOn,@RowData)>0)
Begin
Insert Into @RtnValue (data)
Select
Data = ltrim(rtrim(Substring(@RowData,1,Charindex(@SplitOn,@RowData)-1)))
Set @RowData = Substring(@RowData,Charindex(@SplitOn,@RowData)+1,len(@RowData))
Set @Cnt = @Cnt + 1
End
Insert Into @RtnValue (data)
Select Data = ltrim(rtrim(@RowData))
Return
END
答案 3 :(得分：7)
您可以使用CHARINDEX并指定起始位置：
DECLARE @x VARCHAR(32) = 'MS-SQL-Server';
SELECT
STUFF(STUFF(@x,3 , 0, '/'), 8, 0, '/') InsertString
,CHARINDEX('-',LTRIM(RTRIM(@x))) FirstIndexOf
,CHARINDEX('-',LTRIM(RTRIM(@x)), (CHARINDEX('-', LTRIM(RTRIM(@x)) )+1)) SecondIndexOf
,CHARINDEX('-',@x,CHARINDEX('-',@x, (CHARINDEX('-',@x)+1))+1) ThirdIndexOf
,CHARINDEX('-',REVERSE(LTRIM(RTRIM(@x)))) LastIndexOf;
GO
答案 4 :(得分：4)
DECLARE @str AS VARCHAR(100)
SET @str='1,2 , 3, 4, 5,6'
SELECT COALESCE(LTRIM(CAST(('<X>'+REPLACE(@str,',' ,'</X><X>')+'</X>') AS XML).value('(/X)[1]', 'varchar(128)')), ''),
COALESCE(LTRIM(CAST(('<X>'+REPLACE(@str,',' ,'</X><X>')+'</X>') AS XML).value('(/X)[2]', 'varchar(128)')), ''),
COALESCE(LTRIM(CAST(('<X>'+REPLACE(@str,',' ,'</X><X>')+'</X>') AS XML).value('(/X)[3]', 'varchar(128)')), ''),
COALESCE(LTRIM(CAST(('<X>'+REPLACE(@str,',' ,'</X><X>')+'</X>') AS XML).value('(/X)[4]', 'varchar(128)')), ''),
COALESCE(LTRIM(CAST(('<X>'+REPLACE(@str,',' ,'</X><X>')+'</X>') AS XML).value('(/X)[5]', 'varchar(128)')), ''),
COALESCE(LTRIM(CAST(('<X>'+REPLACE(@str,',' ,'</X><X>')+'</X>') AS XML).value('(/X)[6]', 'varchar(128)')), ''),
COALESCE(LTRIM(CAST(('<X>'+REPLACE(@str,',' ,'</X><X>')+'</X>') AS XML).value('(/X)[7]', 'varchar(128)')), ''),
COALESCE(LTRIM(CAST(('<X>'+REPLACE(@str,',' ,'</X><X>')+'</X>') AS XML).value('(/X)[8]', 'varchar(128)')), ''),
COALESCE(LTRIM(CAST(('<X>'+REPLACE(@str,',' ,'</X><X>')+'</X>') AS XML).value('(/X)[9]', 'varchar(128)')), '')
答案 5 :(得分：2)
你可以look for the four underscore in this way：
create table #test
( t varchar(50) );
insert into #test values
( 'abc_1_2_3_4.gif'),
('zzz_12_3_3_45.gif');
declare @t varchar(50);
declare @t_aux varchar(50);
declare @t1 int;
declare @t2 int;
declare @t3 int;
declare @t4 int;
DECLARE t_cursor CURSOR
FOR SELECT t FROM #test
OPEN t_cursor
FETCH NEXT FROM t_cursor into @t;
set @t1 = charindex( '_', @t )
set @t2 = charindex( '_', @t , @t1+1)
set @t3 = charindex( '_', @t , @t2+1)
set @t4 = charindex( '_', @t , @t3+1)
select @t1, @t2, t3, t4
--do a loop to iterate over all table
你可以在这里测试一下。
或者用这种简单的方式：
select
charindex( '_', t ) as first,
charindex( '_', t, charindex( '_', t ) + 1 ) as second,
...
from
#test
答案 6 :(得分：2)
DECLARE @LEN INT
DECLARE @VAR VARCHAR(20)
SET @VAR = 'HELLO WORLD'
SET @LEN = LEN(@VAR)
--SELECT @LEN
SELECT PATINDEX('%O%',SUBSTRING(@VAR,PATINDEX('%O%' ,@VAR) + 1 ,PATINDEX('%O%',@VAR) + 1)) + PATINDEX('%O%',@VAR)
答案 7 :(得分：2)
您可以尝试剥离变量/数组，假设列表中有清晰度
declare @array table ----table of values
(
id int identity(1,1)
,value nvarchar(max)
)
DECLARE @VALUE NVARCHAR(MAX)='val1_val2_val3_val4_val5_val6_val7'----string array
DECLARE @CURVAL NVARCHAR(MAX) ---current value
DECLARE @DELIM NVARCHAR(1)='_' ---delimiter
DECLARE @BREAKPT INT ---current index of the delimiter
WHILE EXISTS (SELECT @VALUE)
BEGIN
SET @BREAKPT=CHARINDEX(@DELIM,@VALUE) ---set the current index
---
If @BREAKPT<> 0 ---index at 0 breaks the loop
begin
SET @CURVAL=SUBSTRING(@VALUE,1,@BREAKPT-1) ---current value
set @VALUE=REPLACE(@VALUE,SUBSTRING(@VALUE,1,@BREAKPT),'') ---current value and delimiter, replace
insert into @array(value) ---insert data
select @CURVAL
end
else
begin
SET @CURVAL=@VALUE ---current value now last value
insert into @array(value) ---insert data
select @CURVAL
break ---break loop
end
end
select * from @array ---find nth occurance given the id
答案 8 :(得分：1)
DECLARE @T AS TABLE(pic_name VARCHAR(100));
INSERT INTO @T VALUES ('abc_1_2_3_4.gif'),('zzz_12_3_3_45.gif');
SELECT A.pic_name, P1.D, P2.D, P3.D, P4.D
FROM @T A
CROSS APPLY (SELECT NULLIF(CHARINDEX('_', A.pic_name),0) AS D) P1
CROSS APPLY (SELECT NULLIF(CHARINDEX('_', A.pic_name, P1.D+1), 0) AS D) P2
CROSS APPLY (SELECT NULLIF(CHARINDEX('_', A.pic_name, P2.D+1),0) AS D) P3
CROSS APPLY (SELECT NULLIF(CHARINDEX('_', A.pic_name, P3.D+1),0) AS D) P4
答案 9 :(得分：1)
我决定使用递归函数，因为对我而言，遵循逻辑更容易。请注意，SQL Server的默认函数递归限制为32，因此这仅适用于较小的工作负载。
create function dbo._charindex_nth (
@FindThis varchar(8000),
@InThis varchar(max),
@StartFrom int,
@NthOccurence tinyint
)
returns bigint
as
begin
/*
Recursive helper used by dbo.charindex_nth to return the position of the nth occurance of @FindThis in @InThis
Who When What
PJR 160421 Initial
*/
declare @Pos bigint
if isnull(@NthOccurence, 0) <= 0 or isnull(@StartFrom, 0) <= 0
begin
select @Pos = 0
end else begin
if @NthOccurence = 1
begin
select @Pos = charindex(@FindThis, @InThis, @StartFrom)
end else begin
select @Pos = dbo._charindex_nth(@FindThis, @InThis, nullif(charindex(@FindThis, @InThis, @StartFrom), 0) + 1, @NthOccurence - 1)
end
end
return @Pos
end
create function dbo.charindex_nth (
@FindThis varchar(8000),
@InThis varchar(max),
@NthOccurence tinyint
)
returns bigint
as
begin
/*
Returns the position of the nth occurance of @FindThis in @InThis
Who When What
PJR 160421 Initial
*/
return dbo._charindex_nth(@FindThis, @InThis, 1, @NthOccurence)
end
declare @val varchar(max) = 'zzz_12_3_3_45.gif'
select dbo.charindex_nth('_', @val, 1) Underscore1
, dbo.charindex_nth('_', @val, 2) Underscore2
, dbo.charindex_nth('_', @val, 3) Underscore3
, dbo.charindex_nth('_', @val, 4) Underscore4
答案 10 :(得分：1)
我这样做了几个单独的自定义函数，一个用于搜索字符的每个位置，即第二个，第三个：
创建功能[dbo]。[fnCHARPOS2] （@SEARCHCHAR VARCHAR（255）， @SEARCHSTRING VARCHAR（255）） 退货INT 如 开始 返回CHARINDEX（@ SEARCHCHAR，@ SEARCHSTRING（CHARINDEX（@ SEARCHCHAR，@ SEARCHSTRING，0）+1））;
CREATE FUNCTION [dbo].[fnCHARPOS3]
(@SEARCHCHAR VARCHAR(255),
@SEARCHSTRING VARCHAR(255))
RETURNS INT
AS
BEGIN
RETURN CHARINDEX(@SEARCHCHAR,@SEARCHSTRING, (CHARINDEX(@SEARCHCHAR,@SEARCHSTRING, (CHARINDEX(@SEARCHCHAR,@SEARCHSTRING,0)+1)))+1);
然后，您可以将要搜索的字符和要搜索的字符串作为参数传入：
所以，如果你正在寻找＆＃39; f＆＃39;并想知道前3次出现的位置：
select
database.dbo.fnCHARPOS2('f',tablename.columnname),
database.dbo.fnCHARPOS3('f',tablename.columnname)
from tablename
它对我有用！
答案 11 :(得分：1)
我的SQL支持substring_Index的功能，它将返回n出现的字符串中值的位置。可以编写类似的用户定义函数来实现此目的。 link
中的示例或者你可以使用charindex函数调用x次来报告每个_的位置，给出先前找到的实例的起始位置+1。直到找到0
编辑：NM Charindex是正确的功能
答案 12 :(得分：0)
我用更快的方式来做这件事，而不是简单地遍历字符串。
CREATE FUNCTION [ssf_GetNthSeparatorPosition] ( @TargetString VARCHAR(MAX)
, @Sep VARCHAR(25)
, @n INTEGER )
RETURNS INTEGER
/****************************************************************************************
--#############################################################################
-- Returns the position of the Nth Charactor sequence
-- 1234567890123456789
-- Declare @thatString varchar(max) = 'hi,there,jay,yo'
Select dbo.ssf_GetNthSeparatorPosition(@thatString, ',', 3) --would return 13
--############################################################################
****************************************************************************************/
AS
BEGIN
DECLARE @Retval INTEGER = 0
DECLARE @CurPos INTEGER = 0
DECLARE @LenSep INTEGER = LEN(@Sep)
SELECT @CurPos = CHARINDEX(@Sep, @TargetString)
IF ISNULL(@LenSep, 0) > 0
AND @CurPos > 0
BEGIN
SELECT @CurPos = 0
;with lv0 AS (SELECT 0 g UNION ALL SELECT 0)
,lv1 AS (SELECT 0 g FROM lv0 a CROSS JOIN lv0 b) -- 4
,lv2 AS (SELECT 0 g FROM lv1 a CROSS JOIN lv1 b) -- 16
,lv3 AS (SELECT 0 g FROM lv2 a CROSS JOIN lv2 b) -- 256
,lv4 AS (SELECT 0 g FROM lv3 a CROSS JOIN lv3 b) -- 65,536
,lv5 AS (SELECT 0 g FROM lv4 a CROSS JOIN lv4 b) -- 4,294,967,296
,Tally (n) AS (SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) FROM lv5),
results
AS ( SELECT n - LEN(@Sep) AS Nth
, row_number() OVER ( ORDER BY n ) - 1 AS Position
FROM Tally t
WHERE n BETWEEN 1
AND DATALENGTH(@TargetString) + DATALENGTH(@Sep)
AND SUBSTRING(@Sep + @TargetString, n, LEN(@Sep)) = @Sep)
SELECT @CurPos = Nth
FROM results
WHERE results.Position = @n
END
RETURN @CurPos
END
GO
答案 13 :(得分：0)
使用xml转换执行此操作的简单示例：
SELECT 'A|B|C'
, concat('<x>', REPLACE('A|B|C', '|', '</x><x>'), '</x>')
, cast(concat('<x>', REPLACE('A|B|C', '|', '</x><x>'), '</x>') as xml).query('/x[2]')
, cast(concat('<x>', REPLACE('A|B|C', '|', '</x><x>'), '</x>') as xml).value('/x[2]',
'varchar');
这里有您的样本的翻译：
SELECT gifname
,cast(concat('<x>', REPLACE(gifname, '_', '</x><x>'), '</x>') as xml).query('/x[2]') as xmlelement
, cast(concat('<x>', REPLACE(gifname, '_', '</x><x>'), '</x>') as xml).value('/x[2]', 'varchar(10)') as result
FROM (
SELECT 'abc_1_2_3_4.gif' as gifname
UNION ALL
SELECT 'zzz_12_3_3_45.gif'
) tmp
答案 14 :(得分：0)
我使用了一个函数从分隔的字符串字段中获取“nth”元素并取得了巨大的成功。如上所述，它不是处理事物的“快速”方式，但确实很方便。
create function GetArrayIndex(@delimited nvarchar(max), @index int, @delimiter nvarchar(100) = ',') returns nvarchar(max)
as
begin
declare @xml xml, @result nvarchar(max)
set @xml = N'<root><r>' + replace(@delimited, @delimiter,'</r><r>') + '</r></root>'
select @result = r.value('.','varchar(max)')
from @xml.nodes('//root/r[sql:variable("@index")]') as records(r)
return @result
end
答案 15 :(得分：0)
declare @a nvarchar(50)='Enter Your string '
declare @character char='e'
declare @nthoccurence int = 2
declare @i int = 1
declare @j int =0
declare @count int = len(@a)-len(replace(@a,@character,''))
if(@count >= @nthoccurence)
begin
while (@I <= @nthoccurence)
begin
set @j= CHARINDEX(@character,@a,@j+1)
set @i= @i+1
end
print @j
end
else
Print 'you have only '+convert(nvarchar ,@count)+' occurrences of '+@character
end
答案 16 :(得分：0)
DECLARE @x VARCHAR(32) = 'MS-SQL-Server';
SELECT
SUBSTRING(@x,0,CHARINDEX('-',LTRIM(RTRIM(@x)))) A,
SUBSTRING(@x,CHARINDEX('-',LTRIM(RTRIM(@x)))+1,CHARINDEX('-'
,LTRIM(RTRIM(@x)))) B,
SUBSTRING(@x,CHARINDEX('-',REVERSE(LTRIM(RTRIM(@x))))+1,LEN(@x)-1) C
A B C
MS SQL Server
答案 17 :(得分：0)
尝试一下
CREATE FUNCTION [dbo].[CHARINDEX2] (
@expressionToFind VARCHAR(MAX),
@expressionToSearch VARCHAR(MAX),
@occurrenceIndex INT,
@startLocation INT = 0
)
RETURNS INT
AS BEGIN
IF @occurrenceIndex < 1 BEGIN
RETURN CAST('The argument @occurrenceIndex must be a positive integer.' AS INT)
END
IF @startLocation < 0 BEGIN
RETURN CAST('The argument @startLocation must be a non negative integer.' AS INT)
END
DECLARE @returnIndex INT
SET @returnIndex = CHARINDEX(@expressionToFind, @expressionToSearch, @startLocation)
IF (@occurrenceIndex = 1) BEGIN
RETURN @returnIndex
END
DECLARE @target_length INT
SET @target_length = LEN(@expressionToFind)
SET @occurrenceIndex += -1
WHILE (@occurrenceIndex > 0 AND @returnIndex > 0) BEGIN
SET @returnIndex = CHARINDEX(@expressionToFind, @expressionToSearch, @returnIndex + @target_length);
SET @occurrenceIndex += -1
END
RETURN @returnIndex
END
GO