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別の文字列内の文字列のすべての位置を見つける方法

テーブルまたは変数でpatindexを使用してすべての位置を見つけるにはどうすればよいですか?

declare @name nvarchar(max)
set @name ='ALi reza dar yek shabe barani ba yek  '
  + 'dokhtare khoshkel be disco raft va ALi baraye'
  + ' 1 saat anja bud va sepas... ALi...'
select patindex('%ALi%',@name) as pos 

これは1を返しますが、すべての結果が必要です。例:

pos
===
  1
 74
113
11
coditori
declare @name nvarchar(max)
set @name ='ALi reza dar yek shabe barani ba yek  dokhtare khoshkel be disco raft va ALi baraye 1 saat anja bud va sepas... ALi...'

Declare @a table (pos int)
Declare @pos int
Declare @oldpos int
Select @oldpos=0
select @pos=patindex('%ALi%',@name) 
while @pos > 0 and @oldpos<>@pos
 begin
   insert into @a Values (@pos)
   Select @oldpos=@pos
   select @pos=patindex('%ALi%',Substring(@name,@pos + 1,len(@name))) + @pos
end

Select * from @a

これを再利用可能にするには、テーブル関数で使用して次のように呼び出します。

Select * from  dbo.F_CountPats ('ALi reza dar yek shabe barani ba yek  dokhtare khoshkel be disco raft va ALi baraye 1 saat anja bud va sepas... ALi...','%ALi%')

関数は次のようになります

Create FUNCTION [dbo].[F_CountPats] 
(
@txt varchar(max),
@Pat varchar(max)
)
RETURNS 
@tab TABLE 
(
 ID int
)
AS
BEGIN
Declare @pos int
Declare @oldpos int
Select @oldpos=0
select @pos=patindex(@pat,@txt) 
while @pos > 0 and @oldpos<>@pos
 begin
   insert into @tab Values (@pos)
   Select @oldpos=@pos
   select @pos=patindex(@pat,Substring(@txt,@pos + 1,len(@txt))) + @pos
end

RETURN 
END

GO
9
bummi

これは、選択したループ方式( ここでいくつかの証拠 )よりも少し効率的であり、再帰的CTEよりも明らかに効率的だと思います。

CREATE FUNCTION dbo.FindPatternLocation
(
    @string NVARCHAR(MAX),
    @term   NVARCHAR(255)
)
RETURNS TABLE
AS
    RETURN 
    (
      SELECT pos = Number - LEN(@term) 
      FROM (SELECT Number, Item = LTRIM(RTRIM(SUBSTRING(@string, Number, 
      CHARINDEX(@term, @string + @term, Number) - Number)))
      FROM (SELECT ROW_NUMBER() OVER (ORDER BY [object_id])
      FROM sys.all_objects) AS n(Number)
      WHERE Number > 1 AND Number <= CONVERT(INT, LEN(@string)+1)
      AND SUBSTRING(@term + @string, Number, LEN(@term)) = @term
    ) AS y);

使用例:

DECLARE @name NVARCHAR(MAX);

SET @name = N'ALi reza dar yek shabe barani ba yek'
    + '  dokhtare khoshkel be disco raft va ALi baraye '
    + '1 saat anja bud va sepas... ALi...';

SELECT pos FROM dbo.FindPatternLocation(@name, 'ALi');

結果:

pos
---
  1
 74
113

文字列が2Kより長くなる場合は、sys.all_objectsではなくsys.all_columnsを使用してください。 8Kより長い場合は、クロス結合を追加します。

15
Aaron Bertrand

-再帰CTE

with cte as
(select 'ALi reza dar yek shabe barani ba yek  dokhtare khoshkel be disco raft va ALi baraye 1 saat anja bud va sepas... ALi...' as name
), 
pos as
(select patindex('%ALi%',name) pos, name from cte
union all
select pos+patindex('%ALi%',substring(name, pos+1, len(name))) pos, name from pos
where patindex('%ALi%',substring(name, pos+1, len(name)))>0
)
select pos from pos
2
msi77

アーロン・ベルトランの答えが大好きです。完全にはわかりませんが、とてもエレガントに見えます。

過去に、_sys.objects_を使用するときに、権限に関する問題に遭遇しました。コードのトラブルシューティングを行う必要性と合わせて、アーロンのコードのバリエーションを考え出し、以下に追加しました。

これは私の手順です:

_CREATE PROCEDURE dbo.FindPatternLocations
-- Params
@TextToSearch nvarchar (max),
@TextToFind nvarchar (255)

AS
BEGIN

    declare @Length int
    set @Length = (Select LEN(@TextToSearch))

    declare @LengthSearchString int
    set @LengthSearchString = (select LEN (@TextToFind))

    declare @Index int
    set @Index=1

    create table #Positions (
    [POSID] [int] IDENTITY(0,1) NOT FOR REPLICATION NOT NULL,
    POS int
    )

    insert into #Positions (POS) select 0 -- to return a row even if no findings occur

        set @Index = (select charindex(@TextToFind, @TextToSearch, @Index))
                    if @Index = 0 goto Ende -- TextToFind is not in TextToSearch

        insert into #Positions (POS) select @Index


        set @Index = @Index + @LengthSearchString

while @Index <= @Length - @LengthSearchString   
    Begin
            set @Index = (select charindex(@TextToFind, @TextToSearch, @Index) )
            if @Index = 0 goto Ende -- no findings anymore
            insert into #Positions (POS) select @Index
            set @Index = @Index + @LengthSearchString
    end
Ende:
if (select MAX(posid) from #Positions) > 0 delete from #Positions where POSID = 0 -- row is not needed if TextToFind occurs
select * from #Positions
END
GO
_

MAX(posid)値は、見つかった一致の数でもあります。

0
Nils
Declare @search varchar(5)
    sET @search='a'
    Declare @name varchar(40)
    Set @name='AmitabhBachan'
    Declare @init int
    Set @init=1
    Declare @hold int
    Declare @table table (POSITION Int)
    While( @init<= LEn(@name))
    Begin
   Set @hold=(Select CHARINDEX(@search,@name,@init))
   If (@hold!=0)
   BEgin 
   --Print @hold
   Insert into @table
   Select @hold
   Set @init=@hold+1
   End 
   Else
   If (@hold=0)
   BEgin
   Break
   End
  End
  Select * from @table
0
Raghu

これは Aaronの答え に基づく単純なコードです。

  • Sys.all_objectsのサイズに限定されない
  • 最後の「X」をお見逃しなく

コード:

DECLARE @termToFind CHAR(1) = 'X'
DECLARE @string VARCHAR(40) = 'XX XXX  X   XX'

SET @string += '.' --Add any data here (different from the one searched) to get the position of the last character

DECLARE @stringLength BIGINT = len(@string)

SELECT pos = Number - LEN(@termToFind)
FROM (
    SELECT Number
        , Item = LTRIM(RTRIM(SUBSTRING(@string, Number, CHARINDEX(@termToFind, @string + @termToFind, Number) - Number)))
    FROM (
        --All numbers between 1 and the lengh of @string. Better than use sys.all_objects
        SELECT TOP (@stringLength) row_number() OVER (
                ORDER BY t1.number
                ) AS N
        FROM master..spt_values t1
        CROSS JOIN master..spt_values t2
        ) AS n(Number)
    WHERE Number > 1
        AND Number <= CONVERT(INT, LEN(@string))
        AND SUBSTRING(@termToFind + @string, Number, LEN(@termToFind)) = @termToFind
    ) AS y

[〜#〜]結果[〜#〜]

pos
--------------------
1
2
4
5
6
9
13
14

(8 row(s) affected)
0
JotaPardo

遅くなって申し訳ありませんが、これを拡張したい人のために物事を簡単にしたいと思います。私はこれらの各実装を調べていて、私にとって最良と思われる実装(Aaron Bertrand)を採用し、それを簡略化しました。これで「テンプレート」ができました。賢く使ってください。

CREATE FUNCTION dbo.CHARINDICES (
    @search_expression NVARCHAR(4000),
    @expression_to_be_searched NVARCHAR(MAX)
) RETURNS TABLE AS RETURN (
    WITH tally AS (
        SELECT Number = ROW_NUMBER() OVER (ORDER BY [object_id])
        FROM sys.all_objects)
    SELECT DISTINCT n = subIdx -- (4) if we don't perform distinct we'll get result for each searched substring, and we don't want that
    FROM 
        tally 
        CROSS APPLY (SELECT subIdx = CHARINDEX(@search_expression, @expression_to_be_searched, Number)) x -- (2) subIdx is found in the rest of the substring 
    WHERE 
        Number BETWEEN 1 AND LEN(@expression_to_be_searched) -- (1) run for each substring once
        AND SubIdx != 0  -- (3) we care only about the indexes we've found, 0 stands for "not found"
)

SELECT CHARINDEX('C', 'BACBABCBABBCBACBBABC')
SELECT * FROM dbo.CHARINDICES('C', 'BACBABCBABBCBACBBABC')

参考として-PATINDEX()を展開するなど、これから他の動作を派生させることができます。

CREATE FUNCTION dbo.PATINDICES (
    @search_expression NVARCHAR(4000) = '%[cS]%',
    @expression_to_be_searched NVARCHAR(MAX) = 'W3Schools.com'
) RETURNS TABLE AS RETURN (
    WITH tally AS (
        SELECT num = ROW_NUMBER() OVER (ORDER BY [object_id])
        FROM sys.all_objects)
    SELECT DISTINCT n = subIdx + num - 1
    FROM 
        tally 
        CROSS APPLY (SELECT numRev = LEN(@expression_to_be_searched) - num + 1) x
        CROSS APPLY (SELECT subExp = RIGHT(@expression_to_be_searched, numRev)) y
        CROSS APPLY (SELECT subIdx = PATINDEX(@search_expression, subExp)) z
    WHERE 
        num BETWEEN 1 AND LEN(@expression_to_be_searched)
        AND SubIdx != 0
)

SELECT PATINDEX('%[cS]%', 'W3Schools.com')
SELECT * FROM dbo.PATINDICES('%[cS]%', 'W3Schools.com')