web-dev-qa-db-ja.com

複数の文字列の置換

置換マップテーブルを持っています

CREATE TABLE #ReplacementMap (old NVARCHAR(10), new NVARCHAR(10))
INSERT INTO #ReplacementMap VALUES ('A',5)
INSERT INTO #ReplacementMap VALUES ('C',9)
INSERT INTO #ReplacementMap VALUES ('D',4)

と文字列のテーブル

CREATE TABLE #String1 (name NVARCHAR(50), string1 NVARCHAR(100))
INSERT INTO #String1 VALUES ('John','AB')
INSERT INTO #String1 VALUES ('Kyle','ABC')
INSERT INTO #String1 VALUES ('Steven','ABCD')

以下の結果が得られるように、置換マップテーブルに基づいて文字列のビットを置き換える必要があります。

John,5B
Kyle,5B9
Steven,5B94

現在の解決策はREPLACE関数をネストすることですが、実行する必要のある置換の数が多いため、エレガントな方法ではありません。

sql-servert-sqlsql-server-2014replace
4
Przemyslaw Wojda

SQLCLR関数を使用して、SQL Server 2017の新しい TRANSLATE Transact-SQL関数 をエミュレートできます。

関数定義

CREATE Assembly [Translate] AUTHORIZATION [dbo]
FROM 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WITH PERMISSION_SET = SAFE;
GO
CREATE FUNCTION [dbo].[Translate]
(
    @Input nvarchar(4000), 
    @Find nvarchar(4000), 
    @Replace nvarchar(4000)
)
RETURNS nvarchar(4000)
AS EXTERNAL NAME 
    [Translate].[UserDefinedFunctions].[Translate];

使用法

SELECT
    S.[name],
    S.string1,
    Result = dbo.Translate(S.string1, N'ACD', N'594')
FROM #String1 AS S;

╔════════╦═════════╦════════╗
║  name  ║ string1 ║ Result ║
╠════════╬═════════╬════════╣
║ John   ║ AB      ║ 5B     ║
║ Kyle   ║ ABC     ║ 5B9    ║
║ Steven ║ ABCD    ║ 5B94   ║
╚════════╩═════════╩════════╝

この単純なデモの実装では、大文字と小文字を区別する比較を使用します。

ソースコード

using Microsoft.SqlServer.Server;
using System;
using System.Data.SqlTypes;

public partial class UserDefinedFunctions
{
    [SqlFunction(
        DataAccess = DataAccessKind.None,
        IsDeterministic = true,
        IsPrecise = true,
        SystemDataAccess = SystemDataAccessKind.None
    )]
    [return: SqlFacet(IsFixedLength = false, IsNullable = false, MaxSize = 4000)]
    public static SqlChars Translate
        (
            [SqlFacet(IsFixedLength = false, IsNullable = false, MaxSize = 4000)]
            SqlChars Input,
            [SqlFacet(IsFixedLength = false, IsNullable = false, MaxSize = 4000)]
            SqlChars Find,
            [SqlFacet(IsFixedLength = false, IsNullable = false, MaxSize = 4000)]
            SqlChars Replace
        )
    {
        if (Input.IsNull || Find.IsNull || Replace.IsNull)
        {
            // Return unchanged input for any NULL parameters
            return Input;
        }

        if (Find.Length != Replace.Length)
        {
            throw new ArgumentException("Find and Replace parameters must have the same length.");
        }

        // For each character in the input string
        for (int i = 0; i < Input.Length; i++)
        {
            // For each character in the Find string
            for (int j = 0; j < Find.Length; j++)
            {
                // If the character matches...
                if (Input[i] == Find[j])
                {
                    // ...replace it
                    Input[i] = Replace[j];
                }
            }
        }
        return Input;
    }
}
3
Paul White 9

このジョブには再帰CTEを使用できます。 string1列の値は、#ReplacementMapの項目ごとに置き換えられます

CREATE TABLE #ReplacementMap (old NVARCHAR(10), new NVARCHAR(10))
INSERT INTO #ReplacementMap VALUES ('A',5)
INSERT INTO #ReplacementMap VALUES ('C',9)
INSERT INTO #ReplacementMap VALUES ('D',4)

CREATE TABLE #String1 (name NVARCHAR(50), string1 NVARCHAR(100))
INSERT INTO #String1 VALUES ('John','AB')
INSERT INTO #String1 VALUES ('Kyle','ABC')
INSERT INTO #String1 VALUES ('Steven','ABCD')

DECLARE @MaxNumber int = (SELECT count(*) FROM #ReplacementMap)

;with temp AS
(
   SELECT *, row_number() over(order by rm.old) AS Rn
   FROM #ReplacementMap rm
)
,cte AS
(
   SELECT s.name, s.string1, CASt(0 AS int) AS Rn FROM #String1 s    
   UNION ALL
   SELECT cte.name, CAST(Replace(cte.string1,t.old, t.new) AS nvarchar(100))  , cte.Rn + 1 
   FROM cte 
   INNER JOIN temp t ON cte.Rn = t.Rn - 1
)
SELECT * FROM cte c
WHERE rn = @MaxNumber
OPTION (MAXRECURSION 0)


DROP TABLE #ReplacementMap
DROP TABLE #String1
2
TriV

これは再帰的なSQLでも実行できますが、そうするのが良いかどうかはわかりません。置換マップテーブルにID列を追加しました。コードをテストするために、456976の4つの文字列を生成しました。

_CREATE TABLE #ReplacementMap (
ID INT NOT NULL IDENTITY (1, 1), 
old NVARCHAR(10),
new NVARCHAR(10),
PRIMARY KEY (ID)
);

INSERT INTO #ReplacementMap VALUES ('A',5);
INSERT INTO #ReplacementMap VALUES ('C',9);
INSERT INTO #ReplacementMap VALUES ('D',4);


CREATE TABLE #String1 (
ID INT NOT NULL IDENTITY (1, 1),
string1 NVARCHAR(100)
);

WITH ALL_LETTERS AS (
    SELECT distinct CHAR(number) LETTER
    FROM master..spt_values
    WHERE number >= 65 AND number <= 90
)
INSERT INTO #String1 WITH (TABLOCK)
SELECT a1.LETTER + a2.LETTER + a3.LETTER + a4.LETTER
FROM ALL_LETTERS a1
CROSS JOIN ALL_LETTERS a2
CROSS JOIN ALL_LETTERS a3
CROSS JOIN ALL_LETTERS a4;
_

変換を行うコードは次のとおりです。

_WITH rec_cte AS (
    SELECT 
    s.ID
    , REPLACE(s.string1, rm.old, rm.new) new_string1
    , 1 replace_id
    FROM #String1 s
    INNER JOIN #ReplacementMap rm ON rm.ID = 1

    UNION ALL

    SELECT 
    s.ID
    , REPLACE(s.new_string1, rm.old, rm.new) new_string1
    , replace_id + 1
    FROM rec_cte s
    INNER JOIN #ReplacementMap rm ON rm.ID = replace_id + 1
)
SELECT ID, new_string1
FROM rec_cte
WHERE replace_id = (SELECT COUNT(*) FROM #ReplacementMap);
_

_#String1_にS行、_#ReplacementMap_にR行があるとします。 _#ReplacementMap_の各行について、テーブルへの結合を行い、次の行にフィルターをかけ、その行を使用してREPLACE()を実行します。 _#ReplacementMap_に行がなくなると、S X R行の結果セット全体が返されます。これは、サブクエリによって最終的な翻訳にフィルターされます。コードはS X R REPLACE()操作を実行し、R + 1は、いくつかの内部tempdb操作と共に、単一の行の結果セットに結合します。

置換文字列が101個未満である限り、これは変更なしで機能します。このコードは、AdánBucioが投稿したソリューションと同様に機能するようです。私のマシンでは、このクエリは約10秒で終了し、彼のソリューションは20秒で終了しました。ただし、それに基づいてソリューションを選択しないでください。応答時間の要件を満たしている限り、最も使いやすいコードを使用する必要があります。

SQL Server 2017には、この種の操作を簡単にする組み込み関数があることに注意してください [〜＃〜] translate [〜＃〜] 。

2
Joe Obbish

単一の文字を置き換えるので、文字列を単一の文字に分割し、それらを置換マップと結合してから、元に戻すことができます。

SELECT  ref.[name],
        ref.string1 AS original,
        rep.string  AS replaced
FROM    #String1 ref
        CROSS APPLY (
            SELECT  ISNULL(rm.new, 
                        SUBSTRING(ref.string1, num.number, 1)) AS [text()]
            FROM    master.dbo.spt_values num
                    LEFT JOIN #ReplacementMap rm
                        ON rm.old = SUBSTRING(ref.string1, num.number, 1)
            WHERE   num.number > 0 AND  num.number <= LEN(ref.string1)
                    AND num.[type] = 'P'
            FOR XML PATH('')
        ) rep(string);
1
Adán Bucio