Question

Word Replacement Table ScreenShot

您好，

我有一个表＆＃34; tbl_address_replacement＆＃34;，附件是其中数据的屏幕截图。

我的输入字符串是@input =＆＃39; Auriga Building 4＆＃39;

我想通过匹配＆＃34; word_contains＆＃34;中的值来替换上面的输入字符串。上表中的列，并用＆＃34; replace_word＆＃34;替换匹配的值列值。

例如：

所需的o / p：Auriga Bldg。 4

THANKYOU！

Answer 1

这对您有所帮助。

示例：SELECT dbo。[fnSplitString1]（'Auriga Building 4'）

CREATE FUNCTION [dbo].[fnSplitString1] 
( 
     @InputString NVARCHAR(MAX)
 )
 RETURNS VARCHAR(50)
AS
BEGIN
    DECLARE @delimiter CHAR(1) 
    DECLARE @start INT, @end INT 
    DECLARE @Result NVARCHAR(MAX)
    DECLARE @Temp AS NVARCHAR(MAX)

    SET @delimiter = ' '
    SET @Result = ''
    ---SET @InputString = 'Auriga Building 4'

    SELECT @start = 1, @end = CHARINDEX(@delimiter, @InputString) 
    WHILE @start < LEN(@InputString) + 1 
    BEGIN 
        IF @end = 0  
            SET @end = LEN(@InputString) + 1      

            SET @Temp = (SELECT TOP 1 replaceword
                        FROM tbl_address_replacement 
                        WHERE word_contains = SUBSTRING(@InputString, @start, @end - @start))


            IF (@Temp IS NULL)
                SET @Result = @Result + SUBSTRING(@InputString, @start, @end - @start) + ' '
            ELSE
                SET @Result = @Result + @Temp + ' '

        SET @start = @end + 1 
        SET @end = CHARINDEX(@delimiter, @InputString, @start)        
    END 

    RETURN @Result
END

Answer 2

试试这个..

 DECLARE @string AS VARCHAR(MAX)
DECLARE @sCurrentString AS VARCHAR(MAX)
DECLARE @sResultString AS VARCHAR(MAX)
DECLARE @Split AS CHAR(1)
DECLARE @X AS xml
DECLARE @sTemp AS VARCHAR(MAX)

SET @sResultString = ''
SET @string = 'Auriga Building 4'
SET @Split = ' '
SET @X = CONVERT(xml,'<root><s>' + REPLACE(@string,@Split,'</s><s>') + '</s></root>')

DECLARE LoopCursor CURSOR FOR
        SELECT T.c.value('.','varchar(max)') FROM @X.nodes('/root/s') T(c)

OPEN LoopCursor

    FETCH NEXT FROM LoopCursor INTO @sCurrentString

    WHILE @@FETCH_STATUS = 0
    BEGIN  
        SET @sTemp = NULL

        SELECT @sTemp = replace_word FROM tbl_address_replacement WHERE word_contains = @sCurrentString

         IF (ISNULL(@sTemp, '') = '')
             SET @sTemp = @sCurrentString

        SET @sResultString = @sResultString + @Split + @sTemp 

        FETCH NEXT FROM LoopCursor INTO @sCurrentString
    END

SELECT @sResultString

Answer 3

我在做系统迁移的同时写了这样的东西。

以下是适用的代码：

DECLARE @rowcount int;

-- Repeat until there are no iterations left (lazy recursive method)
SET @rowcount = 937;
WHILE (@rowcount > 0)
  BEGIN
    UPDATE addresses
    SET    address_line_1 = Replace(x.address_line_1, ' ' + y.word_contains + ' ', ' ' + y.replace_word + ' ')
    FROM   addresses As x
     INNER
      JOIN tbl_address_replacement As y
        ON x.address_line_1 LIKE '% ' + y.word_contains + ' %'
    ;

    SET @rowcount = @@RowCount;
  END
;

我发现这是最有效的方法，因为它减少了所需的迭代次数。

基本上它会一直循环，直到update语句不再进行更改！

Answer 4

你需要一个游标来循环tbl_address_replacement

中的每一行

这是函数

create function CorrectifyInput(@input nvarchar(max)) returns nvarchar(max)
as
begin
declare @wordContains nvarchar(100), @replaceWord nvarchar(100)
declare cur cursor for select * from tbl_address_replacement
open cur
fetch next from cur into @wordContains, @replaceWord
while @@FETCH_STATUS = 0 begin
    set @input = replace(@input, @wordContains, @replaceWord)
    fetch next from cur into @wordContains, @replaceWord
end
return @input
end

Answer 5

它绝对不是最好的方式，但是从我的头脑中我想出了这个。

SET NOCOUNT ON
DECLARE @Input NVARCHAR(50)
DECLARE @AdressReplacement TABLE([Contains] NVARCHAR(50), [Replace] NVARCHAR(50))

INSERT INTO @AdressReplacement VALUES
    ('Building','Bldg.'),
    ('Building.','Bldg.')

SET @Input = 'Auriga Building. 4'

IF NOT EXISTS(SELECT [Replace] FROM @AdressReplacement WHERE @Input LIKE '%'+[Replace]+'%')
SET @Input = 
    REPLACE --After replacing old value with new value replace all '..' with '.'
    (
        REPLACE 
        (
            @Input,--What to replace
            (SELECT TOP 1[Contains] FROM @AdressReplacement WHERE @Input LIKE '%'+[Contains]+'%'),--Old value
            (SELECT TOP 1[Replace] FROM @AdressReplacement WHERE @Input LIKE '%'+[Contains]+'%')--New value
        ),
    '..',
    '.'
    )
PRINT @Input

Answer 6

如果您在地址中只有一个要替换的字符串，则可以使用我的示例：

UPDATE A 
SET A.address = REPLACE(A.address, AR2.word_contains, AR2.replace_word)
FROM #tbl_address AS A
CROSS APPLY (
    SELECT A.address, word_contains, AR.replace_word 
    FROM #tbl_address_replacement AS AR
    WHERE PATINDEX('%'+AR.word_contains+'%', A.address) > 0
        ) AS AR2;

对于变量替换，您可以使用：

SELECT @input = Replace(@input, word_contains, replace_word) FROM #tbl_address_replacement

如果有人会寻找其他选择（如果有人真的不喜欢循环）。替换我们可以与CTE一起使用：

;with Cte(a)  AS (
    SELECT address As [a] 
    FROM #tbl_address
    UNION ALL
    SELECT CAST(REPLACE(A.a, AR2.word_contains, AR2.replace_word) AS VARCHAR(200)) AS [a]
    FROM Cte AS A
    CROSS APPLY (
        SELECT A.a, word_contains, AR.replace_word 
        FROM #tbl_address_replacement AS AR
        WHERE PATINDEX('%'+AR.word_contains+'%', A.[a]) != 0
            ) AS AR2

)
SELECT A.a 
FROM Cte AS A
WHERE NOT EXISTS (
    SELECT 1 
    FROM #tbl_address_replacement AS AR
    WHERE PATINDEX('%'+AR.word_contains+'%', A.a) != 0
        ) ;

用于根据表中的值替换inputtring中的单词的函数

6 个答案: