Question

SELECT REPLACE('<strong>100</strong><b>.00 GB', '%^(^-?\d*\.{0,1}\d+$)%', '');

我想用上面的正则表达式替换数字的两个部分之间的任何标记，但它似乎不起作用。我不确定它是否是正则表达式语法是错误的，因为我尝试了更简单的语法，如'%[^0-9]%'只是为了测试，但它也没有用。有谁知道我怎么能实现这个目标？

Answer 1

您可以使用PATINDEX 找到模式（字符串）出现的第一个索引。然后使用STUFF将另一个字符串填充到匹配的模式（字符串）中。

遍历每一行。用你想要的东西替换每个非法字符。在您的情况下，将非数字替换为空白。内部循环是指当前单元格中有多个非法字符。

DECLARE @counter int

SET @counter = 0

WHILE(@counter < (SELECT MAX(ID_COLUMN) FROM Table))
BEGIN  

    WHILE 1 = 1
    BEGIN
        DECLARE @RetVal varchar(50)

        SET @RetVal =  (SELECT Column = STUFF(Column, PATINDEX('%[^0-9.]%', Column),1, '')
        FROM Table
        WHERE ID_COLUMN = @counter)

        IF(@RetVal IS NOT NULL)       
          UPDATE Table SET
          Column = @RetVal
          WHERE ID_COLUMN = @counter
        ELSE
            break
    END

    SET @counter = @counter + 1
END

警告：虽然这很慢！拥有varchar列可能会产生影响。所以使用LTRIM RTRIM可能会有所帮助。无论如何，它很慢。

积分转到this StackOverFlow回答。

EDIT 还可以转到@srutzky

编辑（由@Tmdean撰写）这个答案可以适应更基于集合的解决方案，而不是一次一行。它仍然迭代一行中非数字字符数的最大值，因此它并不理想，但我认为在大多数情况下它应该是可接受的。

WHILE 1 = 1 BEGIN
    WITH q AS
        (SELECT ID_Column, PATINDEX('%[^0-9.]%', Column) AS n
        FROM Table)
    UPDATE Table
    SET Column = STUFF(Column, q.n, 1, '')
    FROM q
    WHERE Table.ID_Column = q.ID_Column AND q.n != 0;

    IF @@ROWCOUNT = 0 BREAK;
END;

如果您在表中维护一个表示该字段是否已被清除的列，您还可以提高效率。（在我的示例中，NULL表示“未知”，应该是列默认值。）

DECLARE @done bit = 0;
WHILE @done = 0 BEGIN
    WITH q AS
        (SELECT ID_Column, PATINDEX('%[^0-9.]%', Column) AS n
        FROM Table
        WHERE COALESCE(Scrubbed_Column, 0) = 0)
    UPDATE Table
    SET Column = STUFF(Column, q.n, 1, ''),
        Scrubbed_Column = 0
    FROM q
    WHERE Table.ID_Column = q.ID_Column AND q.n != 0;

    IF @@ROWCOUNT = 0 SET @done = 1;

    -- if Scrubbed_Column is still NULL, then the PATINDEX
    -- must have given 0
    UPDATE table
    SET Scrubbed_Column = CASE
        WHEN Scrubbed_Column IS NULL THEN 1
        ELSE NULLIF(Scrubbed_Column, 0)
    END;
END;

如果您不想更改架构，则很容易适应将中间结果存储在表值变量中，该变量将在最后应用于实际表。

Answer 2

一般而言，SQL Server不支持正则表达式，您不能在本机T-SQL代码中使用它们。

您可以编写CLR函数来执行此操作。例如，请参阅here。

Answer 3

使用Replace(Column, BadFoundCharacter, '')而不是通过其唯一位置剥离找到的角色可能会快得多。此外，不是仅替换每列中的下一个坏字符，而是替换所有找到的字符。

WHILE 1 = 1 BEGIN
    UPDATE dbo.YourTable
    SET Column = Replace(Column, Substring(Column, PatIndex('%[^0-9.-]%', Column), 1), '')
    WHERE Column LIKE '%[^0-9.-]%'
    If @@RowCount = 0 BREAK;
END;

我确信这比接受的答案更有效，只是因为它的操作更少。还有其他方法也可能更快，但我现在没时间探索这些方法。

Answer 4

这是我根据以前的答案写的一个函数来完成这个。

CREATE FUNCTION dbo.RepetitiveReplace
(
    @P_String VARCHAR(MAX),
    @P_Pattern VARCHAR(MAX),
    @P_ReplaceString VARCHAR(MAX),
    @P_ReplaceLength INT = 1
)
RETURNS VARCHAR(MAX)
BEGIN
    DECLARE @Index INT;

    -- Get starting point of pattern
    SET @Index = PATINDEX(@P_Pattern, @P_String);

    while @Index > 0
    begin
        --replace matching charactger at index
        SET @P_String = STUFF(@P_String, PATINDEX(@P_Pattern, @P_String), @P_ReplaceLength, @P_ReplaceString);
        SET @Index = PATINDEX(@P_Pattern, @P_String);
    end

    RETURN @P_String;
END;

Gist

编辑：

最初我在这里有一个递归函数，它不能很好地与sql server一起使用，因为它有32个嵌套级别限制，当你尝试用函数进行32次替换时会导致类似下面的错误。而不是尝试进行服务器级别更改以允许更多嵌套（这可能是危险的，如允许永不结束循环）切换到while循环更有意义。

超出最大存储过程，函数，触发器或视图嵌套级别（限制32）。

Answer 5

我偶然发现了这篇文章寻找其他的东西，但我想我会提到一个更有效的解决方案 - 当与基于集合的查询一起使用时，它应该是任何函数的默认实现 - 这是使用交叉应用表函数。似乎该主题仍然有效，所以希望这对某人有用。

基于运行基于递归集的查询或标量函数的一些答案的示例运行时，基于1m行测试集从随机newid中删除字符，WHILE循环示例的范围从34s到2m05s函数示例的1m3s到{forever}。

使用具有交叉应用的表格功能可在 10s 中实现相同的目标。您可能需要调整它以满足您的需要，例如它处理的最大长度。

功能：

CREATE FUNCTION [dbo].[RemoveChars](@InputUnit VARCHAR(40))
RETURNS TABLE
AS
RETURN
    (
        WITH Numbers_prep(Number) AS
            (
                SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1
            )
        ,Numbers(Number) AS
            (
                SELECT TOP (ISNULL(LEN(@InputUnit),0))
                    row_number() OVER (ORDER BY (SELECT NULL))
                FROM Numbers_prep a
                    CROSS JOIN Numbers_prep b
            )
        SELECT
            OutputUnit
        FROM
            (
                SELECT
                    substring(@InputUnit,Number,1)
                FROM  Numbers
                WHERE substring(@InputUnit,Number,1) like '%[0-9]%'
                ORDER BY Number
                FOR XML PATH('')
            ) Sub(OutputUnit)
    )

用法：

UPDATE t
SET column = o.OutputUnit
FROM ##t t
CROSS APPLY [dbo].[RemoveChars](t.column) o

Answer 6

如果要重用它，将解决方案包装在SQL函数中可能很有用。我甚至在细胞水平上做这个，这就是为什么我把它作为一个不同的答案：

SELECT
  u.user_id as user_id,
  date(u.created) as signup_date,
  cal.date as date,
from (select date(dt) as date from [dw.calendar] where date(dt) < 
CURRENT_DATE() ) cal
  cross join each dw.user u
where
  date(u.created) <= cal.date

Answer 7

对于那些寻求高性能，简单解决方案并愿意启用CLR的人：

create database TestSQLFunctions
go
use TestSQLFunctions
go
alter database TestSQLFunctions set trustworthy on

EXEC sp_configure 'clr enabled', 1
RECONFIGURE WITH OVERRIDE
go

CREATE ASSEMBLY [SQLFunctions]
AUTHORIZATION [dbo]
FROM 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WITH PERMISSION_SET = SAFE

go

CREATE FUNCTION RegexReplace(
    @input nvarchar(max),
    @pattern nvarchar(max),
    @replacement nvarchar(max)
) RETURNS nvarchar  (max)
AS EXTERNAL NAME SQLFunctions.[SQLFunctions.Regex].Replace; 

go

-- outputs This is a test 
select dbo.RegexReplace('This is a test 12345','[0-9]','')

DLL的内容：

Answer 8

我认为这个解决方案更快更简单。我总是使用 CTE/递归，因为 mssql 上的速度太慢了。我在我处理的项目和大型数据库中使用它。

/*
Function:           dbo.kSql_ReplaceRegExp
Create Date:        20.02.2021
Author:             Karcan Ozbal

Description:        The given string value will be replaced according to the given regexp/pattern.

Parameter(s):       @Value       : Value/Text to REPLACE.
                    @RegExp      : The regexp/pattern to be used for REPLACE operation.

Usage:              select dbo.kSql_ReplaceRegExp('2T3EST5','%[0-9]%')
Output:             'TEST'
*/
ALTER FUNCTION [dbo].[kSql_ReplaceRegExp](
    @Value nvarchar(max),
    @RegExp nvarchar(50)
)
RETURNS nvarchar(max)
AS
BEGIN
    DECLARE @Result nvarchar(max)

    ;WITH CTE AS (
        SELECT NUM = 1, VALUE = @Value, IDX = PATINDEX(@RegExp, @Value)
        UNION ALL
        SELECT NUM + 1, VALUE = REPLACE(VALUE, SUBSTRING(VALUE,IDX,1),''), IDX = PATINDEX(@RegExp, REPLACE(VALUE, SUBSTRING(VALUE,IDX,1),'')) 
        FROM CTE
        WHERE IDX > 0
    )
    SELECT TOP(1) @Result = VALUE 
    FROM CTE 
    ORDER BY NUM DESC
    OPTION (maxrecursion 0)

    RETURN @Result
END

Answer 9

如果您只是为进入存储过程的参数执行此操作，则可以使用以下命令：

while PatIndex('%[^0-9]%', @Param) > 0
    select  @Param = Replace(@Param, Substring(@Param, PatIndex('%[^0-9]%', @Param), 1), '')

Answer 10

我认为更简单，更快捷的方法是按字母表中的每个字符进行迭代：

DECLARE @i int
SET @i = 0

WHILE(@i < 256)
BEGIN  

    IF char(@i) NOT IN ('0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '.')      

      UPDATE Table SET Column = replace(Column, char(@i), '')

    SET @i = @i + 1

END

Answer 11

我创建了此函数来清理在时间字段中包含非数字字符的字符串。他们未添加分钟数时，时间中包含问号，例如20：??。函数遍历每个字符并替换？带有0：

 CREATE FUNCTION [dbo].[CleanTime]
(
    -- Add the parameters for the function here
    @intime nvarchar(10) 
)
RETURNS nvarchar(5)
AS
BEGIN
    -- Declare the return variable here
    DECLARE @ResultVar nvarchar(5)
    DECLARE @char char(1)
    -- Add the T-SQL statements to compute the return value here
    DECLARE @i int = 1
    WHILE @i <= LEN(@intime)
    BEGIN
    SELECT @char =  CASE WHEN substring(@intime,@i,1) like '%[0-9:]%' THEN substring(@intime,@i,1) ELSE '0' END
    SELECT @ResultVar = concat(@ResultVar,@char)   
    set @i  = @i + 1       
    END;
    -- Return the result of the function
    RETURN @ResultVar

END

Answer 12

我认为这更清楚：

ALTER FUNCTION [dbo].[func_ReplaceChars](
    @Value nvarchar(max),
    @Chars nvarchar(50)
)
RETURNS nvarchar(max)
AS
BEGIN
    DECLARE @cLen int = len(@Chars);
    DECLARE @curChar int = 0;

    WHILE @curChar<@cLen
    BEGIN
        set @Value = replace(@Value,substring(@Chars,@curChar,1),'');

        set @curChar = @curChar + 1;
    END;

    RETURN @Value
END

SQL替换函数里面的正则表达式模式？

12 个答案:

编辑：