SQL Server 2014 替换为正则表达式

时间:2023-02-23
本文介绍了SQL Server 2014 替换为正则表达式的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

替换列中字符串的最佳方法是什么:

假设我有一个列类型为 varchar 的表,可能的值包含:

'sample text min(my value) 用 ) 和 ('

注意:我的价值可能会有所不同.

现在我想将其替换为:

min(max(my value))

所以在这种情况下的最终值是:

'sample text min(max(my value)) 用 ) 和 ('

我想对整个表执行更新.

是否可以使用纯 T-SQL?

所以 => 之前和转换之后的两个示例行:

<代码>1.值为:min(10).=>值为: min(max(10))2. 样本 min(cat) =>样本最小值(最大值(猫))

基本上用min(max(value))替换所有出现的min(value),其中'value'可以是任何字符串

解决方案

首先你需要这个用户定义的函数来搜索用字符串替换模式:

CREATE FUNCTION dbo.PatternReplace(@InputString VARCHAR(4000),@Pattern VARCHAR(100),@ReplaceText VARCHAR(4000))返回 VARCHAR(4000)作为开始声明 @Result VARCHAR(4000) SET @Result = ''-- 匹配中的第一个字符声明@First INT-- 下一个开始搜索的字符声明 @Next INT SET @Next = 1-- 总字符串的长度 -- 如果@InputString 为 NULL,则为 8001声明@Len INT SET @Len = COALESCE(LEN(@InputString), 8001)-- 模式结束声明 @EndPattern INT而 (@Next <= @Len)开始SET @First = PATINDEX('%' + @Pattern + '%', SUBSTRING(@InputString, @Next, @Len))IF COALESCE(@First, 0) = 0 -- 不匹配 - 返回开始SET @Result = @Result +CASE --return NULL, 就像 REPLACE, 如果输入是 NULL当@InputString 为空时或 @Pattern 为空或 @ReplaceText 为 NULL THEN NULLELSE SUBSTRING(@InputString, @Next, @Len)结尾休息结尾别的开始-- 将匹配之前的字符连接到结果SET @Result = @Result + SUBSTRING(@InputString, @Next, @First - 1)SET @Next = @Next + @First - 1设置@EndPattern = 1-- 查找结束模式范围的开始而 PATINDEX(@Pattern, SUBSTRING(@InputString, @Next, @EndPattern)) = 0设置@EndPattern = @EndPattern + 1-- 查找模式范围的结尾而 PATINDEX(@Pattern, SUBSTRING(@InputString, @Next, @EndPattern)) >0AND @Len >= (@Next + @EndPattern - 1)SET @EndPattern = @EndPattern + 1-- 在模式末尾或@Next + @EndPattern = @LenSET @Result = @Result + @ReplaceTextSET @Next = @Next + @EndPattern - 1结尾结尾返回(@结果)结尾

阅读更多

min(xxx) 出现不止一次:

最后,您可以简单地更新您的表格,如下所示:

更新你的表SET YourColumn = REPLACE(dbo.PatternReplace(YourColumn, '%min(', 'min(max('),'min(max(' + SUBSTRING(SUBSTRING(YourColumn,CHARINDEX('min(', YourColumn)+ 4,LEN(YourColumn)- CHARINDEX('min(',YourColumn)), 1,CHARINDEX(')', YourColumn)- ( CHARINDEX('min(', YourColumn) + 4 ))+ ')','min(max(' + SUBSTRING(SUBSTRING(YourColumn,CHARINDEX('min(', YourColumn)+ 4,LEN(YourColumn)- CHARINDEX('min(',YourColumn)), 1,CHARINDEX(')', YourColumn)- ( CHARINDEX('min(', YourColumn) + 4 ))+ '))');

What is the best way to replace string in column like:

Let's say I have table with column type varchar and with the possible value that contains:

'sample text min(my value) continue with sample text with ) and ('

Note: my value can vary.

Now I would like to replace it with:

min(max(my value))

so that the final value would in that case be:

'sample text min(max(my value)) continue with sample text with ) and ('

I would like to perform update on the whole table.

Is it possible using pure T-SQL?

So two sample rows before => and after transformation:

1. the value is: min(10). => the value is: min(max(10))
2. sample min(cat) => sample min(max(cat)) 

Basically replace all occurrences of min(value) with min(max(value)) where 'value' can be any string

解决方案

First of all you need this user defined function to search for replacing a pattern with string:

CREATE FUNCTION dbo.PatternReplace
(
   @InputString VARCHAR(4000),
   @Pattern VARCHAR(100),
   @ReplaceText VARCHAR(4000)
)
RETURNS VARCHAR(4000)
AS
BEGIN
   DECLARE @Result VARCHAR(4000) SET @Result = ''
   -- First character in a match
   DECLARE @First INT
    -- Next character to start search on
    DECLARE @Next INT SET @Next = 1
    -- Length of the total string -- 8001 if @InputString is NULL
    DECLARE @Len INT SET @Len = COALESCE(LEN(@InputString), 8001)
    -- End of a pattern
    DECLARE @EndPattern INT

     WHILE (@Next <= @Len) 
     BEGIN
     SET @First = PATINDEX('%' + @Pattern + '%', SUBSTRING(@InputString, @Next, @Len))
      IF COALESCE(@First, 0) = 0 --no match - return
       BEGIN
          SET @Result = @Result + 
             CASE --return NULL, just like REPLACE, if inputs are NULL
                WHEN  @InputString IS NULL
                 OR @Pattern IS NULL
                 OR @ReplaceText IS NULL THEN NULL
           ELSE SUBSTRING(@InputString, @Next, @Len)
        END
     BREAK
  END
  ELSE
  BEGIN
     -- Concatenate characters before the match to the result
     SET @Result = @Result + SUBSTRING(@InputString, @Next, @First - 1)
     SET @Next = @Next + @First - 1

     SET @EndPattern = 1
     -- Find start of end pattern range
     WHILE PATINDEX(@Pattern, SUBSTRING(@InputString, @Next, @EndPattern)) = 0
        SET @EndPattern = @EndPattern + 1
     -- Find end of pattern range
     WHILE PATINDEX(@Pattern, SUBSTRING(@InputString, @Next, @EndPattern)) > 0
           AND @Len >= (@Next + @EndPattern - 1)
        SET @EndPattern = @EndPattern + 1

     --Either at the end of the pattern or @Next + @EndPattern = @Len
     SET @Result = @Result + @ReplaceText
     SET @Next = @Next + @EndPattern - 1
  END
      END
      RETURN(@Result)
   END

Read more here

After creating this function you can try this:

DECLARE @x VARCHAR(max)
SET @x = 'sample text min(my value) continue with sample text with ) and ('

DECLARE @val VARCHAR(max)

SET @val =  SUBSTRING(SUBSTRING(@x,CHARINDEX('min(',@x)+4,LEN(@x)-CHARINDEX('min(',@x)),1,CHARINDEX(')',@x)-(CHARINDEX('min(',@x)+4))

SELECT REPLACE(dbo.PatternReplace(@x,'%min(','min(max('),'min(max('+@val+')','min(max('+@val+'))')

And you can see that the output is:

Occurrence of min(xxx) more than once:

Finally you can simply update your table as below:

UPDATE  YourTable
SET     YourColumn = REPLACE(dbo.PatternReplace(YourColumn, '%min(', 'min(max('),
                   'min(max(' + SUBSTRING(SUBSTRING(YourColumn,CHARINDEX('min(', YourColumn)+ 4,LEN(YourColumn)- CHARINDEX('min(',YourColumn)), 1,CHARINDEX(')', YourColumn)- ( CHARINDEX('min(', YourColumn) + 4 ))+ ')',
                   'min(max(' + SUBSTRING(SUBSTRING(YourColumn,CHARINDEX('min(', YourColumn)+ 4,LEN(YourColumn)- CHARINDEX('min(',YourColumn)), 1,CHARINDEX(')', YourColumn)- ( CHARINDEX('min(', YourColumn) + 4 ))+ '))');

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