SQL - 从 varchar 字符串中删除模式

Question

我们的一个应用程序在我们的一个列的末尾附加了注释，我们现在已将其从应用程序中删除，并希望从数据中删除现有的。 我在 Varchar (2000) String 中有一个模式：

• 2019 年 8 月 11 日收到信用（USERNAME @ 08/11/2019）未删除记录
• 请求删除发往地址的记录未由 (USERNAME @ 20/09/2019) 删除

注意事项：

• 模式可以在字符串中的任何位置
• 模式可以在字符串中多次出现
• 用户名不同
• 日期变更

不确定最好的方法是什么？

Answer 1

正如我所说，使用这样的 UDF：

CREATE FUNCTION dbo.F_REMOVE_RECORD_UNDELETE (@ROW VARCHAR(max))
RETURNS VARCHAR(max)
AS
BEGIN
   DECLARE @RESULT VARCHAR(max) = '';
   DECLARE @POSITION INT = CHARINDEX('record undeleted by (', @ROW);
   WHILE @POSITION > 0
   BEGIN
      SET @RESULT = @RESULT + LEFT(@ROW, @POSITION - 1);
      SET @ROW = RIGHT(@ROW, LEN(@ROW) - CHARINDEX(')', @ROW, @POSITION + 1));
      SET @POSITION = CHARINDEX('record undeleted by (', @ROW);
   END
   RETURN @RESULT;
END;
GO

修改很容易与 LIKE 组合：

CREATE TABLE T_TEST_STRING_TST
(TST_ID       INT IDENTITY PRIMARY KEY,
 TST_STRING   VARCHAR(8000));
GO

INSERT INTO T_TEST_STRING_TST VALUES
(' noting '),
('Credit received 08/11/2019 record undeleted by (USERNAME @ 08/11/2019) bla bla bla'),
('anyting record undeleted ()'),
('Request removal of ship to adress record undeleted by (USERNAME @ 20/09/2019)bolo bolo'),
('Request removal of ship to adress record undeleted by (USERNAME @ 20/09/2019)bolo bolo record undeleted by (USERNAME @ 11/12/2020) bolo bolo'),
('anyting record undeleted ( @ 08/11/2019)');

UPDATE T_TEST_STRING_TST
SET    TST_STRING =  dbo.F_REMOVE_RECORD_UNDELETE(TST_STRING)
WHERE  TST_STRING LIKE '%record undeleted by (% @ [0-3][0-9]/[0-1][0-9]/[1-2][0-9][0-9][0-9])%';

对于 SQL 2005，您必须将 UDF 重写为：

CREATE FUNCTION dbo.F_REMOVE_RECORD_UNDELETE (@ROW VARCHAR(max))
RETURNS VARCHAR(max)
AS
BEGIN
   DECLARE @RESULT VARCHAR(max);
   SET @RESULT = '';
   DECLARE @POSITION INT;
   SET POSITION = CHARINDEX('record undeleted by (', @ROW)
   WHILE @POSITION > 0
   BEGIN
      SET @RESULT = @RESULT + LEFT(@ROW, @POSITION - 1);
      SET @ROW = RIGHT(@ROW, LEN(@ROW) - CHARINDEX(')', @ROW, @POSITION + 1));
      SET @POSITION = CHARINDEX('record undeleted by (', @ROW);
   END
   RETURN @RESULT;
END;
GO

Answer 2

你已经被告知，SQL Server v2005 是你真正应该改变的东西。

我不确定这是否适用于 v2005，没有机会在任何地方进行测试，但您可以尝试一下：

DECLARE @tbl TABLE(ID INT IDENTITY, YourString NVARCHAR(MAX));
INSERT INTO @tbl(YourString) VALUES
 ('Credit received 08/11/2019 record undeleted by (USERNAME @ 08/11/2019) Some more to come record undeleted by (SOMEONEELSE @ 10/11/2019)')
,('Request removal of ship to adress record undeleted by (ONEMORE @ 20/09/2019)');

SELECT t.ID
      ,CAST(REPLACE(REPLACE(YourString,' record undeleted by','<!--'),')','-->') AS XML).value('.','nvarchar(max)')
FROM @tbl t;

简而言之：

• 我们使用CAST() 和一些字符串方法来
  • 将其转换为 XML 并
  • 将不需要的文本包装为 XML 注释。
• 现在我们可以使用.value() 来读取XML 的内容。注释被省略。

一个中间结果如下所示

Credit received 08/11/2019<!-- (USERNAME @ 08/11/2019--> Some more to come<!-- (SOMEONEELSE @ 10/11/2019-->

您可以看到，XML 注释中包含任何不需要的内容。

提示 1：假设使用 ) 作为替换字符是不够的。但在结果字符串中用) 重新替换--> 可能就足够了。

提示 2：从 v2008 开始，STUFF() 可以用注释结束标记替换（=覆盖）@ 之后给定数量的字符。在 v2005 中，您将不得不循环...

提示 3：如果 立即 CAST().value() 语法在 v2005 中不受支持，请尝试在没有 .value() 的情况下运行它。然后你可以分两步进行...

更新

对于我的“提示 1”，这将有所帮助：

SELECT t.ID
      ,     REPLACE(
       CAST(REPLACE(
            REPLACE(YourString,' record undeleted by','<!--')
                                                     ,')','-->') AS XML).value('.','nvarchar(max)')
                                                     ,'-->',')')
FROM @tbl t;

Answer 3

第二次尝试更新： 这次我创建了以下 UDF，它试图找出用户名并基于它替换模式。由于它是 UDF，因此它适用于模式的多次出现。我认为从速度上来说这不是一个好的解决方案。

GO

CREATE FUNCTION dbo.test(@input varchar(2000))
RETRUNS varchar(2000)
AS
BEGIN

DECLARE
@start_pos int = 0
,@end_pos1 int = 0
,@r varchar(2000) = ''
,@r1 varchar(2000) = ''
,@to_del varchar(2000) = ''
,@user_name varchar(2000) = ''

set @r = @input

WHILE patindex('%record undeleted by_(%_@_[0-3][0-9]/[0-1][0-9]/[0-9][0-9][0-9][0-9])%', @r) > 0
BEGIN
    set @start_pos = 0; set @r1 = ''; set @user_name = ''; set @end_pos1 = 0 set @to_del = '';
    
    set @start_pos = len('record undeleted by_(') + patindex('%record undeleted by_(%_@_[0-3][0-9]/[0-1][0-9]/[0-9][0-9][0-9][0-9])%', @r) 
    set @r1 = substring(@r, @start_pos, len(@r))
    set @user_name = ltrim(rtrim(substring(@r1, 1, patindex('%_@_[0-3][0-9]/[0-1][0-9]/[0-9][0-9][0-9][0-9])%', @r1))))
    set @end_pos1 = LEN(@user_name) + 15
    set @to_del = 'record undeleted by ('+ ltrim(rtrim(substring(@r1, 1, @end_pos1)))
    set @r = replace(@r, @to_del, '')
END
    RETURN @r
END

GO

以下是使用 UDF 的测试表和 SELECT 语句的代码。

declare @t table
(
id int identity(1,1)
,dscr varchar(2000)
)

insert into @t(dscr)
select 'Credit received 08/11/2019 record undeleted by (@ahmedhuq @ 08/11/2019)' union
select 'Request removal of ship to address record undeleted by (group1\user1 @ 20/09/2019)' union
select 'Request removal of ship to address record undeleted by (group1\user1 @ 20/09/2019) and the quick brown fox' union
select 'Request removal of ship to address record undeleted by (dir1\user2 @ 20/09/2019) and Credit received 08/11/2019 record undeleted by (@ahmedhuq @ 08/11/2019)' union
select 'Request removal of ship to address record undeleted' 

select * , dbo.test(dscr)
from @t where dscr like '%record undeleted by_(%_@_[0-3][0-9]/[0-1][0-9]/[0-9][0-9][0-9][0-9])%'

结果如下：

忽略尝试 1： 下面的脚本不能用作工作解决方案。如果模式只出现一次，它就可以正常工作。对于多次出现的模式，数据会损坏！

declare @t table
(
id int identity(1,1)
,dscr varchar(2000)
)

insert into @t(dscr)
select 'Credit received 08/11/2019 record undeleted by (@ahmedhuq @ 08/11/2019)' union
select 'Request removal of ship to address record undeleted by (group1\user1 @ 20/09/2019)' union
select 'Request removal of ship to address record undeleted by (group1\user1 @ 20/09/2019) and the quick brown fox' union

select 'Request removal of ship to address record undeleted by (dir1\user2 @ 20/09/2019) and Credit received 08/11/2019 record undeleted by (@ahmedhuq @ 08/11/2019)' union
select 'Request removal of ship to address record undeleted' 

select * from @t
where dscr like '%record undeleted by_(%_@___/__/____)%'

;with a as
(
    select * 
    ,patindex( '%record undeleted by_(%_@___/__/____)%', dscr) as start_pos
    ,patindex( reverse('%record undeleted by_(%_@___/__/____)%'), reverse(dscr)) as end_pos_reverse
    ,len(dscr) - patindex( reverse('%record undeleted by_(%_@___/__/____)%'), reverse(dscr)) + 2 as end_pos
    from 
    @t
)
,a1 as
(
    select *
    ,end_pos - start_pos as start_end_len
    from
    a
)
,a2 as
(
    select 
    *,substring(dscr, start_pos, start_end_len) replace_str
    from
    a1
)
select
*
,replace(dscr, replace_str, '') as result1
from
a2