如何替换字符串中的奇怪模式？
https://stackoverflow.com//questions/25048739
-
21-12-2019 - |
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我正在用 SQL 创建一个临时过程，因为我有一个用 markdown 编写的表的值，因此它在 Web 浏览器中显示为呈现的 HTML （Markdown 到 HTML 的转换）.
该列的字符串当前如下所示：
Questions about **general computing hardware and software** are off-topic for Stack Overflow unless they directly involve tools used primarily for programming. You may be able to get help on [Super User](http://superuser.com/about)
我目前正在处理粗体和斜体文本。这个意思 （在粗体文本的情况下） 我需要替换奇数 N 次的模式**和<b>甚至有时与</b>.
我看见 代替（） 但它对字符串的所有模式执行替换。
那么，如何仅在奇数或偶数时替换子字符串呢？
更新： 有些人想知道我正在使用什么模式，所以看看吧 这里.
如果你愿意的话，还有一个额外的： Markdown 风格的超链接到 html 超链接看起来并不那么简单。
解决方案
使用theSTUFFfenction和simpleWHILEloop：
CREATE FUNCTION dbo.fn_OddEvenReplace(@text nvarchar(500),
@textToReplace nvarchar(10),
@oddText nvarchar(10),
@evenText nvarchar(500))
RETURNS varchar(max)
AS
BEGIN
DECLARE @counter tinyint
SET @counter = 1
DECLARE @switchText nvarchar(10)
WHILE CHARINDEX(@textToReplace, @text, 1) > 0
BEGIN
SELECT @text = STUFF(@text,
CHARINDEX(@textToReplace, @text, 1),
LEN(@textToReplace),
IIF(@counter%2=0,@evenText,@oddText)),
@counter = @counter + 1
END
RETURN @text
END
，您可以使用它：
SELECT dbo.fn_OddEvenReplace(column, '**', '<b>', '</b>')
FROM table
更新：
这将重新编写为sp：
CREATE PROC dbo.##sp_OddEvenReplace @text nvarchar(500),
@textToReplace nvarchar(10),
@oddText nvarchar(10),
@evenText nvarchar(10),
@returnText nvarchar(500) output
AS
BEGIN
DECLARE @counter tinyint
SET @counter = 1
DECLARE @switchText nvarchar(10)
WHILE CHARINDEX(@textToReplace, @text, 1) > 0
BEGIN
SELECT @text = STUFF(@text,
CHARINDEX(@textToReplace, @text, 1),
LEN(@textToReplace),
IIF(@counter%2=0,@evenText,@oddText)),
@counter = @counter + 1
END
SET @returnText = @text
END
GO
并执行：
DECLARE @returnText nvarchar(500)
EXEC dbo.##sp_OddEvenReplace '**a** **b** **c**', '**', '<b>', '</b>', @returnText output
SELECT @returnText
其他提示
根据OP的请求，我修改了我的早期答案以执行临时存储过程。我留下了早期的答案，因为我相信对琴弦表的用法也很有用。
如果已知具有至少8000个值的计数器（或数字）表，则可以省略CTE的标记部分，并且CTE引用 Tally 替换为名称现有的Tally表。
create procedure #HtmlTagExpander(
@InString varchar(8000)
,@OutString varchar(8000) output
)as
begin
declare @Delimiter char(2) = '**';
create table #t(
StartLocation int not null
,EndLocation int not null
,constraint PK unique clustered (StartLocation desc)
);
with
-- vvv Only needed in absence of Tally table vvv
E1(N) as (
select 1 from (values
(1),(1),(1),(1),(1),
(1),(1),(1),(1),(1)
) E1(N)
), --10E+1 or 10 rows
E2(N) as (select 1 from E1 a cross join E1 b), --10E+2 or 100 rows
E4(N) As (select 1 from E2 a cross join E2 b), --10E+4 or 10,000 rows max
tally(N) as (select row_number() over (order by (select null)) from E4),
-- ^^^ Only needed in absence of Tally table ^^^
Delimiter as (
select len(@Delimiter) as Length,
len(@Delimiter)-1 as Offset
),
cteTally(N) AS (
select top (isnull(datalength(@InString),0))
row_number() over (order by (select null))
from tally
),
cteStart(N1) AS
select
t.N
from cteTally t cross join Delimiter
where substring(@InString, t.N, Delimiter.Length) = @Delimiter
),
cteValues as (
select
TagNumber = row_number() over(order by N1)
,Location = N1
from cteStart
),
HtmlTagSpotter as (
select
TagNumber
,Location
from cteValues
),
tags as (
select
Location = f.Location
,IsOpen = cast((TagNumber % 2) as bit)
,Occurrence = TagNumber
from HtmlTagSpotter f
)
insert #t(StartLocation,EndLocation)
select
prev.Location
,data.Location
from tags data
join tags prev
on prev.Occurrence = data.Occurrence - 1
and prev.IsOpen = 1;
set @outString = @Instring;
update this
set @outString = stuff(stuff(@outString,this.EndLocation, 2,'</b>')
,this.StartLocation,2,'<b>')
from #t this with (tablockx)
option (maxdop 1);
end
go
像这样调用：
declare @InString varchar(8000)
,@OutString varchar(8000);
set @inString = 'Questions about **general computing hardware and software** are off-topic **for Stack Overflow.';
exec #HtmlTagExpander @InString,@OutString out; select @OutString;
set @inString = 'Questions **about** general computing hardware and software **are off-topic** for Stack Overflow.';
exec #HtmlTagExpander @InString,@OutString out; select @OutString;
go
drop procedure #HtmlTagExpander;
go
它会产生输出：
Questions about <b>general computing hardware and software</b> are off-topic **for Stack Overflow.
Questions <b>about</b> general computing hardware and software <b>are off-topic</b> for Stack Overflow.
一种选择是使用正则表达式，因为它使替换此类模式变得非常简单。RegEx 函数未内置于 SQL Server 中，因此您需要使用 SQL CLR（由您编译或从现有库编译）。
对于这个例子，我将使用 SQL#（SQLsharp） 库（我是它的作者），但 RegEx 函数在免费版本中可用。
SELECT SQL#.RegEx_Replace
(
N'Questions about **general computing hardware and software** are off-topic\
for Stack Overflow unless **they** directly involve tools used primarily for\
**programming. You may be able to get help on [Super User]\
(https://superuser.com/about)', -- @ExpressionToValidate
N'\*\*([^\*]*)\*\*', -- @RegularExpression
N'<b>$1</b>', -- @Replacement
-1, -- @Count (-1 = all)
1, - @StartAt
'IgnoreCase' -- @RegEx options
);
上面的图案 \*\*([^\*]*)\*\* 只是寻找任何被双星号包围的东西。在这种情况下，您无需担心奇数/偶数。这也意味着你不会得到一个格式不良的 <b>- 仅当由于某种原因有额外的标签 ** 在字符串中。我在原始字符串中添加了两个额外的测试用例：一套完整的 ** 围绕这个词 they 以及一套无与伦比的 ** 就在这个词之前 programming. 。输出是：
Questions about <b>general computing hardware and software</b> are off-topicfor Stack Overflow unless <b>they</b> directly involve tools used primarily for **programming. You may be able to get help on [Super User](https://superuser.com/about)
呈现为：
有关问题 通用计算硬件和软件 与 Stack Overflow 无关，除非 他们 直接涉及主要用于**编程的工具。您也许可以获得以下方面的帮助 超级用户
此解决方案利用Jeff Moden中描述的技术在本文上的运行和SQL 中的问题。此解决方案冗长，但通过在SQL Server中使用群集索引的SQL Server中的 Quirky更新，持有比基于光标的解决方案更高的数据集更有效的承诺。
更新 - 修改下面以执行字符串
假设像这样创建的Tally表的存在（至少8000行）：
create table dbo.tally (
N int not null
,unique clustered (N desc)
);
go
with
E1(N) as (
select 1 from (values
(1),(1),(1),(1),(1),
(1),(1),(1),(1),(1)
) E1(N)
), --10E+1 or 10 rows
E2(N) as (select 1 from E1 a cross join E1 b), --10E+2 or 100 rows
E4(N) As (select 1 from E2 a cross join E2 b) --10E+4 or 10,000 rows max
insert dbo.tally(N)
select row_number() over (order by (select null)) from E4;
go
和 htmltagpotter 函数如下所定义：
create function dbo.HtmlTagSPotter(
@pString varchar(8000)
,@pDelimiter char(2))
returns table with schemabinding as
return
WITH
Delimiter as (
select len(@pDelimiter) as Length,
len(@pDelimiter)-1 as Offset
),
cteTally(N) AS (
select top (isnull(datalength(@pstring),0))
row_number() over (order by (select null))
from dbo.tally
),
cteStart(N1) AS (--==== Returns starting position of each "delimiter" )
select
t.N
from cteTally t cross join Delimiter
where substring(@pString, t.N, Delimiter.Length) = @pDelimiter
),
cteValues as (
select
ItemNumber = row_number() over(order by N1)
,Location = N1
from cteStart
)
select
ItemNumber
,Location
from cteValues
go
然后运行以下SQL将执行所需的替换。请注意，端部内连接可防止任何尾随“奇数”标记转换为：
create table #t(
ItemNo int not null
,Item varchar(8000) null
,StartLocation int not null
,EndLocation int not null
,constraint PK unique clustered (ItemNo,StartLocation desc)
);
with data(i,s) as ( select i,s from (values
(1,'Questions about **general computing hardware and software** are off-topic **for Stack Overflow.')
,(2,'Questions **about **general computing hardware and software** are off-topic **for Stack Overflow.')
--....,....1....,....2....,....3....,....4....,....5....,....6....,....7....,....8....,....9....,....0
)data(i,s)
),
tags as (
select
ItemNo = data.i
,Item = data.s
,Location = f.Location
,IsOpen = cast((TagNumber % 2) as bit)
,Occurrence = TagNumber
from data
cross apply dbo.HtmlTagSPotter(data.s,'**') f
)
insert #t(ItemNo,Item,StartLocation,EndLocation)
select
data.ItemNo
,data.Item
,prev.Location
,data.Location
from tags data
join tags prev
on prev.ItemNo = data.ItemNo
and prev.Occurrence = data.Occurrence - 1
and prev.IsOpen = 1
union all
select
i,s,8001,8002
from data
;
declare @ItemNo int
,@ThisStting varchar(8000);
declare @s varchar(8000);
update this
set @s = this.Item = case when this.StartLocation > 8000
then this.Item
else stuff(stuff(@s,this.EndLocation, 2,'</b>')
,this.StartLocation,2,'<b>')
end
from #t this with (tablockx)
option (maxdop 1);
select
Item
from (
select
Item
,ROW_NUMBER() over (partition by ItemNo order by StartLocation) as rn
from #t
) t
where rn = 1
go
屈服：
Item
------------------------------------------------------------------------------------------------------------
Questions about <b>general computing hardware and software</b> are off-topic **for Stack Overflow.
Questions <b>about </b>general computing hardware and software<b> are off-topic </b>for Stack Overflow.
不隶属于 StackOverflow
