3

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，所以它在 Web 浏览器中显示为呈现的 HTML （markdown 到 HTML 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)

我目前正在使用粗体和斜体文本。这意味着（在粗体文本的情况下）我需要将奇数 N 次替换**为<b>，偶数次替换为</b>。
我看到了replace()但它对字符串的所有模式执行替换。

那么我如何才能仅在奇数或偶数时替换子字符串？

更新：有些人想知道我使用的是什么模式，所以看看这里。

如果您愿意，还可以多加一点：指向 html 超链接的 markdown 样式超链接看起来并不那么简单。

4

4 回答 4

2

使用STUFF函数和一个简单的WHILE循环：

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

你可以像这样使用它：

SELECT dbo.fn_OddEvenReplace(column, '**', '<b>', '</b>')
FROM table

更新：

这被重写为 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
于 2014-07-31T01:19:03.487 回答
1

根据 OP 的要求，我修改了我之前的答案以作为临时存储过程执行。我已经留下了我之前的答案，因为我相信对字符串表的使用也很有用。

如果已知一个 Tally（或 Numbers）表已存在至少 8000 个值，则可以省略 CTE 的标记部分，并将 CTE 参考计数替换为现有 Tally 表的名称。

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

它作为输出产生：

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.
于 2014-07-31T22:55:33.167 回答
0

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 函数未内置到 SQL Server 中，因此您需要使用 SQL CLR，无论是由您编译还是从现有库编译。

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
);

上面的模式\*\*([^\*]*)\*\*只是寻找被双星号包围的任何东西。在这种情况下，您无需担心奇数/偶数。这也意味着<b>如果由于某种原因**在字符串中有多余的标签，您将不会得到格式不正确的 -only 标记。我在原始字符串中添加了两个额外的测试用例：**围绕单词的完整集和在单词之前的they不匹配集。输出是：**programming

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：

除非直接涉及主要用于**编程的工具，否则有关通用计算硬件和软件的问题对于 Stack Overflow 来说是无关紧要的。您也许可以获得有关超级用户的帮助

于 2014-07-31T03:43:20.150 回答
0

此解决方案利用了 Jeff Moden 在这篇关于 SQL 中的运行总和问题的文章中描述的技术。这个解决方案很长，但是通过在聚集索引上使用 SQL Server 中的Quirky 更新，有望比基于游标的解决方案更高效地处理大型数据集。

更新- 在下面修改以操作字符串表

假设存在这样创建的计数表（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

和一个像这样定义的HtmlTagSpotter函数：

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，末尾的内部连接可防止转换任何尾随的“奇数”标签：

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

产生：

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.
于 2014-07-31T03:06:34.750 回答