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问题:
I'm in the process of creating a temporary procedure in SQL because I have a value of a table which is written in markdown, so it appear as rendered HTML in the web browser (markdown to HTML conversion).
String of the column currently look like this:
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)
I'm currently working with bold and italic text. This mean (in the case of bold text) I will need to replace odd N times the pattern**
with<b>
and even times with</b>
.
I saw replace() but it perform the replacement on all the patterns of the string.
So How I can replace a sub-string only if it is odd or only it is even?
Update: Some peoples wonder what schemas I'm using so just take a look here.
One more extra if you want: The markdown style hyperlink to html hyperlink doesn't look so simple.
回答1:
Using theSTUFF
function and a simpleWHILE
loop:
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
And you can use it like this:
SELECT dbo.fn_OddEvenReplace(column, '**', '<b>', '</b>')
FROM table
UPDATE:
This is re-written as an 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
And to execute:
DECLARE @returnText nvarchar(500)
EXEC dbo.##sp_OddEvenReplace '**a** **b** **c**', '**', '<b>', '</b>', @returnText output
SELECT @returnText
回答2:
As per OP's request I have modified my earlier answer to perform as a temporary stored procedure. I have left my earlier answer as I believe the usage against a table of strings to be useful also.
If a Tally (or Numbers) table is known to already exist with at least 8000 values, then the marked section of the CTE can be omitted and the CTE reference tally replaced with the name of the existing Tally table.
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
Invoked like this:
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
It yields as output:
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.
回答3:
One option is to use a Regular Expression as it makes replacing such patterns very simple. RegEx functions are not built into SQL Server so you need to use SQL CLR, either compiled by you or from an existing library.
For this example I will use the SQL# (SQLsharp) library (which I am the author of) but the RegEx functions are available in the Free version.
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
);
The above pattern \*\*([^\*]*)\*\*
just looks for anything surrounded by double-asterisks. In this case you don't need to worry about odd / even. It also means that you won't get a poorly-formed <b>
-only tag if for some reason there is an extra **
in the string. I added two additional test cases to the original string: a complete set of **
around the word they
and an unmatched set of **
just before the word programming
. The output is:
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)
which renders as:
Questions about general computing hardware and software are off-topicfor Stack Overflow unless they directly involve tools used primarily for **programming. You may be able to get help on Super User
回答4:
This solution makes use of techniques described by Jeff Moden in this article on the Running Sum problem in SQL. This solution is lengthy, but by making use of the Quirky Update in SQL Server over a clustered index, holds the promise of being much more efficient over large data sets than cursor-based solutions.
Update - amended below to operate off a table of strings
Assuming the existence of a tally table created like this (with at least 8000 rows):
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
and a HtmlTagSpotter function defined like this:
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
then running the following SQL will perform the required substitution. Note that the inner join at the end prevents any trailing "odd" tags from being converted:
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
yielding:
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.