remove word to left and right of character - sql-server-2008

Question

Using MS SQL 2008, I need to remove email addresses contained within a string of text. i.e.:
"This is a sample line of text with multiple email addresses like fred#example.com or I could also have bert#home.co.uk or even someone#somewhere.pl to mix things up".
The desired result is:
"This is a sample line of text with multiple email addresses like or I could also have or even to mix things up"
or even:
"This is a sample line of text with multiple email addresses like fred#***** or I could also have bert#***** or even someone#***** to mix things up".
There are many examples of removing everything to the left or right of a certain character, but not the removal of the text to the left or right of a fixed character up to the first space. Any help is appreciated.

Answer 1

Assuming the string of text is within a table column you can do this in SQL Server 2008 with the help of a string spitting function and string concatenation via for xml. Within newer versions of SQL Server - which you should really try to migrate towards - this is made easier with functions like string_agg instead of the stuff( ... for xml ) combo.
This implentation is greedy in that it will mask any combination of characters between two spaces (or the start/end of the string) that include at least one # character. So fullsentences-like^this'one%will£be$included,so\long)as/there>is.a=#*character&somewhere as well as strings made up of just one or more # character.
I'll leave how to deal with strings that aren't email addresses up to you.
Query
with t as
(
select *
from(values('This is a sample line of text with multiple email addresses like fred#example.com or I could also have bert#home.co.uk or even someone#somewhere.pl to mix things up')
,('This is another sample line of text with multiple email addresses like fred2#example2.com or I could also have bert2#home2.co.uk or even someone2#somewhere2.pl to mix things up')
,('fullsentences-like^this''one%will£be$included,so\long)as/there>is.a=#*character&somewhere')
,('#')
,('a ##### b')
,('# # # # #')
,('Let''s meet # the beach')
) as s(s)
)
,s as
(
select t.s
,s.rn
,case when charindex('#',s.item) > 0 then '***' else s.item end as item
from t
cross apply dbo.fn_StringSplit4k(t.s,' ',null) as s
)
select t.s
,stuff((select ' ' + s.item
from s
where t.s = s.s
order by s.rn
for xml path('')
)
,1,1,''
) as s
from t;
Output
s
s
This is a sample line of text with multiple email addresses like fred#example.com or I could also have bert#home.co.uk or even someone#somewhere.pl to mix things up
This is a sample line of text with multiple email addresses like *** or I could also have *** or even *** to mix things up
This is another sample line of text with multiple email addresses like fred2#example2.com or I could also have bert2#home2.co.uk or even someone2#somewhere2.pl to mix things up
This is another sample line of text with multiple email addresses like *** or I could also have *** or even *** to mix things up
fullsentences-like^this'one%will£be$included,so\long)as/there>is.a=#*character&somewhere
***
#
***
a ##### b
a *** b
# # # # #
*** *** *** *** ***
Let's meet # the beach
Let's meet *** the beach
String Splitting Function
create function [dbo].[fn_StringSplit4k]
(
#str nvarchar(4000) = ' ' -- String to split.
,#delimiter as nvarchar(1) = ',' -- Delimiting value to split on.
,#num as int = null -- Which value to return.
)
returns table
as
return
-- Start tally table with 10 rows.
with n(n) as (select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1)
-- Select the same number of rows as characters in #str as incremental row numbers.
-- Cross joins increase exponentially to a max possible 10,000 rows to cover largest #str length.
,t(t) as (select top (select len(isnull(#str,'')) a) row_number() over (order by (select null)) from n n1,n n2,n n3,n n4)
-- Return the position of every value that follows the specified delimiter.
,s(s) as (select 1 union all select t+1 from t where substring(isnull(#str,''),t,1) = #delimiter)
-- Return the start and length of every value, to use in the SUBSTRING function.
-- ISNULL/NULLIF combo handles the last value where there is no delimiter at the end of the string.
,l(s,l) as (select s,isnull(nullif(charindex(#delimiter,isnull(#str,''),s),0)-s,4000) from s)
select rn
,item
from(select row_number() over(order by s) as rn
,substring(#str,s,l) as item
from l
) a
where rn = #num
or #num is null;