SQL trying to replace middle characters with * - sql

Question

I am trying to replace SQL results with all the middle values with asterix, *. All results are words. I am using SSMS.
The words that are 4-5 letters, it should only show 1 letter in the beginning, one to the end.
6 letters and more, it it should only show 2 letter in the beginning, 2 letters in the end.
1-3 letters, no replacement.
For example:
(I am now using - instead of * so it does not make the text bold).
"Banana" 6 letters should become ba--na
"False" 5 letters should become F---e
"a" stays the same
"Selin is a vegetable and banana is a fruit" becomes "S---n is a ve-----le and ba--na is a f---t."
What I have done so far, is to make this for emails, after the #. But now I want it to happen with every word of the result.
What I've done:
DECLARE #String VARCHAR(100) = 'sample#gmail.com'
SELECT STUFF(STUFF(#STring,
CHARINDEX('#',#String)+2,
(CHARINDEX('.',#String, CHARINDEX('#',#String))-CHARINDEX('#',#String)-3),
REPLICATE('*',CHARINDEX('.',#String, CHARINDEX('#',#String))-CHARINDEX('#',#String)))
,2
,CHARINDEX('#',#String)-3
,REPLICATE('*',CHARINDEX('#',#String)-3))```
With result s----e#g------l.com
instead of -
And I tried the mask method
Select
--select first character from Email and use replicate
SUBSTRING(Sxolia,1,1) + REPLICATE('*',5)+
--function to put asterisks
SUBSTRING(Sxolia,CHARINDEX('#',Sxolia),len(Sxolia)-CHARINDEX('#',Sxolia)+1)
--at this statement i select this part #gmail,com and to first part to become like this A*****#gmail.com
as Emailmask
From [mytable]
With result
B***** Bana is a fruit
And
declare #str nvarchar(max)
select #str = '123456'
select '****' + substring(#str, 5, len(#str) - 3)
Result: ****56
Not what I am looking for.
How should I look into this?

Answer 1

If I had to deal with this in SQL Server I'd operate on each word as a row, however using string_split is not (currently) an option since it does not guarantee ordering.
The following uses json to split the string as an array and provides a key value for ordering, which allows the words to be aggregated in the correct order:
select t.Sentence,
String_Agg( masked, ' ') within group(order by seq) Masked
from t
cross apply (
select seq, [value] word,
case
when l<=3 then [value]
when l<=5 then Stuff([value],2,l-2,Replicate('*',l-2))
else
Stuff([value],3,l-4,Replicate('*',l-4))
end Masked
from (
select j.[value], 1 + Convert(tinyint,j.[key]) Seq
from OpenJson(Concat('["',replace(t.Sentence,' ', '","'),'"]')) j
)w
cross apply (values(Len([value])))x(l)
)w
group by t.Sentence;
See working demo
Result:

Answer 2

I'm not sure how e-mail fits into all this because you're asking for word masks, so I'm going to assume you actually want this. Use divide and conquer to implement this, so first implement an expression that would do this for simplest cases (e.g. single words). Then if you need it for e-mails, just split the e-mails however you see fit and then apply the same expression.
The expression itself is rather simple:
SELECT *
FROM (VALUES
('banana'),
('selin'),
('vegetable')
) words(word)
CROSS
APPLY (SELECT CASE
WHEN ln BETWEEN 4 AND 5
THEN LEFT(word, 1) + REPLICATE('*', ln-2) + RIGHT(word, 1)
WHEN ln >= 6
THEN LEFT(word, 2) + REPLICATE('*', ln-4) + RIGHT(word, 2)
ELSE word
END as result
FROM (VALUES (LEN(words.word))) x(ln)
) calc
This already provides the expected result. You could define a function out of this, if you have the permissions, and use it like so:
SELECT *
FROM (VALUES
('banana'),
('selin'),
('vegetable')
) words(word)
CROSS
APPLY fnMaskWord(word)
Here's a working demo on dbfiddle, it includes the statement to create the function.

Answer 3

Expanding on a few answers:
select case when len(#String) <= 3 then #String
when len(#String) > 3 AND len(#String) <= 5 then
substring(#String, 1, 2) +
REPLICATE('*', Len(#String) - 2) +
substring(#String, Len(#String) - 1, 2)
when len(#String) >= 6 then
substring(#String, 1, 2) +
REPLICATE('*', Len(#String) - 2) +
substring(#String, Len(#String) - 1, 2)
else 'unrecognized length!'
If the length of the string is less than or equal to 3, return the string.
If the length of the string is more than 3 and less than or equal to 5 then create a substring starting at position 1, then replicate * by the length of the string -2 and finally add another substring -1 from the end of the string.
Similar for if the result is over 6 characters.
Else unrecognized length!
Hope this helps understand what's going on!

Answer 4

Maybe this can help
declare #t table (word varchar(50))
insert into #t values ('banana'), ('selin'), ('vegetable')
select case when len(t.word) < 3 then t.word
else left(t.word, 1) + -- take first char from left
replicate('*', Len(t.word) - 2) + -- fill middle with *
right(t.word, 1) -- take last char from right
end
from #t t
this returns
COLUMN1
b****a
s***l
v*******e
If you want to keep 2 chars left and right when the len > 5 then maybe this
select case when len(t.word) < 3 then t.word
when len(t.word) < 6 then
left(t.word, 1) +
replicate('*', len(t.word) - 2) +
right(t.word, 1)
else left(t.word, 2) +
replicate('*', len(t.word) - 4) +
right(t.word, 2)
end
from #t t
The result
COLUMN1
ba**na
s***l
ve*****le
EDIT: What if there is a whole sentence ?
Well then we first split the sentence in words,
and then concat the individual words back together while putting the ** in them
declare #t table (word varchar(50))
insert into #t values ('banana'), ('selin'), ('vegetable'), ('Banana is a fruit')
select t.word,
-- put the words back togheter into the sentence, and ** them while we are at it
( select string_agg(case when len(value) < 3 then value
when len(value) < 6 then
left(value, 1) +
replicate('*', len(value) - 2) +
right(value, 1)
else left(value, 2) +
replicate('*', len(value) - 4) +
right(value, 2)
end,
' ')
)
from #t t
cross apply string_split(t.word, ' ') s -- split the sentence into words
group by t.word
the result is
word COLUMN1
---- -------
banana ba**na
Banana is a fruit Ba**na is a f***t
selin s***n
vegetable ve*****le