PATINDEX pattern to replace character that is *not* first character

Question

I'm trying to paritally 'leet-ify' a word, by taking a single character from a pre-defined set of replacable characters, and replacing it with a number/special character that resembles the letter being replaced. I don't want to replace the first character though.

Earlier I get a word that I've ensured has at least one letter after the first that is within the set of replaceable letters, and place it in the @Word variable. I then use the following to replace one letter. (Using 'loveable' as an example)

DECLARE @Word varchar(8) = 'lovable';
DECLARE @ReplaceChar varchar(1);


SET @ReplaceChar = SUBSTRING(@Word,PATINDEX('[a-z][abeilost]',@Word)+1,1);
SET @ReplaceChar = 
    (SELECT CASE @ReplaceChar
            WHEN 'a' THEN '@'
            WHEN 'b' THEN '8'
            WHEN 'e' THEN '3'
            WHEN 'i' THEN '!'
            WHEN 'l' THEN '1'
            WHEN 'o' THEN '0'
            WHEN 's' THEN '$'
            WHEN 't' THEN '+'
        END 
    );

SET @Word = STUFF(@Word,PATINDEX('[a-z][abeilost]',@Word)+1,1,@ReplaceChar);

As I understand it, PATINDEX should be finding the starting location of the first any-letter-followed-by-a-matching-letter string, and SUBSTRING/STUFF explicitly adds 1 to that number before collecting or replacing, so I should never get 'lovable' changed to '1oveable'... But that's what I'm getting. What am I missing?

To clarify expected results:

Input	Output
loveable	l0veable
give	g1ve
shelter	sh3lter
grams	gr@ms
phrygian	phryg!an

Answer 1

After our talk in the comments above I'd suggest this approach:

DECLARE @Word varchar(8) = 'truth';

DECLARE @toBeReplaced VARCHAR(10)='abeilost';
DECLARE @replaceWith VARCHAR(10)='@83!10$+';

DECLARE @position INT=PATINDEX(CONCAT('%[',@toBeReplaced,']%'),SUBSTRING(@word,2,8000))+1;

SELECT STUFF(@word,@position,1,TRANSLATE(SUBSTRING(@word,@position,1),@toBeReplaced,@replaceWith));

The idea in short:

• We define your translate parameters.
• We find the position using PATINDEX() behind the first character.
• Now we can use STUFF() to replace exactly one character at the given position by its translation.

For the next time: It would help a lot if you'd provided some samples with the expected result.

UPDATE

Using this at a tabular result, you can avoid the declared variable and do this inline:

DECLARE @WordTable TABLE(SomeText varchar(8));
INSERT INTO @WordTable VALUES('truth'),('loveable');

DECLARE @toBeReplaced VARCHAR(10)='abeilost';
DECLARE @replaceWith VARCHAR(10)='@83!10$+';

--the new query

SELECT STUFF(wt.SomeText,pos,1,TRANSLATE(SUBSTRING(wt.SomeText,pos,1),@toBeReplaced,@replaceWith))
FROM @WordTable wt
CROSS APPLY(SELECT PATINDEX(CONCAT('%[',@toBeReplaced,']%'),SUBSTRING(wt.SomeText,2,8000))+1) A(pos);

Answer 2

SELECT PATINDEX('[a-z][abeilost]', @Word)

Is returning 0. According to the documentation 0 means not found. The reason it is not being found it that your search criteria is for 2 characters only, the first being a single character which matches [a-z], and the second being a single character which matches [abeilost]. If you want to allow more characters you need to extend the search expression e.g.

SELECT PATINDEX('[a-z][abeilost]%', @Word)

Although I expect that isn't doing what you want as I imagine you want to repeat the specified characters any number of times which isn't possible with PATINDEX.