Oracle regular expression replace in string ignoring first and last characters

Question

regular expression replace in string ignoring first and last characters if they matched

my try

SELECT REGEXP_REPLACE ('TLYU TK0T23T', '[^(\A.)]T[^(.\Z)]', 'Qq' ) contain FROM dual; 

result

TLYUQqQq3T

expected result

TLYU QqK0Qq23T

Thanks in advance

Answer 1

This returns the answer you want:

SELECT REGEXP_REPLACE 
('TLYU TK0T23T', '([^(\A.)])T([^(.\Z)])', '1円Qq2円' ) contain 
FROM dual;

Bobby

• Bobby Durrett: You needed to add parentheses around the strings that came before and after the letter T that you were matching. Then in the replace string you refer to the things in parentheses with 1円 and 2円. I think that the parentheses within the square brackets are not used for this purpose. So, ([^(\A.)]) is 1円 and ([^(.\Z)]) is 2円. Is that clear?