Regular expression except first

Question

Hi I have string that could be like this:

%a%%%%%bc%%%d%

I need to replace the first % and the last %, at the start and at the end of the string with "". Also i need to replace all the %+ except the first one for each sequence group of string.

Result should be like this:

a%bc%d

How can I do with regex ? I tried something like this:.*?%(\W+)% but it didn't work

Thanks.

Answer 1

You may use

.replaceAll("^%|%$|(%)+", "$1")

See the regex demo

Details:

• ^% - the % at the start of the string is matched
• | - or
• %$ - a % at the string end
• | - or
• (%)+ - a repeated capturing group that matches 1+ % symbols, but captures only one % at each iteration into Group 1, and later, the $1 backreference to this group replaces multiple % chars with 1 %.