Replace the occurrences except the last one using Regex

Question

This is my regex pattern: [!@#$%^&*().{}/-]. I want to replace all the occurrences of these characters except the last occurrence. Please help me. Thanks.

Look into what a _positive lookahead_ is, you can replace patterns where there may exist a matching pattern infront of it. eg `[!@#$%^&*().{}\/-](?=.*[!@#$%^&*().{}\/-])` — Scuzzy, Oct 04 '19 at 07:33
There is no replacing in pure regex, you need to specify what regex engine you are using. — ruohola, Oct 04 '19 at 07:34

score 0 · Answer 1 · answered Oct 04 '19 at 08:01

Something like that might help:

[!@#$%^&*().\{\}\/\-](?=.*[!@#$%^&*().\{\}\/\-])

(?=.*[!@#$%^&*().\{\}\/-]) is a positive lookahead ((?= starts the positive lookahead). Meaning the pattern only matches if there is another of these characters somewhere ahead.

[!@#$%^&*().\{\}\/\-] your pattern that should be matched (some escapes would be needed)

(?= start of psotive lookahead (we need this after)

.* any number of arbitrary characters.

[!@#$%^&*().\{\}\/\-] again your pattern

) end of positive lookahead

The important part is that the lookahead is not part of the matched string as such.

score 0 · Answer 2 · answered Oct 04 '19 at 08:05

0

For an “universal” solution I propose you to do this:

[a-z]*([a-z])

Replaced by: $1

Adjust this solution for your case and language. You don’t need lookaheads

answered Oct 04 '19 at 08:05

YOGO

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Replace the occurrences except the last one using Regex

2 Answers2