如何在要删除的句子末尾删除带有 # 或 $ 的单词（正则表达式）

Question

I have the following:

The moment #BTC, we have all $BTC been waiting for:We are happy to announce NIX Platform is rebranding to.. $NBT > $VOICE $NIX > $MUTE $ETH $BTC #BTC

I would like to remove only the remove the words with # or $ to be removed from the end, not the middle, so the above string would look like this

The moment #BTC, we have all $BTC been waiting for:We are happy to announce NIX Platform is rebranding to..  >  > 

Currently I have #(??(:?hashtag)\b)[\w-]+(?=(::\s+#[\w-]+)*\s*$) as regex that removes words with # from end but not $, not sure what I need to include those as well

Answer 1

An alternative regex with a word boundary and negative lookahead to disallow any word character following a whitespace:

[#$@]\w+\b(?!\W*\s\w)

RegEx Demo

RegEx Explanation:

• [#$@] : Match # or $ or @
• \w+b : Match 1+ word characters followed by word boundary
• (?!\W*\s\w) : Negative lookahead to assert that we don't have a word character after a whitespace ahead of current position

Alternatively , you can also use an *atomic group` to disallow backtracking:

[#$@](?>\w+)(?!\W*\s\w)

Answer 2

You could use a positive lookahead and assert that until the end of the string all words start with a non word character.

[#$][^\s$#]+(?=(?:\s+[^\w\s]\S*)*\s*$)

• [#$][^\s$#]+ Match either # or $ and 1+ occurrences of any char except # or $
• (?= Positive lookahead, assert on the right
  • (?:\s+[^\w\s]\S*)* Optional repetitions of 1 or more whitespace chars followed by any char except a whitspace or char char
  • \s* Match optional trailing whitspace chars
  • $ End of string
• ) Close lookahead

Regex demo

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If the start with $ or # can contain the other one, another option could be using a tempered greedy token approach with a capture group and a backreference \1 .

To prevent partial matches, you might add asserting a whitespace boundary at the left.

(?<!\S)([#$])(?:(?!\1)\S)*(?=(?:\s+[^\w\s]\S*)*\s*$)

Regex demo