正则表达式从字符串中删除前导和尾随特殊字符和空格

Question

I'm trying to find a regex to remove all leading/trailing spaces as well as all leading/trailing special characters.

If I have the string:

test = '~!#@$@  hello this is a #! test  ^^#!^^    '

I'd like it to return as:

'hello this is a #! test'

So special characters and spaces in between the first and last letter are preserved, but all leading/trailing ones are cut out. Right now I have this:

test.replace(/[^a-zA-Z ]/g,"").replace(/^\s+|\s+$/g, "")

which returns:

'hello this is a  test'

so it is removing all special characters and leading/trailing spaces. How can I preserve that "#?" between "a" and "test"?

Answer 1

By using the unicode flag u you can use the Unicode Categories for Letters \p{L} and Numbers \p{N} in a negative character class [^...] .
And match them at the start ^ or | at the end $ to get anything that's not a letter or number and is trailing or leading.

Pattern:

^[^\p{L}\p{N}]+|[^\p{L}\p{N}]+$

Test Snippet:

 let test = '~;#@$@ hello this is a #. test ^^#,^^ '; test = test.replace(/^[^\p{L}\p{N}]+|[^\p{L}\p{N}]+$/gu; ''); document.write('['+test+']');