What regex trims arbitrary punctuation from the start and end of a string in Javascript?

Question

I have an input line that accepts a string of arbitrary length containing two entities: an identifier type from a small list, and an identifier that can contain letters, numbers, and punctuation. The two entities are separated by punctuation that might be, but is not limited to: | : / \\ , | : / \\ ,

I have some code that finds the identifier type in the supplied string:

 for(var i = 0; i < idTypes.length; i++){ var search = rawInput.toUpperCase().search(idTypes[i]); if (search >= 0){ var inputType = idTypes[i]; var regEx = new RegExp(inputType, "i") var inputContents = rawInput.replace(regEx,""); console.log("This is type " + inputType + " with contents " + inputContents); return [inputType,inputContents]; } } 

However, this does not capture the punctuation serving as separators:

if

rawinput = "T14 11/15/11 | WPK |"

then

inputContents == "T14 11/15/11 | |"

whereas I would like

inputContents == "T14 11/15/11"

Is there a regex that will strip out all leading or trailing punctuation and white space, but will preserve the punctuation in the middle?

Answer 1

(.+?)\\s+[^a-zA-Z0-9]+.*$

You can see its test at: https://regex101.com/r/stDlXi/1

If you wish to specifically look for punctuation marks, then this will also do: (.+?)\\s+[\\|\\\\/\\?!\\.,;:-]+.*$ . I have omitted brackets, guillemets, ellipsis, apostrophe, quotation marks etc. If you need to consider them you can include them inside the square brackets.