How to remove specific character surrounding a string?
Search for character mask and return the rest without.
This proposal the use of the bitwise not ~ operator for checking.
~is a bitwise not operator. It is perfect for use withindexOf(), becauseindexOfreturns if found the index0 ... nand if not-1:value ~value boolean -1 => 0 => false 0 => -1 => true 1 => -2 => true 2 => -3 => true and so on
function trim(s, mask) { while (~mask.indexOf(s[0])) { s = s.slice(1); } while (~mask.indexOf(s[s.length - 1])) { s = s.slice(0, -1); } return s;}console.log(trim('??? this is a ? test ?', '? '));console.log(trim('abc this is a ? test abc', 'cba '));
Simply use:
let text = '?? something ? really ??'text = text.replace(/^([?]*)/g, '')text = text.replace(/([?]*)$/g, '')console.log(text)
A possible solution would be to use recursive functions to remove the unwanted leading and trailing characters. This doesn't use regular expressions.
function ltrim(char, str) { if (str.slice(0, char.length) === char) { return ltrim(char, str.slice(char.length)); } else { return str; }}function rtrim(char, str) { if (str.slice(str.length - char.length) === char) { return rtrim(char, str.slice(0, 0 - char.length)); } else { return str; }}Of course this is only one of many possible solutions. The function trim would use both ltrim and rtrim.
The reason that char is the first argument and the string that needs to be cleaned the second, is to make it easier to change this into a functional programming style function, like so (ES 2015):
function ltrim(char) { (str) => { <body of function> }}// No need to specify str herefunction ltrimSpaces = ltrim(' ');