Is there any function that can do .trim for specified characters or string?
Something like:
var x = '@@@hello world@@';
console.log(x.trim('@')); // prints 'hello world'
var y = 'hellohellohelloworld';
console.log(y.trim('hello')); // prints ' world'
var z = '@@hello@world@@';
console log(z.trim('@')); // prints 'hello@world'
Even tho I can do without this, it would be way less efficient and not as clean
You can use a pattern with an alternation | to match either what you want to remove at the start or at the end of the string by repeating it 1 or more time in a non capture group.
The repetition looks like this (?:@)+ for a single @ char, or like this (?:hello)+ for the word hello
If you want to make a function for it and want to pass any string, you have to escape the regex meta characters with a \
var x = '@@@hello world@@';
var y = 'hellohellohelloworld';
var z = '@@hello@world@@';
var a = '*+hello*+'
const customTrim = (strSource, strToRemove) => {
let escaped = strToRemove.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&');
return strSource.replace(new RegExp(`^(?:${escaped})+|(?:${escaped})+$`, 'g'), "")
};
console.log(customTrim(x, "\\s"));
console.log(customTrim(y, "hello"));
console.log(customTrim(z, "@"));
console.log(customTrim(a, "*+"));export function trim(str: string, char: string): string {
const leading = new RegExp(`^[${char}]+`);
const trailing = new RegExp(`[${char}]+$`);
return str.replace(leading, "").replace(trailing, "");
}