I have an expression in JS to match all the words containing (for example) 'foo+', but I would like to discard those of them who end with some other pattern.
Is there a way to make this in only one regular expression?
example:
myExp = /foo+\w*/; /*should add something to avoid ending with 'ba+r'*/
myExp.matchAll('boo, fo, food, foobaaar, foobr');
should result in ["food", "foobr"] not including 'foobaaar'
You want to use a negative lookahead like (?!not_this), of course replace not_this with the appropriate pattern to exclude using word-break characters \b, or whatever fits your logic.
let myExp = /\bfoo+(?!\w*ba+r\b)\w*/g;
console.log(...' boo fo food foobaaar foobr foobaardd boofoo foodbar '.matchAll(myExp));You can use negative look ahead. To exclude cases where ba+r is not the ending, use \b in that look ahead.
let myExp = /foo+(?!\w*ba+r\b)\w*/g;
let input = 'boo, fo, food, foobaaar, foobr, foobaardd';
console.log(...input.match(myExp));