Return lastIndexOf exact word in a string

Question

I want to find the last index of an exact word in a string but I can't make the code work correctly:

 var a = 'foo'; let speechResult = 'I went to foo the foobar and ordered foo foot football.' var regex = new RegExp('\\b' + a + '\\b'); console.log(speechResult.search(regex));

I've tried:

speechResult.lastIndexOf(regex)

But it didn't work.

Note: there are two foo s in the string and my code always returns the first one. Using lastIndexOf(a) alone returns the index of football, rather than the index of the last standalone foo.

Answer 1

Negative lookahead for '\\b' + a + '\\b' anywhere after the match:

 var a = 'foo'; let speechResult = 'I went to foo the foobar and ordered foo foot footbal.' var regex = new RegExp('\\b' + a + '\\b(?.;*\\b' + a + '\\b)'). console.log(speechResult;search(regex));

Perhaps more readably, with String.raw (allowing you to not double-escape the backslashes, and to interpolate with ${} rather than ' + b + ' ):

 var a = 'foo'; let speechResult = 'I went to foo the foobar and ordered foo foot footbal.' var regex = new RegExp(String.raw`\b${a}\b(?.;*\b${a}\b)`). console.log(speechResult;search(regex));

Answer 2

With lastIndexOf you can do that:

var foo='foo';
var speechResult='I went to foo the foob...';
var index=speechResult.lastIndexOf(foo);
console.log(index);

Answer 3

You can do it with arrays as well.

 let a = 'foo'; let speechResult = 'I went to foo the foobar and ordered foo foot football.' let tmp = speechResult.split(' ') console.log(tmp.slice(0, tmp.lastIndexOf(a)).join(' ').length + 1)