Well, if the string really ends with the pattern, you could do this:

str = str.replace(new RegExp(list[i] + '$'), 'finish');

You can use a regular expression.

This will remove the last underscore:

var str = 'a_b_c';
console.log(  str.replace(/_([^_]*)$/, '$1')  ) //a_bc

This will replace it with the contents of the variable replacement:

var str = 'a_b_c',
    replacement = '!';

console.log(  str.replace(/_([^_]*)$/, replacement + '$1')  ) //a_b!c

Here is an idea ....

This is a case-sensitive string search version

var str = 'abc def abc xyz';
var word = 'abc';
var newWord = 'test';

// find the index of last time word was used
// please note lastIndexOf() is case sensitive
var n = str.lastIndexOf(word);

// slice the string in 2, one from the start to the lastIndexOf
// and then replace the word in the rest
str = str.slice(0, n) + str.slice(n).replace(word, newWord);
// result abc def test xyz

If you want a case-insensitive version, then the code has to be altered. Let me know and I can alter it for you. (PS. I am doing it so I will post it shortly)

Update: Here is a case-insensitive string search version

var str = 'abc def AbC xyz';
var word = 'abc';
var newWord = 'test';

// find the index of last time word was used
var n = str.toLowerCase().lastIndexOf(word.toLowerCase());

// slice the string in 2, one from the start to the lastIndexOf
// and then replace the word in the rest
var pat = new RegExp(word, 'i')
str = str.slice(0, n) + str.slice(n).replace(pat, newWord);
// result abc def test xyz

N.B. Above codes looks for a string. not whole word (ie with word boundaries in RegEx). If the string has to be a whole word, then it has to be reworked.

Update 2: Here is a case-insensitive whole word match version with RegEx

var str = 'abc def AbC abcde xyz';
var word = 'abc';
var newWord = 'test';

var pat = new RegExp('(\\b' + word + '\\b)(?!.*\\b\\1\\b)', 'i');
str = str.replace(pat, newWord);
// result abc def test abcde xyz

Good luck :)

You can use this replace:

var str = '12-44-12-1564';
str = str.replace(/12(?![\s\S]*12)/, 'aa');
console.log(str);

explanations:

(?!            # open a negative lookahead (means not followed by)
   [\s\S]*     # all characters including newlines (space+not space)
               # zero or more times
   12
)              # close the lookahead

In other words the pattern means: 12 not followed by another 12 until the end of the string.

foo.replace(/,([^,]*)$/, ' and $1')

use the $ (end of line) anchor to give you your position, and look for a pattern to the right of the comma index which does not include any further commas.

Edit:

The above works exactly for the requirements defined (though the replacement string is arbitrarily loose) but based on criticism from comments the below better reflects the spirit of the original requirement.

console.log( 
   'test1, test2, test3'.replace(/,\s([^,]+)1') 
)

Why last 'AND' is not removed/replaced?

Because you are trying to replace the output of lastIndexOf which is an integer

Use replace with regex

obj.where_str = obj.where_str.replace( /(AND)\s*$/, "" )

/(AND)\s*$/ will replace AND near the end of the string may (or may not) be followed by spaces.

Please try below solution this should solve your problem:

function replaceLastMatch(path, regex, replacement) {
  const matches = path.match(regex);
  if (matches && matches.length > 0) {
    const lastMatch = matches[matches.length - 1];
    const lastIndex = path.lastIndexOf(lastMatch);
    return path.substring(0, lastIndex) + path.substring(lastIndex).replace(lastMatch, replacement);
  }
  return path;
}

const path = '/:demo/:demo1/:demo2';
const regex = /\/:[^/]+(?=\/|$)/g;
const replacement = '/(?<pathinfo>[^/]+)';
const result = replaceLastMatch(path, regex, replacement);
console.log(result); // Output: /:demo/:demo1/(?<pathinfo>[^/]+)