See my code snippet below:
var list = [\'one\', \'two\', \'three\', \'four\'];
var str = \'one two, one three, one four, one\';
for ( var i = 0; i < list.
Simple solution would be to use substring method.
Since string is ending with list element, we can use string.length and calculate end index for substring without using lastIndexOf
method
str = str.substring(0, str.length - list[i].length) + "finish"
I would suggest using the replace-last npm package.
var str = 'one two, one three, one four, one';
var result = replaceLast(str, 'one', 'finish');
console.log(result);
<script src="https://unpkg.com/replace-last@latest/replaceLast.js"></script>
This works for string and regex replacements.
I know this is silly, but I'm feeling creative this morning:
'one two, one three, one four, one'
.split(' ') // array: ["one", "two,", "one", "three,", "one", "four,", "one"]
.reverse() // array: ["one", "four,", "one", "three,", "one", "two,", "one"]
.join(' ') // string: "one four, one three, one two, one"
.replace(/one/, 'finish') // string: "finish four, one three, one two, one"
.split(' ') // array: ["finish", "four,", "one", "three,", "one", "two,", "one"]
.reverse() // array: ["one", "two,", "one", "three,", "one", "four,", "finish"]
.join(' '); // final string: "one two, one three, one four, finish"
So really, all you'd need to do is add this function to the String prototype:
String.prototype.replaceLast = function (what, replacement) {
return this.split(' ').reverse().join(' ').replace(new RegExp(what), replacement).split(' ').reverse().join(' ');
};
Then run it like so:
str = str.replaceLast('one', 'finish');
One limitation you should know is that, since the function is splitting by space, you probably can't find/replace anything with a space.
Actually, now that I think of it, you could get around the 'space' problem by splitting with an empty token.
String.prototype.reverse = function () {
return this.split('').reverse().join('');
};
String.prototype.replaceLast = function (what, replacement) {
return this.reverse().replace(new RegExp(what.reverse()), replacement.reverse()).reverse();
};
str = str.replaceLast('one', 'finish');
Well, if the string really ends with the pattern, you could do this:
str = str.replace(new RegExp(list[i] + '$'), 'finish');
Old fashioned and big code but efficient as possible:
function replaceLast(origin,text){
textLenght = text.length;
originLen = origin.length
if(textLenght == 0)
return origin;
start = originLen-textLenght;
if(start < 0){
return origin;
}
if(start == 0){
return "";
}
for(i = start; i >= 0; i--){
k = 0;
while(origin[i+k] == text[k]){
k++
if(k == textLenght)
break;
}
if(k == textLenght)
break;
}
//not founded
if(k != textLenght)
return origin;
//founded and i starts on correct and i+k is the first char after
end = origin.substring(i+k,originLen);
if(i == 0)
return end;
else{
start = origin.substring(0,i)
return (start + end);
}
}
I did not like any of the answers above and came up with the below
function replaceLastOccurrenceInString(input, find, replaceWith) {
if (!input || !find || !replaceWith || !input.length || !find.length || !replaceWith.length) {
// returns input on invalid arguments
return input;
}
const lastIndex = input.lastIndexOf(find);
if (lastIndex < 0) {
return input;
}
return input.substr(0, lastIndex) + replaceWith + input.substr(lastIndex + find.length);
}
Usage:
const input = 'ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen twenty';
const find = 'teen';
const replaceWith = 'teenhundred';
const output = replaceLastOccurrenceInString(input, find, replaceWith);
console.log(output);
// output: ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteenhundred twenty
Hope that helps!