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What is the regex to remove last + sign from a string

I'm trying to generate a link using jQuery and need to trim the last '+' sign off the end. Is there a way to detect if there is one there, and then trim it off?
So far the code removes the word 'hotel' and replaces spaces with '+', I think I just need another replace for the '+' that shows up sometimes but not sure how to be super specific with it.
var nameSearch = name.replace("Hotel", "");
nameSearch = nameSearch.replace(/ /g, "+");

The answer to
What is the regex to remove last + sign from a string
is this
const str = "Hotel+"
const re = /\+$/; // remove the last plus if present. $ is "end of string"
console.log(str.replace(re,""))
The question is however if this is answering the actual problem at hand
If you have the string
"Ritz Hotel"
and you want to have
https://www.ritz.com
then you could trim the string:
const fullName = "Ritz Hotel",
name = fullName.replace("Hotel", "").trim().toLowerCase(),
link = `https://www.${name}.com`;
console.log(link)
// or if you want spaces to be converted in url safe format
const fullName1 = "The Ritz Hotel",
name1 = fullName1.replace("Hotel", "").trim().toLowerCase(),
link1 = new URL(`https://www.hotels.com/search?${name1}`).toString()
console.log(link1)

As an alternative to mplungjan's answer, you can use str.endsWith() for the check. If it ends on the + it will be cut out. There is no need for regex. If you can avoid regex you definitely should.
let str = "Hotel+";
if (str.endsWith("+")) {
str = str.substr(0, str.length - 1);
}
console.log(str);
Below you can find a function to replace all the whitespace characters with + excluding the last one:
const raw = "My Ho te l ";
function replaceSpacesWithPlus(raw) {
let rawArray = Array.from(raw);
let replArray = [];
for (let i = 0; i < rawArray.length; i++) {
const char = rawArray[i];
// handle characters 0 to n-1
if (i < rawArray.length - 1) {
if (char === ' ') {
replArray.push('+');
} else {
replArray.push(char);
}
} else {
// handle last char
if (char !== ' ' && char !== '+') {
replArray.push(char);
}
}
}
return replArray;
}
console.log(replaceSpacesWithPlus(raw));

The below snippet will remove all the existing + symbols from string.
let str = 'abcd + efg + hij';
str = str.replace(/\+/gm, '');
//output: abcd efg hij
For trim use the below snippet. It will remove the spaces from around the string.
let str = " Hello World!! "
str = str.trim();
// output: Hello World!!
If you want to replace the last + symbol only.
let str = 'abcd + efg + hij';
let lastIndex = str.lastIndexOf('+');
if (lastIndex > -1) {
let nextString = str.split('');
nextString.splice(lastIndex, 1, '');
str = nextString.join('');
}
// output: abcd + efg hij

How to replace a character with an specific indexOf to uppercase in a string?

Im trying to replace a character at a specific indexOf to uppercase.
My string is a surname plus the first letter in the last name,
looking like this: "lovisa t".
I check the position with this and it gives me the right place in the string. So the second gives me 8(in this case).
first = texten.indexOf(" ");
second = texten.indexOf(" ", first + 1);
And with this I replace the first letter to uppercase.
var name = texten.substring(0, second);
name=name.replace(/^./, name[0].toUpperCase());
But how do I replace the character at "second" to uppercase?
I tested with
name=name.replace(/.$/, name[second].toUpperCase());
But it did´t work, so any input really appreciated, thanks.

Your error is the second letter isn't in position 8, but 7.
Also this second = texten.indexOf(" ", first + 1); gives -1, not 8, because you do not have a two spaces in your string.
If you know that the string is always in the format surname space oneLetter and you want to capitalize the first letter and the last letter you can simply do this:
var name = 'something s';
name = name[0].toUpperCase() + name.substring(1, name.length - 1) + name[name.length -1].toUpperCase();
console.log(name)

Here's a version that does exactly what your question title asks for: It uppercases a specific index in a string.
function upperCaseAt(str, i) {
return str.substr(0, i) + str.charAt(i).toUpperCase() + str.substr(i + 1);
}
var str = 'lovisa t';
var i = str.indexOf(' ');
console.log(upperCaseAt(str, i + 1));
However, if you want to look for specific patterns in the string, you don't need to deal with indices.
var str = 'lovisa t';
console.log(str.replace(/.$/, function (m0) { return m0.toUpperCase(); }));
This version uses a regex to find the last character in a string and a replacement function to uppercase the match.
var str = 'lovisa t';
console.log(str.replace(/ [a-z]/, function (m0) { return m0.toUpperCase(); }));
This version is similar but instead of looking for the last character, it looks for a space followed by a lowercase letter.
var str = 'lovisa t';
console.log(str.replace(/(?:^|\s)\S/g, function (m0) { return m0.toUpperCase(); }));
Finally, here we're looking for (and uppercasing) all non-space characters that are preceded by the beginning of the string or a space character; i.e. we're uppercasing the start of each (space-separated) word.

All can be done by regex replace.
"lovisa t".replace(/(^|\s)\w/g, s=>s.toUpperCase());

Try this one (if it will be helpfull, better move constants to other place, due performance issues(yes, regexp creation is not fast)):
function normalize(str){
var LOW_DASH = /\_/g;
var NORMAL_TEXT_REGEXP = /([a-z])([A-Z])/g;
if(!str)str = '';
if(str.indexOf('_') > -1) {
str = str.replace(LOW_DASH, ' ');
}
if(str.match(NORMAL_TEXT_REGEXP)) {
str = str.replace(NORMAL_TEXT_REGEXP, '$1 $2');
}
if(str.indexOf(' ') > -1) {
var p = str.split(' ');
var out = '';
for (var i = 0; i < p.length; i++) {
if (!p[i])continue;
out += p[i].charAt(0).toUpperCase() + p[i].substring(1) + (i !== p.length - 1 ? ' ' : '');
}
return out;
} else {
return str.charAt(0).toUpperCase() + str.substring(1);
}
}
console.log(normalize('firstLast'));//First Last
console.log(normalize('first last'));//First Last
console.log(normalize('first_last'));//First Last

Javascript - Regex to replace last 4 digit

I have a number variable at JavaScript and i want it replaced in last 4 character. Example:
I have a number 123456789 and i want it to be replaced like this 12345****
Is there any regex to do that in JavaScript?

Use replace() with regex /\d{4}$/
var res = '123456789'.replace(/\d{4}$/, '****');
document.write(res);
Regex explanation
Or using substring() or substr()
var str = '123456789',
res = str.substr(0, str.length - 4) + '****';
document.write(res);

You could use substring as well:
var s = '123456789';
var ns = s.substring(0, s.length - 4) + '****';
document.write(ns);

How to add newline after a repeating pattern

Assume that there is a string like this:
var content = "1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.17.18.19.20";
I want to add <br /> after every 5 dots.
So, the result should be:
1.2.3.4.5.<br />
6.7.8.9.10.<br />
11.12.13.14.15.<br />
16.17.18.19.20.<br />
I want to do this without a for loop. Is it possible with just regex?
i'm doing this with this code;
regenerate:function(content,call){
var data2;
var brbr = content.replace(/[\u0250-\ue007]/g, '').match(/(\r\n)/g);
if (brbr !== "") {
data2 = content.replace(/[\u0250-\ue007]/g, '').replace(/(\r\n)/gm, "<br><br>");
} else {
data2 = content.replace(/[\u0250-\ue007]/g, '');
}
var dataArr = data2.split(".");
for (var y = 10; y < dataArr.length - 10; y += 10) {
var dataArrSpecific1 = dataArr[y] + ".";
var dataArrSpecific2 = dataArr[y] + ".<br>";
var dataArrSpecificBosluk = dataArr[y + 1];
var data3 = data2.replace(new RegExp(dataArrSpecific1.replace(/[\u0250-\ue007]/g, ''), "g"), "" + dataArrSpecific2.replace(/[\u0250-\ue007]/g, '') + "");
data3 = data3.replace(new RegExp(dataArrSpecificBosluk.replace(/[\u0250-\ue007]/g, ''), "g"), " " + dataArrSpecificBosluk.replace(/[\u0250-\ue007]/g, '') + "");
data2 = data3;
}
call(data2.replace(/[\u0250-\ue007]/g, ''));
}
Actually , i want to refactoring this code

Working bin:http://jsbin.com/dikifipelo/1/
var string = "1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.17.18.19.20." ;
string = string.replace(/(([^\.]+\.){5})/g, "$1<br/>");
Works with any type and length of characters between the dots.
Explanation:
The pattern /(([^.]+.){5})/g can be broken down as such:
[^\.] - any character that is not a dot
[^\.]+ - any character that is not a dot, one or more times
[^\.]+\. - any character that is not a dot, one or more times, followed by a dot
([^\.]+\.){5} - any character....dot, appearing five times
(([^\.]+\.){5}) - any...five times, capture this (all round brackets capture unless told not to, with a ?: as the first thing inside them)
the /g/ flag makes it so that the whole string is matched - ie, all matches are found
"$1" represents the results of the first group (or bracket)
so, the replace function finds all instances of the pattern in the string, and replaces them with the match itself + a line break (br).
Once you learn regular expressions, life is never the same.

Get a number after a Specific Symbol in JQuery / Javascript

I have a String Like This:
"Dark Bronze - add $120.00"
I need to pull the 120 into a float number variable.
How would I do that?

var str = "Dark Bronze - add $120.00";
var val = str.match(/\$[0-9]*\.[0-9]*/)[0];
var f = Number(val.substring(1));
// (f is a number, do whatever you want with it)

var input = 'Dark Bronze - add $120.00',
toParse = input.substring(input.indexOf('$') + 1),
dollaz = parseFloat(toParse);
alert(dollaz);
Demo →

var str="Dark Bronze - add $120.00", val;
val = parseFloat(str.slice(str.indexOf('$')));
alert('The value is ' + val);

var str = 'Dark Bronze - add $120.00';
var pos = str.indexOf('$');
if (pos < 0) {
// string doesn't contain the $ symbol
}
else {
var val = parseFloat(str.substring(pos + 1));
// do something with val
}

var str = "Dark Bronze - add $120.00";
/*
[\$£¥€] - a character class with all currencies you are looking for
( - capture
\d+ - at least one digit
\. - a literal point character
\d{2} - exactly 2 digits
) - stop capturing
*/
var rxp = /[\$£¥€](\d+\.\d{2})/;
// the second member of the array returned by `match` contains the first capture
var strVal = str.match( rxp )[1];
var floatVal = parseFloat( strVal );
console.log( floatVal ); //120