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Remove a character at a certain position in a string - javascript [duplicate]

This question already has answers here:
How can I remove a character from a string using JavaScript?
(22 answers)
Closed 5 years ago.
Is there an easy way to remove the character at a certain position in javascript?
e.g. if I have the string "Hello World", can I remove the character at position 3?
the result I would be looking for would the following:
"Helo World"
This question isn't a duplicate of How can I remove a character from a string using JavaScript?, because this one is about removing the character at a specific position, and that question is about removing all instances of a character.

It depends how easy you find the following, which uses simple String methods (in this case slice()).
var str = "Hello World";
str = str.slice(0, 3) + str.slice(4);
console.log(str)

You can try it this way:
var str = "Hello World";
var position = 6; // its 1 based
var newStr = str.substring(0, position - 1) + str.substring(position, str.length);
alert(newStr);
Here is a live example: http://jsbin.com/ogagaq

Turn the string into array, cut a character at specified index and turn back to string
let str = 'Hello World'.split('')
str.splice(3, 1)
str = str.join('')
// str = 'Helo World'.

If you omit the particular index character then use this method
function removeByIndex(str,index) {
return str.slice(0,index) + str.slice(index+1);
}
var str = "Hello world", index=3;
console.log(removeByIndex(str,index));
// Output: "Helo world"

var str = 'Hello World';
str = setCharAt(str, 3, '');
alert(str);
function setCharAt(str, index, chr)
{
if (index > str.length - 1) return str;
return str.substr(0, index) + chr + str.substr(index + 1);
}

you can use substring() method. ex,
var x = "Hello world"
var x = x.substring(0, i) + 'h' + x.substring(i+1);

Hi starbeamrainbowlabs ,
You can do this with the following:
var oldValue = "pic quality, hello" ;
var newValue = "hello";
var oldValueLength = oldValue.length ;
var newValueLength = newValue.length ;
var from = oldValue.search(newValue) ;
var to = from + newValueLength ;
var nes = oldValue.substr(0,from) + oldValue.substr(to,oldValueLength);
console.log(nes);
I tested this in my javascript console so you can also check this out
Thanks

var str = 'Hello World',
i = 3,
result = str.substr(0, i-1)+str.substring(i);
alert(result);
Value of i should not be less then 1.

How can i match and replace a string and its before one letter in javascript?

var str="itss[BACK][BACK][BACK][BACK][BACK][BACK] it's a test stringgg[BACK][BACK]";
var word = '[BACK]';
var substrings = str.split(word);
var cnt= substrings.length - 1;
for(var i = 0;i<cnt;i++){
str = str.replace(/.{1}\[BACK\]{1}/i,""); //remove backspace and one character before it.
}
The above script returns something like "[BACK it's a test string" I need to get this result as "it's a test string" please help me....

It's easier to do this without a regex actually.
String.prototype.replaceFromIndex=function(index, length, replace) {
return this.substr(0, index) + replace + this.substr(index+length);
}
var search = '[BACK]';
var str="itss[BACK][BACK][BACK][BACK][BACK][BACK] it's a test stringgg[BACK][BACK]";
while((index = str.indexOf(search)) >= 0){
str = str.replaceFromIndex(index-1, search.length+1, '');
}
alert(str);
Check http://jsfiddle.net/fRThH/2/ for a working example.
Wrap it in a function and you are ready to go!
Courtesy to Cem Kalyoncu ( https://stackoverflow.com/a/1431113/187018 ) for a slightly modified version of String.prototype.replaceAt

My idea is to count all the backspaces [BACK] and then replace them with an empty string one by one:
var str="itss[BACK][BACK][BACK][BACK][BACK][BACK] it's a test stringgg[BACK][BACK]";
var backspaces = str.match(/\[BACK\]/g).length;
for(i=0; i<backspaces; i++)
{
str = str.replace(/.?\[BACK\]/, '');
}
document.write( str );
working example: jsFiddle

If I understood correctly
var dat = str.split('[BACK]').filter(function(e){return e})[1];
here is the working demo.

One of the problems that I found out was that you didn't set a condition in which you would not have to remove the first character when the string '[BACK]' is in position zero.
Well, the solution I am posting here first search for the position of the first '[BACK]' string, and then creates a substring of the characters that we want to remove, so, if there is a character before the string '[BACK]', it is included in the substring. Then, the substring is removed from the main string, and it continues looping until all the '[BACK]' s are removed.
var str = "itss[BACK][BACK][BACK][BACK][BACK][BACK] it's a test stringgg[BACK][BACK]";
var word = '[BACK]';
var substrings = str.split(word);
var cnt = substrings.length - 1;
for (i = 0; i < cnt; i++) {
pos = str.search("[BACK]");
if (pos - 1 > 0) {
str = str.replace(str.substring(pos - 2, pos + 5), '');
} else {
str = str.replace(str.substring(pos - 1, pos + 5), '');
}
}
Here is the code in jsfiddle:

Remove/ truncate leading zeros by javascript/jquery

Suggest solution for removing or truncating leading zeros from number(any string) by javascript,jquery.

You can use a regular expression that matches zeroes at the beginning of the string:
s = s.replace(/^0+/, '');

I would use the Number() function:
var str = "00001";
str = Number(str).toString();
>> "1"
Or I would multiply my string by 1
var str = "00000000002346301625363";
str = (str * 1).toString();
>> "2346301625363"

Maybe a little late, but I want to add my 2 cents.
if your string ALWAYS represents a number, with possible leading zeros, you can simply cast the string to a number by using the '+' operator.
e.g.
x= "00005";
alert(typeof x); //"string"
alert(x);// "00005"
x = +x ; //or x= +"00005"; //do NOT confuse with x+=x, which will only concatenate the value
alert(typeof x); //number , voila!
alert(x); // 5 (as number)
if your string doesn't represent a number and you only need to remove the 0's use the other solutions, but if you only need them as number, this is the shortest way.
and FYI you can do the opposite, force numbers to act as strings if you concatenate an empty string to them, like:
x = 5;
alert(typeof x); //number
x = x+"";
alert(typeof x); //string
hope it helps somebody

Since you said "any string", I'm assuming this is a string you want to handle, too.
"00012 34 0000432 0035"
So, regex is the way to go:
var trimmed = s.replace(/\b0+/g, "");
And this will prevent loss of a "000000" value.
var trimmed = s.replace(/\b(0(?!\b))+/g, "")
You can see a working example here

parseInt(value) or parseFloat(value)
This will work nicely.

I got this solution for truncating leading zeros(number or any string) in javascript:
<script language="JavaScript" type="text/javascript">
<!--
function trimNumber(s) {
while (s.substr(0,1) == '0' && s.length>1) { s = s.substr(1,9999); }
return s;
}
var s1 = '00123';
var s2 = '000assa';
var s3 = 'assa34300';
var s4 = 'ssa';
var s5 = '121212000';
alert(s1 + '=' + trimNumber(s1));
alert(s2 + '=' + trimNumber(s2));
alert(s3 + '=' + trimNumber(s3));
alert(s4 + '=' + trimNumber(s4));
alert(s5 + '=' + trimNumber(s5));
// end hiding contents -->
</script>

Simply try to multiply by one as following:
"00123" * 1; // Get as number
"00123" * 1 + ""; // Get as string

1. The most explicit is to use parseInt():
parseInt(number, 10)
2. Another way is to use the + unary operator:
+number
3. You can also go the regular expression route, like this:
number.replace(/^0+/, '')

Try this,
function ltrim(str, chars) {
chars = chars || "\\s";
return str.replace(new RegExp("^[" + chars + "]+", "g"), "");
}
var str =ltrim("01545878","0");
More here

You should use the "radix" parameter of the "parseInt" function :
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/parseInt?redirectlocale=en-US&redirectslug=JavaScript%2FReference%2FGlobal_Objects%2FparseInt
parseInt('015', 10) => 15
if you don't use it, some javascript engine might use it as an octal
parseInt('015') => 0

If number is int use
"" + parseInt(str)
If the number is float use
"" + parseFloat(str)

const number = '0000007457841';
console.log(+number) //7457841;
OR number.replace(/^0+/, '')

Regex solution from Guffa, but leaving at least one character
"123".replace(/^0*(.+)/, '$1'); // 123
"012".replace(/^0*(.+)/, '$1'); // 12
"000".replace(/^0*(.+)/, '$1'); // 0

I wanted to remove all leading zeros for every sequence of digits in a string and to return 0 if the digit value equals to zero.
And I ended up doing so:
str = str.replace(/(0{1,}\d+)/, "removeLeadingZeros('$1')")
function removeLeadingZeros(string) {
if (string.length == 1) return string
if (string == 0) return 0
string = string.replace(/^0{1,}/, '');
return string
}

One another way without regex:
function trimLeadingZerosSubstr(str) {
var xLastChr = str.length - 1, xChrIdx = 0;
while (str[xChrIdx] === "0" && xChrIdx < xLastChr) {
xChrIdx++;
}
return xChrIdx > 0 ? str.substr(xChrIdx) : str;
}
With short string it will be more faster than regex (jsperf)

const input = '0093';
const match = input.match(/^(0+)(\d+)$/);
const result = match && match[2] || input;

Use "Math.abs"
eg: Math.abs(003) = 3;
console.log(Math.abs(003))

Replace last occurrence of character in string

Is there an easy way in javascript to replace the last occurrence of an '_' (underscore) in a given string?

You don't need jQuery, just 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

No need for jQuery nor regex assuming the character you want to replace exists in the string
Replace last char in a string
str = str.substring(0,str.length-2)+otherchar
Replace last underscore in a string
var pos = str.lastIndexOf('_');
str = str.substring(0,pos) + otherchar + str.substring(pos+1)
or use one of the regular expressions from the other answers
var str1 = "Replace the full stop with a questionmark."
var str2 = "Replace last _ with another char other than the underscore _ near the end"
// Replace last char in a string
console.log(
str1.substring(0,str1.length-2)+"?"
)
// alternative syntax
console.log(
str1.slice(0,-1)+"?"
)
// Replace last underscore in a string
var pos = str2.lastIndexOf('_'), otherchar = "|";
console.log(
str2.substring(0,pos) + otherchar + str2.substring(pos+1)
)
// alternative syntax
console.log(
str2.slice(0,pos) + otherchar + str2.slice(pos+1)
)

What about this?
function replaceLast(x, y, z){
var a = x.split("");
a[x.lastIndexOf(y)] = z;
return a.join("");
}
replaceLast("Hello world!", "l", "x"); // Hello worxd!

Another super clear way of doing this could be as follows:
let modifiedString = originalString
.split('').reverse().join('')
.replace('_', '')
.split('').reverse().join('')

Keep it simple
var someString = "a_b_c";
var newCharacter = "+";
var newString = someString.substring(0, someString.lastIndexOf('_')) + newCharacter + someString.substring(someString.lastIndexOf('_')+1);

var someString = "(/n{})+++(/n{})---(/n{})$$$";
var toRemove = "(/n{})"; // should find & remove last occurrence
function removeLast(s, r){
s = s.split(r)
return s.slice(0,-1).join(r) + s.pop()
}
console.log(
removeLast(someString, toRemove)
)
Breakdown:
s = s.split(toRemove) // ["", "+++", "---", "$$$"]
s.slice(0,-1) // ["", "+++", "---"]
s.slice(0,-1).join(toRemove) // "})()+++})()---"
s.pop() // "$$$"

Reverse the string, replace the char, reverse the string.
Here is a post for reversing a string in javascript: How do you reverse a string in place in JavaScript?

// Define variables
let haystack = 'I do not want to replace this, but this'
let needle = 'this'
let replacement = 'hey it works :)'
// Reverse it
haystack = Array.from(haystack).reverse().join('')
needle = Array.from(needle).reverse().join('')
replacement = Array.from(replacement).reverse().join('')
// Make the replacement
haystack = haystack.replace(needle, replacement)
// Reverse it back
let results = Array.from(haystack).reverse().join('')
console.log(results)
// 'I do not want to replace this, but hey it works :)'

This is very similar to mplungjan's answer, but can be a bit easier (especially if you need to do other string manipulation right after and want to keep it as an array)
Anyway, I just thought I'd put it out there in case someone prefers it.
var str = 'a_b_c';
str = str.split(''); //['a','_','b','_','c']
str.splice(str.lastIndexOf('_'),1,'-'); //['a','_','b','-','c']
str = str.join(''); //'a_b-c'
The '_' can be swapped out with the char you want to replace
And the '-' can be replaced with the char or string you want to replace it with

You can use this code
var str="test_String_ABC";
var strReplacedWith=" and ";
var currentIndex = str.lastIndexOf("_");
str = str.substring(0, currentIndex) + strReplacedWith + str.substring(currentIndex + 1, str.length);
alert(str);

This is a recursive way that removes multiple occurrences of "endchar":
function TrimEnd(str, endchar) {
while (str.endsWith(endchar) && str !== "" && endchar !== "") {
str = str.slice(0, -1);
}
return str;
}
var res = TrimEnd("Look at me. I'm a string without dots at the end...", ".");
console.log(res)

Replace all spaces in a string with '+' [duplicate]

This question already has answers here:
How do I replace all occurrences of a string in JavaScript?
(78 answers)
Closed 7 years ago.
I have a string that contains multiple spaces. I want to replace these with a plus symbol. I thought I could use
var str = 'a b c';
var replaced = str.replace(' ', '+');
but it only replaces the first occurrence. How can I get it replace all occurrences?

You need the /g (global) option, like this:
var replaced = str.replace(/ /g, '+');
You can give it a try here. Unlike most other languages, JavaScript, by default, only replaces the first occurrence.

Here's an alternative that doesn't require regex:
var str = 'a b c';
var replaced = str.split(' ').join('+');

var str = 'a b c';
var replaced = str.replace(/\s/g, '+');

You can also do it like:
str = str.replace(/\s/g, "+");
Have a look at this fiddle.

Use global search in the string. g flag
str.replace(/\s+/g, '+');
source: replaceAll function

Use a regular expression with the g modifier:
var replaced = str.replace(/ /g, '+');
From Using Regular Expressions with JavaScript and ActionScript:
/g enables "global" matching. When using the replace() method, specify this modifier to replace all matches, rather than only the first one.

You need to look for some replaceAll option
str = str.replace(/ /g, "+");
this is a regular expression way of doing a replaceAll.
function ReplaceAll(Source, stringToFind, stringToReplace) {
var temp = Source;
var index = temp.indexOf(stringToFind);
while (index != -1) {
temp = temp.replace(stringToFind, stringToReplace);
index = temp.indexOf(stringToFind);
}
return temp;
}
String.prototype.ReplaceAll = function (stringToFind, stringToReplace) {
var temp = this;
var index = temp.indexOf(stringToFind);
while (index != -1) {
temp = temp.replace(stringToFind, stringToReplace);
index = temp.indexOf(stringToFind);
}
return temp;
};

NON BREAKING SPACE ISSUE
In some browsers
(MSIE "as usually" ;-))
replacing space in string ignores the non-breaking space (the 160 char code).
One should always replace like this:
myString.replace(/[ \u00A0]/, myReplaceString)
Very nice detailed explanation:
http://www.adamkoch.com/2009/07/25/white-space-and-character-160/

Do this recursively:
public String replaceSpace(String s){
if (s.length() < 2) {
if(s.equals(" "))
return "+";
else
return s;
}
if (s.charAt(0) == ' ')
return "+" + replaceSpace(s.substring(1));
else
return s.substring(0, 1) + replaceSpace(s.substring(1));
}