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Reverse a string in JavaScript without using an array

A simple way of reversing a string is as below:
const test = 'hello';
let i = 0;
let j = test.length - 1;
while (i < j) {
let temp = test[i];
test[j] = test[i];
test[i] = temp;
i++;
j--;
}
console.log(test);
If we try to access string using an index it works fine. For example console.log(test[2]) returns 'l'
But reversing a string using the method above returns unchanged string 'hello'. We need to use an array, reverse it and then join it to return the reversed string. But in that case we will be using an extra space. Can we do it without using an extra space?

Strings are immutable in JavaScript. Therefore, they cannot be changed in-place. Any new string requires a new memory allocation, even when doing something as simple as
const str1 = "hello";
const str2 = str[0];
Leaves two strings in memory: "hello" and "h".
Since any attempt to produce a string will create at least one new string, it is therefore impossible to reverse a string without allocating space for a new string where the characters are reversed.
The minimum space complexity for this task is thusO(n) - scales linearly with the string length. Creating an array which can be rearranged in-place and then combined back to the reversed string fulfils this.

Here is a recursive way of doing it:
const rev = s => s.length>1 ? s.at(-1)+rev(s.slice(0,-1)) : s;
console.log(rev("This is a test string."))

The final line of your question means that the answer is "no". We cannot do this without using extra space [in userland JS].
We could, however, do this if we relied on a function written in a systems programming language. And this is the C code used by V8 for Array#join. In such a language the binary representation of the reversed string could be constructed step by step and simply cast to be a UTF-16 string in the final step. I presume this approximates what Array#join does under the hood.
If your requirement is simply to avoid using an array, the following simply successively pulls the code units from the end of the input string and builds a new string from them.
This will fail horribly with surrogate pairs (eg emoji) and grapheme clusters.
const reverse = (s) => {
let result = ''
for(let x = s.length-1; x >= 0; x--) {
result += s[x]
}
return result
}
console.log(reverse('hello'))

What about a hacky for loop?
const rev = (str) => {
for(var i = str.length - 1; i >= 0; i--) {
str += str[i];
}
return str.slice(str.length / 2, str.length);
}
console.log(rev("t"));
console.log(rev("te"));
console.log(rev("tes"));
console.log(rev("test"));

OP
"Can we do it without using an extra space."
nope.
Anyhow ... the OP's while based approached which this time does not try to change characters in place but programmatically a) removes character by character from the input value while b) concatenating the reversed result string ...
function reverseStringWithoutHelpOfArray(value) {
value = String(value); // let i = 0;
let result = ''; // let j = test.length - 1;
// while (i < j) {
while (value) { // let temp = test[i];
result = result + value.slice(-1); // test[j] = test[i];
value = value.substring(0, value.length - 1); // test[i] = temp;
} // i++; j--;
return result; // }
}
console.log(
reverseStringWithoutHelpOfArray('hallo')
);

Javascript: converting string to an array

This is one of my function for a calculator project. First I needed to convert the input string into an array, and do the operation later. (assuming that input has only numbers and '+' sign for now.
My question here is, how do I improve this code? What are the other ways to deal with this problem? (Time complexity, cleanness, shorter code.......whatever)
function convertArray(input) {
let array = [];
let num = "";
for (let i = 0; i < input.length; i++) {
if (input.charAt(i) == '+') {
array.push(input.charAt(i));
} else {
do {
num += input.charAt(i);
i++;
} while (i < input.length && input.charAt(i) !== '+');
array.push(num);
num = "";
i--;
}
}
return array;
}
console.log(convertArray("10+2+3000+70+1"));

You could split with a group. this add the group as well to the array.
For other calculation signs, you could add them to the brackets.
const convertArray = string => string.split(/([+])/);
console.log(convertArray("10+2+3000+70+1"));

const q = prompt('Sum?');
alert('Answer: ' + eval(q));
Would not recommend using eval, but if all you need is a quick and dirty trick, it works.
Personally, I'd recommend a library such as Math.js, but any will do.
If you really need to do this by yourself for a project, I'd recommend checking out the answers here: Evaluating a string as a mathematical expression in JavaScript.
Hope you succeed in whatever you're planning on doing.

It seems the complexity must have something to do with your wish to determing operators. In your code you just push them all into the array. To do that is like
const re = /((\d+)|([^\d]+))/g
const convertArray = str => {
let match, arr=[];
while (match = re.exec(str)) {
arr.push(match[1]) // here you can determine if you have an operator
console.log(match[1],"Operator?",!/^\d+$/.test(match[1]))
}
return arr
}
const str = "10+2+3000+70+1";
console.log(convertArray(str));

How make a more concise code using the For statement?

I'm new in StackOverflow and JavaScript, I'm trying to get the first letter that repeats from a string considering both uppercase and lowercase letters and counting and obtaining results using the for statement. The problem is that the form I used is too long Analyzing the situation reaches such a point that maybe you can only use a "For" statement for this exercise, which I get to iterate, but not with a cleaner and reduced code has me completely blocked, this is the reason why I request help to understand and continue with the understanding and use of this sentence. In this case, the result was tested in a JavaScript script inside a function and 3 "For" sentences obtaining quite positive results, but I can not create it in 1 only For (Sorry for my bad english google translate)
I making in HTML with JavasScript
var letter = "SYAHSVCXCyXSssssssyBxAVMZsXhZV";
var contendor = [];
var calc = [];
var mycalc = 0;
letter = letter.toUpperCase()
console.log(letter)
function repeats(){
for (var i = 0; i < letter.length; i++) {
if (contendor.includes(letter[i])) {
}else{
contendor.push(letter[i])
calc.push(0)
}
}
for (var p = 0; p < letter.length; p++) {
for (var l = 0; l < contendor.length; l++) {
if (letter[p] == contendor[l]) {
calc [l]= calc [l]+1
}
}
}
for (var f = 0; f < calc.length; f++) {
if ( calc[f] > calc[mycalc]) {
mycalc = f
}
}
}
repeats()
console.log("The most repeated letter its: " + contendor[mycalc]);
I Expected: A result with concise code

It would probably be a lot more concise to use a regular expression: match a character, then lookahead for more characters until you can match that first character again:
var letter = "SYAHSVCXCyXSssssssyBxAVMZsXhZV";
const firstRepeatedRegex = /(.)(?=.*\1)/;
console.log(letter.match(firstRepeatedRegex)[1]);
Of course, if you aren't sure whether a given string contains a repeated character, check that the match isn't null before trying to extract the character:
const input = 'abcde';
const firstRepeatedRegex = /(.)(?=.*\1)/;
const match = input.match(firstRepeatedRegex);
if (match) {
console.log(match[0]);
} else {
console.log('No repeated characters');
}
You could also turn the input into an array and use .find to find the first character whose lastIndexOf is not the same as the index of the character being iterated over:
const getFirstRepeatedCharacter = (str) => {
const chars = [...str];
const char = chars.find((char, i) => chars.lastIndexOf(char) !== i);
return char || 'No repeated characters';
};
console.log(getFirstRepeatedCharacter('abcde'));
console.log(getFirstRepeatedCharacter('SYAHSVCXCyXSssssssyBxAVMZsXhZV'));
If what you're actually looking for is the character that occurs most often, case-insensitive, use reduce to transform the string into an object indexed by character, whose values are the number of occurrences of that character, then identify the largest value:
const getMostRepeatedCharacter = (str) => {
const charsByCount = [...str.toUpperCase()].reduce((a, char) => {
a[char] = (a[char] || 0) + 1;
return a;
}, {});
const mostRepeatedEntry = Object.entries(charsByCount).reduce((a, b) => a[1] >= b[1] ? a : b);
return mostRepeatedEntry[0];
};
console.log(getMostRepeatedCharacter('abcde'));
console.log(getMostRepeatedCharacter('SYAHSVCXCyXSssssssyBxAVMZsXhZV'));

If the first repeated character is what you want, you can push it into an array and check if the character already exists
function getFirstRepeating( str ){
chars = []
for ( var i = 0; i < str.length; i++){
var char = str.charAt(i);
if ( chars.includes( char ) ){
return char;
} else {
chars.push( char );
}
}
return -1;
}
This will return the first repeating character if it exists, or will return -1.
Working
function getFirstRepeating( str ){
chars = []
for ( var i = 0; i < str.length; i++){
var char = str.charAt(i);
if ( chars.includes( char ) ){
return char;
} else {
chars.push( char );
}
}
return -1;
}
console.log(getFirstRepeating("SYAHSVCXCyXSssssssyBxAVMZsXhZV"))

Have you worked with JavaScript objects yet?
You should look into it.
When you loop through your string
let characters = "hemdhdksksbbd";
let charCount = {};
let max = { count: 0, ch: ""}; // will contain max
// rep letter
//Turn string into an array of letters and for
// each letter create a key in the charcount
// object , set it to 1 (meaning that's the first of
// that letter you've found) and any other time
// you see the letter, increment by 1.
characters.split("").forEach(function(character)
{
if(!charCount[character])
charCount[character] = 1;
else
charCount[character]++;
}
//charCount should now contain letters and
// their counts.
//Get the letters from charCount and find the
// max count
Object.keys(charCount). forEach (function(ch){
if(max.count < charCount[ch])
max = { count: charCount[ch], ch: ch};
}
console.log("most reps is: " , max.ch);

This is a pretty terrible solution. It takes 2 loops (reduce) and doesn't handle ties, but it's short and complicated.
Basically keep turning the results into arrays and use array methods split and reduce to find the answer. The first reduce is wrapped in Object.entries() to turn the object back into an array.
let letter = Object.entries(
"SYAHSVCXCyXSssssssyBxAVMZsXhZV".
toUpperCase().
split('').
reduce((p, c) => {
p[c] = isNaN(++p[c]) ? 1 : p[c];
return p;
}, {})
).
reduce((p, c) => p = c[1] > p[1] ? c : p);
console.log(`The most repeated letter is ${letter[0]}, ${letter[1]} times.`);

Javascript: Split a string by comma, except inside parentheses

Given string in the form:
'"abc",ab(),c(d(),e()),f(g(),zyx),h(123)'
How can I split it to get the below array format:
abc
ab()
c(d(),e())
f(g(),zyx)
h(123)
I have tried normal javascript split, however it doesn't work as desired. Trying Regular Expression but not yet successful.

You can keep track of the parentheses, and add those expressions when the left and right parens equalize.
For example-
function splitNoParen(s){
var left= 0, right= 0, A= [],
M= s.match(/([^()]+)|([()])/g), L= M.length, next, str= '';
for(var i= 0; i<L; i++){
next= M[i];
if(next=== '(')++left;
else if(next=== ')')++right;
if(left!== 0){
str+= next;
if(left=== right){
A[A.length-1]+=str;
left= right= 0;
str= '';
}
}
else A=A.concat(next.match(/([^,]+)/g));
}
return A;
}
var s1= '"abc",ab(),c(d(),e()),f(g(),zyx),h(123)';
splitNoParen(s1).join('\n');
/* returned value: (String)
"abc"
ab()
c(d(),e())
f(g(),zyx)
h(123)
*/

This might be not the best or more refined solution, and also maybe won't fit every single possibility, but based on your example it works:
var data = '"abc",ab(),c(d(),e()),f(g(),zyx),h(123)';
// Create a preResult splitting the commas.
var preResult = data.replace(/"/g, '').split(',');
// Create an empty result.
var result = [];
for (var i = 0; i < preResult.length; i++) {
// Check on every preResult if the number of parentheses match.
// Opening ones...
var opening = preResult[i].match(/\(/g) || 0;
// Closing ones...
var closing = preResult[i].match(/\)/g) || 0;
if (opening != 0 &&
closing != 0 &&
opening.length != closing.length) {
// If the current item contains a different number of opening
// and closing parentheses, merge it with the next adding a
// comma in between.
result.push(preResult[i] + ',' + preResult[i + 1]);
i++;
} else {
// Leave it as it is.
result.push(preResult[i]);
}
}
Demo

For future reference, here's another approach to top-level splitting, using string.replace as a control flow operator:
function psplit(s) {
var depth = 0, seg = 0, rv = [];
s.replace(/[^(),]*([)]*)([(]*)(,)?/g,
function (m, cls, opn, com, off, s) {
depth += opn.length - cls.length;
var newseg = off + m.length;
if (!depth && com) {
rv.push(s.substring(seg, newseg - 1));
seg = newseg;
}
return m;
});
rv.push(s.substring(seg));
return rv;
}
console.log(psplit('abc,ab(),c(d(),e()),f(g(),zyx),h(123)'))
["abc", "ab()", "c(d(),e())", "f(g(),zyx)", "h(123)"]
Getting it to handle quotes as well would not be too complicated, but at some point you need to decide to use a real parser such as jison, and I suspect that would be the point. In any event, there's not enough detail in the question to know what the desired handling of double quotes is.

You can't use .split for this, but instead you'll have to write a small parser like this:
function splitNoParen(s){
let results = [];
let next;
let str = '';
let left = 0, right = 0;
function keepResult() {
results.push(str);
str = '';
}
for(var i = 0; i<s.length; i++) {
switch(s[i]) {
case ',':
if((left === right)) {
keepResult();
left = right = 0;
} else {
str += s[i];
}
break;
case '(':
left++;
str += s[i];
break;
case ')':
right++;
str += s[i];
break;
default:
str += s[i];
}
}
keepResult();
return results;
}
var s1= '"abc",ab(),c(d(),e()),f(g(),zyx),h(123)';
console.log(splitNoParen(s1).join('\n'));
var s2='cats,(my-foo)-bar,baz';
console.log(splitNoParen(s2).join('\n'));

Had a similar issue and existing solutions were hard to generalize. So here's another parser that's a bit more readable and easier to extend to your personal needs. It'll also work with curly braces, brackets, normal braces, and strings of any type. License is MIT.
/**
* This function takes an input string and splits it by the given token, but only if the token is not inside
* braces of any kind, or a string.
* #param {string} input The string to split.
* #param {string} split_by Must be a single character.
* #returns {string[]} An array of split parts without the split_by character.
*/
export function parse_split(input:string, split_by:string = ",") : string[]
{
// Javascript has 3 types of strings
const STRING_TYPES = ["'","`","\""] as const;
// Some symbols can be nested, like braces, and must be counted
const state = {"{":0,"[":0,"(":0};
// Some cannot be nested, like a string, and just flip a flag.
// Additionally, once the string flag has been flipped, it can only be unflipped
// by the same token.
let string_state : (typeof STRING_TYPES)[number] | undefined = undefined
// Nestable symbols come in sets, usually in pairs.
// These sets increase or decrease the state, depending on the symbol.
const pairs : Record<string,[keyof typeof state,number]> = {
"{":["{",1],
"}":["{",-1],
"[":["[",1],
"]":["[",-1],
"(":["(",1],
")":["(",-1]
}
let start = 0;
let results = [];
let length = input.length;
for(let i = 0; i < length; ++i)
{
let char = input[i];
// Backslash escapes the next character. We directly skip 2 characters by incrementing i one extra time.
if(char === "\\")
{
i++;
continue;
}
// If the symbol exists in the single/not nested state object, flip the corresponding state flag.
if(char == string_state)
{
string_state = undefined;
console.log("Closed string ", string_state);
}
// if it's not in a string, but it's a string opener, remember the string type in string_state.
else if(string_state === undefined && STRING_TYPES.includes(char as typeof STRING_TYPES[number]))
{
string_state = char as typeof STRING_TYPES[number];
}
// If it's not in a string, and if it's a paired symbol, increase or decrease the state based on our "pairs" constant.
else if(string_state === undefined && (char in pairs) )
{
let [key,value] = pairs[char];
state[key] += value;
}
// If it's our split symbol...
else if(char === split_by)
{
// ... check whether any flags are active ...
if(Object.entries(state).every(([k,v])=>v == 0) && (string_state === undefined))
{
// ... if not, then this is a valid split.
results.push(input.substring(start,i))
start = i+1;
}
}
}
// Add the last segment if the string didn't end in the split_by symbol, otherwise add an empty string
if(start < input.length)
{
results.push(input.substring(start,input.length))
}
else
results.push("");
return results;
}

With this regex, it makes the job:
const regex = /,(?![^(]*\))/g;
const str = '"abc",ab(),c(d(),e()),f(g(),zyx),h(123)';
const result = str.split(regex);
console.log(result);

Javascript
var str='"abc",ab(),c(d(),e()),f(g(),zyx),h(123)'
str.split('"').toString().split(',').filter(Boolean);
this should work

Count the number of occurrences of a character in a string in Javascript

I need to count the number of occurrences of a character in a string.
For example, suppose my string contains:
var mainStr = "str1,str2,str3,str4";
I want to find the count of comma , character, which is 3. And the count of individual strings after the split along comma, which is 4.
I also need to validate that each of the strings i.e str1 or str2 or str3 or str4 should not exceed, say, 15 characters.

I have updated this answer. I like the idea of using a match better, but it is slower:
console.log(("str1,str2,str3,str4".match(/,/g) || []).length); //logs 3
console.log(("str1,str2,str3,str4".match(new RegExp("str", "g")) || []).length); //logs 4
Use a regular expression literal if you know what you are searching for beforehand, if not you can use the RegExp constructor, and pass in the g flag as an argument.
match returns null with no results thus the || []
The original answer I made in 2009 is below. It creates an array unnecessarily, but using a split is faster (as of September 2014). I'm ambivalent, if I really needed the speed there would be no question that I would use a split, but I would prefer to use match.
Old answer (from 2009):
If you're looking for the commas:
(mainStr.split(",").length - 1) //3
If you're looking for the str
(mainStr.split("str").length - 1) //4
Both in #Lo's answer and in my own silly performance test split comes ahead in speed, at least in Chrome, but again creating the extra array just doesn't seem sane.

There are at least five ways. The best option, which should also be the fastest (owing to the native RegEx engine) is placed at the top.
Method 1
("this is foo bar".match(/o/g)||[]).length;
// returns 2
Method 2
"this is foo bar".split("o").length - 1;
// returns 2
Split not recommended as it is resource hungry. It allocates new instances of 'Array' for each match. Don't try it for a >100MB file via FileReader. You can observe the exact resource usage using Chrome's profiler option.
Method 3
var stringsearch = "o"
,str = "this is foo bar";
for(var count=-1,index=-2; index != -1; count++,index=str.indexOf(stringsearch,index+1) );
// returns 2
Method 4
Searching for a single character
var stringsearch = "o"
,str = "this is foo bar";
for(var i=count=0; i<str.length; count+=+(stringsearch===str[i++]));
// returns 2
Method 5
Element mapping and filtering. This is not recommended due to its overall resource preallocation rather than using Pythonian 'generators':
var str = "this is foo bar"
str.split('').map( function(e,i){ if(e === 'o') return i;} )
.filter(Boolean)
//>[9, 10]
[9, 10].length
// returns 2
Share:
I made this gist, with currently 8 methods of character-counting, so we can directly pool and share our ideas - just for fun, and perhaps some interesting benchmarks :)

Add this function to sting prototype :
String.prototype.count=function(c) {
var result = 0, i = 0;
for(i;i<this.length;i++)if(this[i]==c)result++;
return result;
};
usage:
console.log("strings".count("s")); //2

Simply, use the split to find out the number of occurrences of a character in a string.
mainStr.split(',').length // gives 4 which is the number of strings after splitting using delimiter comma
mainStr.split(',').length - 1 // gives 3 which is the count of comma

A quick Google search got this (from http://www.codecodex.com/wiki/index.php?title=Count_the_number_of_occurrences_of_a_specific_character_in_a_string#JavaScript)
String.prototype.count=function(s1) {
return (this.length - this.replace(new RegExp(s1,"g"), '').length) / s1.length;
}
Use it like this:
test = 'one,two,three,four'
commas = test.count(',') // returns 3

You can also rest your string and work with it like an array of elements using
Array.prototype.filter()
const mainStr = 'str1,str2,str3,str4';
const commas = [...mainStr].filter(l => l === ',').length;
console.log(commas);
Or
Array.prototype.reduce()
const mainStr = 'str1,str2,str3,str4';
const commas = [...mainStr].reduce((a, c) => c === ',' ? ++a : a, 0);
console.log(commas);

UPDATE: This might be simple, but it is not the fastest. See benchmarks below.
It's amazing that in 13 years, this answer hasn't shown up. Intuitively, it seems like it should be fastest:
const s = "The quick brown fox jumps over the lazy dog.";
const oCount = s.length - s.replaceAll('o', '').length;
If there are only two kinds of character in the string, then this is faster still:
const s = "001101001";
const oneCount = s.replaceAll('0', '').length;
BENCHMARKS
const { performance } = require('node:perf_hooks');
const ITERATIONS = 10000000;
const TEST_STRING = "The quick brown fox jumps over the lazy dog.";
console.log(ITERATIONS, "iterations");
let sum = 0; // make sure compiler doesn't optimize code out
let start = performance.now();
for (let i = 0; i < ITERATIONS; ++i) {
sum += TEST_STRING.length - TEST_STRING.replaceAll('o', '').length;
}
let end = performance.now();
console.log(" replaceAll duration", end - start, `(sum ${sum})`);
sum = 0;
start = performance.now();
for (let i = 0; i < ITERATIONS; ++i) {
sum += TEST_STRING.split('o').length - 1
}
end = performance.now();
console.log(" split duration", end - start, `(sum ${sum})`);
10000 iterations
replaceAll duration 2.6167500019073486 (sum 40000)
split duration 2.0777920186519623 (sum 40000)
100000 iterations
replaceAll duration 17.563208997249603 (sum 400000)
split duration 8.087624996900558 (sum 400000)
1000000 iterations
replaceAll duration 128.71587499976158 (sum 4000000)
split duration 64.15841698646545 (sum 4000000)
10000000 iterations
replaceAll duration 1223.3415840268135 (sum 40000000)
split duration 629.1629169881344 (sum 40000000)

Here is a similar solution, but it uses Array.prototype.reduce
function countCharacters(char, string) {
return string.split('').reduce((acc, ch) => ch === char ? acc + 1: acc, 0)
}
As was mentioned, String.prototype.split works much faster than String.prototype.replace.

If you are using lodash, the _.countBy method will do this:
_.countBy("abcda")['a'] //2
This method also work with array:
_.countBy(['ab', 'cd', 'ab'])['ab'] //2

ok, an other one with regexp - probably not fast, but short and better readable then others, in my case just '_' to count
key.replace(/[^_]/g,'').length
just remove everything that does not look like your char
but it does not look nice with a string as input

I have found that the best approach to search for a character in a very large string (that is 1 000 000 characters long, for example) is to use the replace() method.
window.count_replace = function (str, schar) {
return str.length - str.replace(RegExp(schar), '').length;
};
You can see yet another JSPerf suite to test this method along with other methods of finding a character in a string.

Performance of Split vs RegExp
var i = 0;
var split_start = new Date().getTime();
while (i < 30000) {
"1234,453,123,324".split(",").length -1;
i++;
}
var split_end = new Date().getTime();
var split_time = split_end - split_start;
i= 0;
var reg_start = new Date().getTime();
while (i < 30000) {
("1234,453,123,324".match(/,/g) || []).length;
i++;
}
var reg_end = new Date().getTime();
var reg_time = reg_end - reg_start;
alert ('Split Execution time: ' + split_time + "\n" + 'RegExp Execution time: ' + reg_time + "\n");

I made a slight improvement on the accepted answer, it allows to check with case-sensitive/case-insensitive matching, and is a method attached to the string object:
String.prototype.count = function(lit, cis) {
var m = this.toString().match(new RegExp(lit, ((cis) ? "gi" : "g")));
return (m != null) ? m.length : 0;
}
lit is the string to search for ( such as 'ex' ), and cis is case-insensitivity, defaulted to false, it will allow for choice of case insensitive matches.
To search the string 'I love StackOverflow.com' for the lower-case letter 'o', you would use:
var amount_of_os = 'I love StackOverflow.com'.count('o');
amount_of_os would be equal to 2.
If we were to search the same string again using case-insensitive matching, you would use:
var amount_of_os = 'I love StackOverflow.com'.count('o', true);
This time, amount_of_os would be equal to 3, since the capital O from the string gets included in the search.

Easiest way i found out...
Example-
str = 'mississippi';
function find_occurences(str, char_to_count){
return str.split(char_to_count).length - 1;
}
find_occurences(str, 'i') //outputs 4

Here is my solution. Lots of solution already posted before me. But I love to share my view here.
const mainStr = 'str1,str2,str3,str4';
const commaAndStringCounter = (str) => {
const commas = [...str].filter(letter => letter === ',').length;
const numOfStr = str.split(',').length;
return `Commas: ${commas}, String: ${numOfStr}`;
}
// Run the code
console.log(commaAndStringCounter(mainStr)); // Output: Commas: 3, String: 4
Here you find my REPL

I just did a very quick and dirty test on repl.it using Node v7.4. For a single character, the standard for loop is quickest:
Some code:
// winner!
function charCount1(s, c) {
let count = 0;
c = c.charAt(0); // we save some time here
for(let i = 0; i < s.length; ++i) {
if(c === s.charAt(i)) {
++count;
}
}
return count;
}
function charCount2(s, c) {
return (s.match(new RegExp(c[0], 'g')) || []).length;
}
function charCount3(s, c) {
let count = 0;
for(ch of s) {
if(c === ch) {
++count;
}
}
return count;
}
function perfIt() {
const s = 'Hello, World!';
const c = 'o';
console.time('charCount1');
for(let i = 0; i < 10000; i++) {
charCount1(s, c);
}
console.timeEnd('charCount1');
console.time('charCount2');
for(let i = 0; i < 10000; i++) {
charCount2(s, c);
}
console.timeEnd('charCount2');
console.time('charCount3');
for(let i = 0; i < 10000; i++) {
charCount2(s, c);
}
console.timeEnd('charCount3');
}
Results from a few runs:
perfIt()
charCount1: 3.301ms
charCount2: 11.652ms
charCount3: 174.043ms
undefined
perfIt()
charCount1: 2.110ms
charCount2: 11.931ms
charCount3: 177.743ms
undefined
perfIt()
charCount1: 2.074ms
charCount2: 11.738ms
charCount3: 152.611ms
undefined
perfIt()
charCount1: 2.076ms
charCount2: 11.685ms
charCount3: 154.757ms
undefined
Update 2021-Feb-10: Fixed typo in repl.it demo
Update 2020-Oct-24: Still the case with Node.js 12 (play with it yourself here)

UPDATE 06/10/2022
So I ran various perf tests and if your use case allows it, it seems that using split is going to perform the best overall.
function countChar(char: string, string: string): number {
return string.split(char).length - 1
}
countChar('x', 'foo x bar x baz x')
I know I am late to the party here but I was rather baffled no one answered this with the most basic of approaches. A large portion of the answers provided by the community for this question are iteration based but all are moving over strings on a per-character basis which is not really efficient.
When dealing with a large string that contains thousands of characters walking over each character to get the occurance count can become rather extraneous not to mention a code-smell. The below solutions take advantage of slice, indexOf and the trusted traditional while loop. These approaches prevent us having to walk over each character and will greatly speed up the time it takes to count occurances. These follow similar logic to that you'd find in parsers and lexical analyzers that require string walks.
Using with Slice
In this approach we are leveraging slice and with every indexOf match we will move our way through the string and eliminate the previous searched potions. Each time we call indexOf the size of the string it searches will be smaller.
function countChar (char: string, search: string): number {
let num: number = 0;
let str: string = search;
let pos: number = str.indexOf(char);
while(pos > -1) {
str = str.slice(pos + 1);
pos = str.indexOf(char);
num++;
}
return num;
}
// Call the function
countChar('x', 'foo x bar x baz x') // 3
Using with IndexOf from position
Similar to the first approach using slice but instead of augmenting the string we are searching it will leverage the from parameter in indexOf method.
function countChar (char: string, str: string): number {
let num: number = 0;
let pos: number = str.indexOf(char);
while(pos > -1) {
pos = str.indexOf(char, pos + 1);
num++;
}
return num;
}
// Call the function
countChar('x', 'foo x bar x baz x') // 3
Personally, I go for the second approach over the first, but both are fine and performant when dealing with large strings but also smaller sized ones too.

s = 'dir/dir/dir/dir/'
for(i=l=0;i<s.length;i++)
if(s[i] == '/')
l++

I was working on a small project that required a sub-string counter. Searching for the wrong phrases provided me with no results, however after writing my own implementation I have stumbled upon this question. Anyway, here is my way, it is probably slower than most here but might be helpful to someone:
function count_letters() {
var counter = 0;
for (var i = 0; i < input.length; i++) {
var index_of_sub = input.indexOf(input_letter, i);
if (index_of_sub > -1) {
counter++;
i = index_of_sub;
}
}
http://jsfiddle.net/5ZzHt/1/
Please let me know if you find this implementation to fail or do not follow some standards! :)
UPDATE
You may want to substitute:
for (var i = 0; i < input.length; i++) {
With:
for (var i = 0, input_length = input.length; i < input_length; i++) {
Interesting read discussing the above:
http://www.erichynds.com/blog/javascript-length-property-is-a-stored-value

What about string.split(desiredCharecter).length-1
Example:
var str = "hellow how is life";
var len = str.split("h").length-1; will give count 2 for character "h" in the above string;

The fastest method seems to be via the index operator:
function charOccurances (str, char)
{
for (var c = 0, i = 0, len = str.length; i < len; ++i)
{
if (str[i] == char)
{
++c;
}
}
return c;
}
console.log( charOccurances('example/path/script.js', '/') ); // 2
Or as a prototype function:
String.prototype.charOccurances = function (char)
{
for (var c = 0, i = 0, len = this.length; i < len; ++i)
{
if (this[i] == char)
{
++c;
}
}
return c;
}
console.log( 'example/path/script.js'.charOccurances('/') ); // 2

function len(text,char){
return text.innerText.split(string).length
}
console.log(len("str1,str2,str3,str4",","))
This is a very short function.

The following uses a regular expression to test the length. testex ensures you don't have 16 or greater consecutive non-comma characters. If it passes the test, then it proceeds to split the string. counting the commas is as simple as counting the tokens minus one.
var mainStr = "str1,str2,str3,str4";
var testregex = /([^,]{16,})/g;
if (testregex.test(mainStr)) {
alert("values must be separated by commas and each may not exceed 15 characters");
} else {
var strs = mainStr.split(',');
alert("mainStr contains " + strs.length + " substrings separated by commas.");
alert("mainStr contains " + (strs.length-1) + " commas.");
}

I'm using Node.js v.6.0.0 and the fastest is the one with index (the 3rd method in Lo Sauer's answer).
The second is:
function count(s, c) {
var n = 0;
for (let x of s) {
if (x == c)
n++;
}
return n;
}

And there is:
function character_count(string, char, ptr = 0, count = 0) {
while (ptr = string.indexOf(char, ptr) + 1) {count ++}
return count
}
Works with integers too!

Here's one just as fast as the split() and the replace methods, which are a tiny bit faster than the regex method (in Chrome and Firefox both).
let num = 0;
let str = "str1,str2,str3,str4";
//Note: Pre-calculating `.length` is an optimization;
//otherwise, it recalculates it every loop iteration.
let len = str.length;
//Note: Don't use a `for (... of ...)` loop, it's slow!
for (let charIndex = 0; charIndex < len; ++charIndex) {
if (str[charIndex] === ',') {
++num;
}
}

var mainStr = "str1,str2,str3,str4";
var splitStr = mainStr.split(",").length - 1; // subtracting 1 is important!
alert(splitStr);
Splitting into an array gives us a number of elements, which will always be 1 more than the number of instances of the character. This may not be the most memory efficient, but if your input is always going to be small, this is a straight-forward and easy to understand way to do it.
If you need to parse very large strings (greater than a few hundred characters), or if this is in a core loop that processes large volumes of data, I would recommend a different strategy.

String.prototype.reduce = Array.prototype.reduce;
String.prototype.count = function(c) {
return this.reduce(((n, x) => n + (x === c ? 1 : 0)), 0)
};
const n = "bugs bunny was here".count("b")
console.log(n)
Similar to the prototype based above, but does not allocate an array for the string. Allocation is the problem of nearly every version above, except the loop variants. This avoids loop code, reusing the browser implemented Array.reduce function.

My solution:
function countOcurrences(str, value){
var regExp = new RegExp(value, "gi");
return str.match(regExp) ? str.match(regExp).length : 0;
}

I know this might be an old question but I have a simple solution for low-level beginners in JavaScript.
As a beginner, I could only understand some of the solutions to this question so I used two nested FOR loops to check each character against every other character in the string, incrementing a count variable for each character found that equals that character.
I created a new blank object where each property key is a character and the value is how many times each character appeared in the string(count).
Example function:-
function countAllCharacters(str) {
var obj = {};
if(str.length!==0){
for(i=0;i<str.length;i++){
var count = 0;
for(j=0;j<str.length;j++){
if(str[i] === str[j]){
count++;
}
}
if(!obj.hasOwnProperty(str[i])){
obj[str[i]] = count;
}
}
}
return obj;
}