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Replace matched regex group by occurence

I'm working on a drag and drop function for SVG path, which lets a user move the co-ordinates of the path.
Please consider the string below:
M162.323 150.513L232.645 8L303.504 149.837L461.168 173.5L347.156 284.5L373.605 440.728L233.5 367.854L91.7415 442L118.424 284.883L5.151 173.549Z
Would it be possible to replace a specific(let's say the 4th) occurence of a matched regex group using the .replace method?
Regex:
[A-Z](-?\d*\.?\d*\s-?\d*\.?\d*)

regex.exec() is a method that is used to find the next match in a string based on a regular expression. It returns an array containing the matched string and any capturing groups, or null if no match is found. This method can be used in a loop to iterate over all matches in a string and adjust the match accordingly.
let string = "M162.323 150.513L232.645 8L303.504 149.837L461.168 173.5L347.156 284.5L373.605 440.728L233.5 367.854L91.7415 442L118.424 284.883L5.151 173.549Z";
let regex = /[A-Z](-?\d*\.?\d*\s-?\d*\.?\d*)/g;
// Replace the 4th match
let newString = "";
let index = 0;
let match;
while (match = regex.exec(string)) {
if (index === 3) {
// Do something to modify the 4th match
newString += match[0].replace(/-?\d*\.?\d*\s-?\d*\.?\d*/, "REPLACED");
} else {
// Leave other matches unchanged
newString += match[0];
}
index++;
}
console.log(newString);

const s = 'M162.323 150.513L232.645 8L303.504 149.837L461.168 173.5L347.156 284.5L373.605 440.728L233.5 367.854L91.7415 442L118.424 284.883L5.151 173.549Z'
let n = 4, regex = /[A-Z](-?\d*\.?\d*\s-?\d*\.?\d*)/gm
console.log(s.replace(regex, m => --n ? m : 'hello'))

What am I missing regarding the toUpperCase method?

I am trying to make the first letter of each word capitalized via toUpperCase method and the rest of the word is in the lower case via the toLowerCase method. But I am missing something... Why temp value is not matching with result[1][0] even if I am using that method for both?
Note: I know about other ways (map, replace, etc) for my solution, but I want to just use a for-loop with toUpperCase and toLowerCase methods.
function titleCase(str) {
let regex = /[^0-9\s]+/g;
var result = str.match(regex);
let temp = "";
for (let i = 0; i < result.length; i++) {
for (let j = 0; j < result[i].length; j++) {
result[1][0] = result[1][0].toUpperCase();
temp = result[1][0].toUpperCase();
}
}
console.log(temp); // Output is 'A'
console.log(result[1][0]); //Output is 'a'
// Normally 'temp' and 'result[1][0]' should be equal, but one returns a lowercase character and the other an uppercase character.
return str;
}
titleCase("I'm a little tea pot");

Your problem is not with the toUppercase(), it is with the reference.
When referencing result[1][0], why are you including the 0? You already have the second character with result[1]
result[1] === 'a'. No need to include the [0] as well.
Change your code so it looks like this:
function titleCase(str) {
let regex = /[^0-9\s]+/g;
var result = str.match(regex);
let temp = "";
result[1] = result[1].toUpperCase();
temp = result[1].toUpperCase();
console.log(temp); // Output is 'A'
console.log(result[1]); //Output is also 'A'
// both now equals capital A
return str;
}
titleCase("I'm a little tea pot");
EDIT:
Updating the function to uppercase the first letter of the word.
We can use ES6, which would make this really simple:
const capitalize = (string = '') => [...string].map((char, index) => index ? char : char.toUpperCase()).join('')
Use it: capitalize("hello") returns 'Hello'.
First we convert the string to an array, using the spread operator, to get each char individually as a string. Then we map each character to get the index to apply the uppercase to it. Index true means not equal 0, so (!index) is the first character. We then apply the uppercase function to it and then return the string.
If you want a more object oriented approach, we can do something like this:
String.prototype.capitalize = function(allWords) {
return (allWords) ?
this.split(' ').map(word => word.capitalize()).join(' ') :
return this.charAt(0).toUpperCase() + this.slice(1);
}
Use it: "hello, world!".capitalize(); returns "Hello, World"
We break down the phrase to words and then recursive calls until capitalising all words. If allWords is undefined, capitalise only the first word meaning the first character of the whole string.

I was tried to change a specific character in the string but strings are immutable in JS so this does not make sense.

Counting brackets in a string

been working on a school problem, and haven't been able to figure it out. Any help is appreciated!
Write a function named countBrackets that accepts a string
and returns the count of square bracket and curly bracket
characters in the string. That is, it should count occurrences of
these four characters “{ } [ ]”. Use function expression syntax.
var countBrackets = function(s){
let sum = 0
for(let i = ""; i == s ; i++ )
if(i ==="{}[]"){
sum+=i
}
return sum}
console.log(countBrackets("[123],{456},[{}]")) //8
console.log(countBrackets("Test string")) // 0
I'm a little confused on how I'm supposed to get it to count a string I guess.

You can use a global regex (regular expression) matching for this. The regex is between / / followed by the g flag to make it global (otherwise it only returns the first result it finds and stops).
Within the regex, | is the OR operator, so you match for /{|}|[|]/
Since [ and ] have special meaning in regular expressions you need to escape those using a \ so your total regex becomes /{|}|\[|\]/g.
This returns an array of matches, I called the function findBrackets.
To get the number of brackets, the function countBrackets just returns the .length of that array.
const findBrackets = str => str.match(/{|}|\[|\]/g);
const countBrackets = str => findBrackets(str) ? findBrackets(str).length : 0;
console.log(findBrackets('qw[e}rt[y]ui{}o{p'));
console.log(countBrackets('qw[e}rt[y]ui{}o{p'));
console.log(countBrackets('no brackets here'));
Edit: seeing the code you posted, you decided to use a for-loop to loop over your string, which is a totally valid solution.
Note that in my above example
const findBrackets = str => str.match(/{|}|\[|\]/g);
Is basically the same but a newer way of writing functions (with a few nuances)
I could have written:
var findBrackets = function(str) {
return str.match(/{|}|\[|\]/g);
}
instead which would be almost the same.
If you want to loop over a string, you can access a letter in the string by using square bracket notation, so for example
const testString = 'hello';
console.log(testString[1]);
Would output the letter 'e'.
So to use this in a function:
const countBrackets = (str) => {
let counter = 0;
for (let i = 0; i < str.length; i++) {
if (str[i] === '[' || str[i] === ']' || str[i] === '{' || str[i] === '}') {
counter++;
}
}
return counter;
}
console.log(countBrackets('qw[e}rt[y]ui{}o{p'));
Here you loop over the string from 0 to < length of the string and check every letter of it, by seeing if str[i] is {, }, [ or ], and if it is you increment a counter. After that's done you return the final count.

Remove all special characters except space from a string using JavaScript

I want to remove all special characters except space from a string using JavaScript.
For example,
abc's test#s
should output as
abcs tests.

You should use the string replace function, with a single regex.
Assuming by special characters, you mean anything that's not letter, here is a solution:
const str = "abc's test#s";
console.log(str.replace(/[^a-zA-Z ]/g, ""));

You can do it specifying the characters you want to remove:
string = string.replace(/[&\/\\#,+()$~%.'":*?<>{}]/g, '');
Alternatively, to change all characters except numbers and letters, try:
string = string.replace(/[^a-zA-Z0-9]/g, '');

The first solution does not work for any UTF-8 alphabet. (It will cut text such as Привіт). I have managed to create a function which does not use RegExp and use good UTF-8 support in the JavaScript engine. The idea is simple if a symbol is equal in uppercase and lowercase it is a special character. The only exception is made for whitespace.
function removeSpecials(str) {
var lower = str.toLowerCase();
var upper = str.toUpperCase();
var res = "";
for(var i=0; i<lower.length; ++i) {
if(lower[i] != upper[i] || lower[i].trim() === '')
res += str[i];
}
return res;
}
Update: Please note, that this solution works only for languages where there are small and capital letters. In languages like Chinese, this won't work.
Update 2: I came to the original solution when I was working on a fuzzy search. If you also trying to remove special characters to implement search functionality, there is a better approach. Use any transliteration library which will produce you string only from Latin characters and then the simple Regexp will do all magic of removing special characters. (This will work for Chinese also and you also will receive side benefits by making Tromsø == Tromso).

search all not (word characters || space):
str.replace(/[^\w ]/, '')

I don't know JavaScript, but isn't it possible using regex?
Something like [^\w\d\s] will match anything but digits, characters and whitespaces. It would be just a question to find the syntax in JavaScript.

I tried Seagul's very creative solution, but found it treated numbers also as special characters, which did not suit my needs. So here is my (failsafe) tweak of Seagul's solution...
//return true if char is a number
function isNumber (text) {
if(text) {
var reg = new RegExp('[0-9]+$');
return reg.test(text);
}
return false;
}
function removeSpecial (text) {
if(text) {
var lower = text.toLowerCase();
var upper = text.toUpperCase();
var result = "";
for(var i=0; i<lower.length; ++i) {
if(isNumber(text[i]) || (lower[i] != upper[i]) || (lower[i].trim() === '')) {
result += text[i];
}
}
return result;
}
return '';
}

const str = "abc's#thy#^g&test#s";
console.log(str.replace(/[^a-zA-Z ]/g, ""));

Try to use this one
var result= stringToReplace.replace(/[^\w\s]/g, '')
[^] is for negation, \w for [a-zA-Z0-9_] word characters and \s for space,
/[]/g for global

With regular expression
let string = "!#This tool removes $special *characters* /other/ than! digits, characters and spaces!!!$";
var NewString= string.replace(/[^\w\s]/gi, '');
console.log(NewString);
Result //This tool removes special characters other than digits characters and spaces
Live Example : https://helpseotools.com/text-tools/remove-special-characters

dot (.) may not be considered special. I have added an OR condition to Mozfet's & Seagull's answer:
function isNumber (text) {
reg = new RegExp('[0-9]+$');
if(text) {
return reg.test(text);
}
return false;
}
function removeSpecial (text) {
if(text) {
var lower = text.toLowerCase();
var upper = text.toUpperCase();
var result = "";
for(var i=0; i<lower.length; ++i) {
if(isNumber(text[i]) || (lower[i] != upper[i]) || (lower[i].trim() === '') || (lower[i].trim() === '.')) {
result += text[i];
}
}
return result;
}
return '';
}

Try this:
const strippedString = htmlString.replace(/(<([^>]+)>)/gi, "");
console.log(strippedString);

const input = `#if_1 $(PR_CONTRACT_END_DATE) == '23-09-2019' #
Test27919<alerts#imimobile.com> #elseif_1 $(PR_CONTRACT_START_DATE) == '20-09-2019' #
Sender539<rama.sns#gmail.com> #elseif_1 $(PR_ACCOUNT_ID) == '1234' #
AdestraSID<hello#imimobile.co> #else_1#Test27919<alerts#imimobile.com>#endif_1#`;
const replaceString = input.split('$(').join('->').split(')').join('<-');
console.log(replaceString.match(/(?<=->).*?(?=<-)/g));

Whose special characters you want to remove from a string, prepare a list of them and then user javascript replace function to remove all special characters.
var str = 'abc'de#;:sfjkewr47239847duifyh';
alert(str.replace("'","").replace("#","").replace(";","").replace(":",""));
or you can run loop for a whole string and compare single single character with the ASCII code and regenerate a new string.

Count number of words in string using JavaScript

I am trying to count the number of words in a given string using the following code:
var t = document.getElementById('MSO_ContentTable').textContent;
if (t == undefined) {
var total = document.getElementById('MSO_ContentTable').innerText;
} else {
var total = document.getElementById('MSO_ContentTable').textContent;
}
countTotal = cword(total);
function cword(w) {
var count = 0;
var words = w.split(" ");
for (i = 0; i < words.length; i++) {
// inner loop -- do the count
if (words[i] != "") {
count += 1;
}
}
return (count);
}
In that code I am getting data from a div tag and sending it to the cword() function for counting. Though the return value is different in IE and Firefox. Is there any change required in the regular expression? One thing that I show that both browser send same string there is a problem inside the cword() function.

[edit 2022, based on comment] Nowadays, one would not extend the native prototype this way. A way to extend the native protype without the danger of naming conflicts is to use the es20xx symbol. Here is an example of a wordcounter using that.
Old answer: you can use split and add a wordcounter to the String prototype:
if (!String.prototype.countWords) {
String.prototype.countWords = function() {
return this.length && this.split(/\s+\b/).length || 0;
};
}
console.log(`'this string has five words'.countWords() => ${
'this string has five words'.countWords()}`);
console.log(`'this string has five words ... and counting'.countWords() => ${
'this string has five words ... and counting'.countWords()}`);
console.log(`''.countWords() => ${''.countWords()}`);

I would prefer a RegEx only solution:
var str = "your long string with many words.";
var wordCount = str.match(/(\w+)/g).length;
alert(wordCount); //6
The regex is
\w+ between one and unlimited word characters
/g greedy - don't stop after the first match
The brackets create a group around every match. So the length of all matched groups should match the word count.

This is the best solution I've found:
function wordCount(str) {
var m = str.match(/[^\s]+/g)
return m ? m.length : 0;
}
This inverts whitespace selection, which is better than \w+ because it only matches the latin alphabet and _ (see http://www.ecma-international.org/ecma-262/5.1/#sec-15.10.2.6)
If you're not careful with whitespace matching you'll count empty strings, strings with leading and trailing whitespace, and all whitespace strings as matches while this solution handles strings like ' ', ' a\t\t!\r\n#$%() d ' correctly (if you define 'correct' as 0 and 4).

You can make a clever use of the replace() method although you are not replacing anything.
var str = "the very long text you have...";
var counter = 0;
// lets loop through the string and count the words
str.replace(/(\b+)/g,function (a) {
// for each word found increase the counter value by 1
counter++;
})
alert(counter);
the regex can be improved to exclude html tags for example

//Count words in a string or what appears as words :-)
function countWordsString(string){
var counter = 1;
// Change multiple spaces for one space
string=string.replace(/[\s]+/gim, ' ');
// Lets loop through the string and count the words
string.replace(/(\s+)/g, function (a) {
// For each word found increase the counter value by 1
counter++;
});
return counter;
}
var numberWords = countWordsString(string);