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JS: given a string, change all the symbols between the first and the last letters of each word of the string if the word starts and ends with 'a'

Example:
aza asssa axxxa rrra -> a!a a!!!a a!!!a rrra
So far I've come up with this solution:
const argument = "aza asssa axxxa rrra";
const amount_of_spaces = [...argument].filter(x => x === " ").length;
let j = 0;
const argument__clone = [...argument];
const space__indices = [];
function do__stuff() {
while (j < amount_of_spaces) {
space__indices.push(argument__clone.indexOf(" ") + j);
argument__clone.splice((argument__clone.indexOf(" ")), 1);
j++;
do__stuff();
}
};
do__stuff();
const words = [];
let word = '';
for (let i = 0; i < argument.length; i++) {
if (!(space__indices.includes(i))) {
word += argument[i];
}
else {
words.push(word);
word = '';
}
}
words.push(word);
let new__word = '';
const new__words = [];
const words__static = [];
for (i of words) {
if (i[0] === 'a' && i[i.length - 1] === 'a') {
for (let j = 1; j < i.length - 1; j++) {
new__word += "!";
}
new__words.push(new__word);
new__word = '';
}
else {
words__static.push(i);
}
}
new__words.map(i => "a" + i + "a");
console.log(new__words);
console.log(words__static);
So one array stores the indices of spaces and the other one stores the words from the given string. We can separate the words because we know when one ends because we have the array with space indices. Then we check for each word whether it starts with 'a' and ends with 'a'. If the requirements are met we change all the letters within the word for "!" (excluding the very first and the very last ones). If the requirements are not met we store the word into the other array.
Eventually we have two arrays that I want to concatenate into one. The problems is if I was given something like this:
aza asssa rrra axxxa
It wouldn't have worked because of the order
Is there any better solution?

A regular expression would be simpler. Match an a after a word boundary, match more non-space characters, and finally match another a followed by a word boundary.
const input = 'aza asssa axxxa rrra';
const output = input.replace(
/(?<=\ba)\S+(?=a\b)/g,
interiorWord => '!'.repeat(interiorWord.length)
);
console.log(output);
For a more manual approach, split the input by spaces so you have an array of words, then for each word, check if it begins and ends with an a - if so, construct a new word by checking the old word's length. Then turn the array back into a single string.
const input = 'aza asssa axxxa rrra';
const words = input.split(' ');
const replacedWords = words.map(word => (
word[0] === 'a' && word[word.length - 1] === 'a' && word.length >= 3
? 'a' + '!'.repeat(word.length - 2) + 'a'
: word
));
const output = replacedWords.join(' ');
console.log(output);

Function to capitalize first and last letter of each word not working

I created a function that given any string will return the string with the first and last letter of each word capitalized. So far it works in some words, not on others, can someone help me figure out why?
function Capitalize(str) {
var spl = str.split(" ");
var words = [];
for (let i = 0; i < spl.length; i++) {
//For every word
for (let j = 0; j < spl[i].length; j++) {
//For every letter in each word
var word = spl[i];
var size = spl[i].length;
var firstLetterCapital = word.replace(word[0], word[0].toUpperCase()); //Creates new array
var LastLetterCapital = firstLetterCapital.replace(
word[size - 1],
word[size - 1].toUpperCase()
);
}
words.push(LastLetterCapital);
}
console.log(words.join(" "));
}
Capitalize("hello there");
It works when I type : Capitalize("my name is john smith"), but not with Capitalize("hello there")
I know it's a complete mess and probably a very bad way to do it, but I started programming a month ago so give me a break :)

#symlink has already explained why it is "HellO ThEre" instead of "Hello TherE". He also has given a solution to explicitly target first and last character of the string. I have accomplished not much different than already posted by members, except for .. "may be" a little more explanation.
You can break the entire problem in these four steps.
Get all the words into an array.
Create a function, that takes each word and targets first and last character, changes it and returns the changed word.
Apply a mapping step using the function created above (in step 2) to the entire array of words (obtained in step 1).
Join the transformed array, obtained in step 3, using a blank space as a separator.
I have written two functions that accomplish this task. I am sorry for long name of functions. It helps me keep track of things in a complex program (especially when I am in a hurry!).
Step 2 function
function Capitalize_FirstAndLast_One_Word(word){
// Split the string in array for easy access/manipulation by indexing
Split_String = word.split("")
// Target the first word
Split_String[0] = Split_String[0].toUpperCase();
// Target the last word
Split_String[Split_String.length - 1] = Split_String[Split_String.length - 1].toUpperCase();
// Join the array into a single word
Joined_Back = Split_String.join("")
return Joined_Back;
}
Step 1, 3 and 4 function
function Capitalize_Entire_String(str){
Regular_Exp = new RegExp(/\w+/g);
//Below is step 1
MatchedArray = str.match(Regular_Exp);
//Below is step 3
ConvertedArray = MatchedArray.map(Capitalize_FirstAndLast_One_Word);
// Below is step 4
ReturnedString = ConvertedArray.join(" ");
console.log(ReturnedString);
return ReturnedString;
}
Now you have everything. You can use the function like below.
Capitalize_Entire_String("hello there");
Capitalize_Entire_String("hello there this is a test");
Hope this helps. I am sorry if this turned out to be a redundant answer for you.

Reason your code don't work is the use of replace(). replace() will always replace the first character found.
There is absolutely no reason to run a nested loop. You can achieve this using a single loop.
function cap(str){
let spl = str.split(' ');
for(let i = 0; i < spl.length; i++){
let temp = spl[i];
temp = temp[0].toUpperCase() + temp.slice(1)
temp = temp.slice(0,-1) + temp[temp.length - 1].toUpperCase();
spl[i] = temp;
}
return spl.join(' ');
}
console.log(cap("a quick brown fox"))
An easier way is to use map() and template strings.
const cap = str => str
.split(' ')
.map(x => (
x.length === 1 ?
x.toUpperCase() :
`${x[0].toUpperCase()}${x.slice(1,-1)}${x[x.length -1].toUpperCase()}`)
)
.join(' ')
console.log(cap("a quick brown fox"))

To simplify the function, you could split the string into an array, map each word to the desired format, and join it together into a string again.
function Capitalize(str){
return str.split(" ").map((word) => word.charAt(0).toUpperCase() +
(word.length > 2 ? word.substring(1, word.length - 1) : "") +
(word.length > 1 ? word.charAt(word.length - 1).toUpperCase() : "")).join(" ");
}
console.log(Capitalize("i want to capitalize first and last letters"));

Congrats on starting out programming...
You can use this to achieve what you want to do
function capitalizeFirstAndLastLetters (str) {
const words = str.split(" "); // Split the string into words
const modified = [];
for (const word of words) {
if (word.length <= 2) {
modified.push(word.toUpperCase()); // If the word less than 3 characters, the whole word is capitalized
continue;
}
var firstCapital = word[0].toUpperCase(); // word[0] gets the first index of the string (I.e. the first letter of the word)
var lastCapital = word.slice(-1).toUpperCase(); // The slice function slices a portion of the word. slice(-1) gets the last letter
var middlePart = word.slice(1, -1); // slice(1, -1) means start slicing from the second index (I.e. 1) and ignore the last index
modified.push(firstCapital + middlePart + lastCapital);
}
return modified.join(" "); // Join each element in the modified array with a space to get the final string with each words first and last letters capitalized
}
capitalizeFirstAndLastLetters("hello there I am a boy"); // "HellO TherE I AM A BoY"

Try this, it worked for hello world because I guess you want the outcome to be HellO TherE right?:
function capitalize(str) {
var spl = str.split(" ");
var words = [];
for (let i = 0; i < spl.length; i++) {
//For every word
let changedWord = "";
for (let j = 0; j < spl[i].length; j++) {
//For every letter in each word
if(j == 0 || j == spl[i].length - 1) {
changedWord += spl[i][j].toUpperCase();
} else {
changedWord += spl[i][j].toLowerCase();
}
}
words.push(changedWord);
console.log(words);
}
console.log(words.join(" "));
}
capitalize("hello there");
ALSO: Make your functions name start with lowercase letter. Thats just how it is. Starting with uppercase letters usually are Classes. Just a quick tip

Maybe this does what you want, don't want to change much from your code:
function Capitalize(str) {
var spl = str.split(" ");
var words = [];
for (let i = 0; i < spl.length; i++) {
var word = spl[i];
var firstCapital = word[0].toUpperCase(); // get first character after capitalizing
var lastCapital = word.slice(-1).toUpperCase(); // get last character after capitalizing
var midOriginal = word.slice(1, -1);
words.push(firstCapital + midOriginal + lastCapital) // concat 3 parts
}
console.log(words.join(" "));
}
Capitalize("hello there");

This expression:
var LastLetterCapital = firstLetterCapital.replace(
word[size - 1],
word[size - 1].toUpperCase()
);
Is replacing the first occurrence of the character "e" in "There" with an uppercase "E".
Explanation
The replace() function first translates the first param: word[size - 1] to the literal character "e", then replaces the first occurrence of that character with the uppercase "E", resulting in the string "ThEre".
Solution
Use a regular expression as your first parameter instead, to ensure that the last character is targeted, regardless of whether or not that same character shows up anywhere else in the word:
var LastLetterCapital = firstLetterCapital.replace(/.$/, word[size - 1].toUpperCase());
function Capitalize(str) {
var spl = str.split(" ");
var words = [];
for (let i = 0; i < spl.length; i++) {
//For every word
var word = spl[i];
var size = spl[i].length;
for (let j = 0; j < size; j++) {
//For every letter in each word
var firstLetterCapital = word.replace(word[0], word[0].toUpperCase()); //Creates new array
var LastLetterCapital = firstLetterCapital.replace(/.$/, word[size - 1].toUpperCase());
}
words.push(LastLetterCapital);
}
console.log(words.join(" "));
}
Capitalize("hello there");

This should do the trick:
function Capitalize(str) {
return str.replace(/(\b\w|\w\b)/g, l => l.toUpperCase())
}
console.log(Capitalize('i want to be capitalized in a rather strange way'))
Explanation:
In the regular expression /(\b\w|\w\b)/g, \b means "word boundary" and \w means "word character", so (\b\w|\w\b) matches a word boundary followed by a word character OR a word character followed by a word boundary (i.e. the first and last character of words).
The matches of this expression are then passed to the inline function l => l.toUpperCase() (which itself is the second argument to replace) that capitalizes the passed letter.

the string type is immutable, so why don't you try to convert the string to an array like y = word.split('') and do y[0] = word.charAt(0).toUpperCase() and then convert back to string with y.join('')

How to print each word in a separate line. Each word is capitalized except the first

I am trying to create a function that prints each word on a new line. The argument given is a string with words that aren't separated by a space but capitalized except the first word i.e. "helloMyNameIsMark". I have something that works but wondering if there's a better way of doing this in javaScript.
separateWords = (string) => {
const letters = string.split('');
let word = "";
const words = letters.reduce((acc, letter, idx) => {
if (letter === letter.toUpperCase()) {
acc.push(word);
word = "";
word = word.concat(letter);
} else if (idx === letters.length - 1) {
word = word.concat(letter);
acc.push(word);
} else {
word = word.concat(letter)
}
return acc
}, []);
words.forEach(word => {
console.log(word)
})
}

You could use the regex [A-Z] and replace each upper case letter with \n prefix
const separateWords = str => str.replace(/[A-Z]/g, m => '\n' + m)
console.log(separateWords('helloMyNameIsMark'))
Or you could use a lookahead (?=[A-Z]) to split at each upper case letter to get an array of words. Then loop through the array to log each word:
const separateWords = str => str.split(/(?=[A-Z])/g)
separateWords('helloMyNameIsMark').forEach(w => console.log(w))

I would separate the breaking of words into an array and the printing of that array into two distinct functions. Regular expressions make that first part much easier than your reduce call. (But that reduce is a good thought if you don't see a regex solution.)
My version might look like this:
const separateWords = (str) => str .replace (/([A-Z])/g, " $1") .split (' ')
const printSeparateWords = (str) => separateWords (str) .forEach (word => console.log (word) )
printSeparateWords ("helloMyNameIsMark")

Very similar to adiga's answer, but it can actually be simpler:
const separateWords = str => str.replace(/[A-Z]/g, '\n$&');
This will also benefit from improved performance (might matter if used at scale).

Here's a more literal interpretation of your stated requirement: prints each word on a new line.
function separateWords(str){
let currentWord = '';
for (let chr of str){
if (chr == chr.toUpperCase()){
console.log(currentWord);
currentWord = chr;
} else {
currentWord += chr;
}
}
if (currentWord)
console.log(currentWord);
}
separateWords('helloMyNameIsMark');

regexp split and keep the seperator [duplicate]

I have a string:
var string = "aaaaaa<br />† bbbb<br />‡ cccc"
And I would like to split this string with the delimiter <br /> followed by a special character.
To do that, I am using this:
string.split(/<br \/>&#?[a-zA-Z0-9]+;/g);
I am getting what I need, except that I am losing the delimiter.
Here is the example: http://jsfiddle.net/JwrZ6/1/
How can I keep the delimiter?

I was having similar but slight different problem. Anyway, here are examples of three different scenarios for where to keep the deliminator.
"1、2、3".split("、") == ["1", "2", "3"]
"1、2、3".split(/(、)/g) == ["1", "、", "2", "、", "3"]
"1、2、3".split(/(?=、)/g) == ["1", "、2", "、3"]
"1、2、3".split(/(?!、)/g) == ["1、", "2、", "3"]
"1、2、3".split(/(.*?、)/g) == ["", "1、", "", "2、", "3"]
Warning: The fourth will only work to split single characters. ConnorsFan presents an alternative:
// Split a path, but keep the slashes that follow directories
var str = 'Animation/rawr/javascript.js';
var tokens = str.match(/[^\/]+\/?|\//g);

Use (positive) lookahead so that the regular expression asserts that the special character exists, but does not actually match it:
string.split(/<br \/>(?=&#?[a-zA-Z0-9]+;)/g);
See it in action:
var string = "aaaaaa<br />† bbbb<br />‡ cccc";
console.log(string.split(/<br \/>(?=&#?[a-zA-Z0-9]+;)/g));

If you wrap the delimiter in parantheses it will be part of the returned array.
string.split(/(<br \/>&#?[a-zA-Z0-9]+);/g);
// returns ["aaaaaa", "<br />†", "bbbb", "<br />‡", "cccc"]
Depending on which part you want to keep change which subgroup you match
string.split(/(<br \/>)&#?[a-zA-Z0-9]+;/g);
// returns ["aaaaaa", "<br />", "bbbb", "<br />", "cccc"]
You could improve the expression by ignoring the case of letters
string.split(/()&#?[a-z0-9]+;/gi);
And you can match for predefined groups like this: \d equals [0-9] and \w equals [a-zA-Z0-9_]. This means your expression could look like this.
string.split(/<br \/>(&#?[a-z\d]+;)/gi);
There is a good Regular Expression Reference on JavaScriptKit.

If you group the split pattern, its match will be kept in the output and it is by design:
If separator is a regular expression with capturing parentheses, then
each time separator matches, the results (including any undefined
results) of the capturing parentheses are spliced into the output
array.
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/split#description
You don't need a lookahead or global flag unless your search pattern uses one.
const str = `How much wood would a woodchuck chuck, if a woodchuck could chuck wood?`
const result = str.split(/(\s+)/);
console.log(result);
// We can verify the result
const isSame = result.join('') === str;
console.log({ isSame });
You can use multiple groups. You can be as creative as you like and what remains outside the groups will be removed:
const str = `How much wood would a woodchuck chuck, if a woodchuck could chuck wood?`
const result = str.split(/(\s+)(\w{1,2})\w+/);
console.log(result, result.join(''));

answered it here also JavaScript Split Regular Expression keep the delimiter
use the (?=pattern) lookahead pattern in the regex
example
var string = '500x500-11*90~1+1';
string = string.replace(/(?=[$-/:-?{-~!"^_`\[\]])/gi, ",");
string = string.split(",");
this will give you the following result.
[ '500x500', '-11', '*90', '~1', '+1' ]
Can also be directly split
string = string.split(/(?=[$-/:-?{-~!"^_`\[\]])/gi);
giving the same result
[ '500x500', '-11', '*90', '~1', '+1' ]

I made a modification to jichi's answer, and put it in a function which also supports multiple letters.
String.prototype.splitAndKeep = function(separator, method='seperate'){
var str = this;
if(method == 'seperate'){
str = str.split(new RegExp(`(${separator})`, 'g'));
}else if(method == 'infront'){
str = str.split(new RegExp(`(?=${separator})`, 'g'));
}else if(method == 'behind'){
str = str.split(new RegExp(`(.*?${separator})`, 'g'));
str = str.filter(function(el){return el !== "";});
}
return str;
};
jichi's answers 3rd method would not work in this function, so I took the 4th method, and removed the empty spaces to get the same result.
edit:
second method which excepts an array to split char1 or char2
String.prototype.splitAndKeep = function(separator, method='seperate'){
var str = this;
function splitAndKeep(str, separator, method='seperate'){
if(method == 'seperate'){
str = str.split(new RegExp(`(${separator})`, 'g'));
}else if(method == 'infront'){
str = str.split(new RegExp(`(?=${separator})`, 'g'));
}else if(method == 'behind'){
str = str.split(new RegExp(`(.*?${separator})`, 'g'));
str = str.filter(function(el){return el !== "";});
}
return str;
}
if(Array.isArray(separator)){
var parts = splitAndKeep(str, separator[0], method);
for(var i = 1; i < separator.length; i++){
var partsTemp = parts;
parts = [];
for(var p = 0; p < partsTemp.length; p++){
parts = parts.concat(splitAndKeep(partsTemp[p], separator[i], method));
}
}
return parts;
}else{
return splitAndKeep(str, separator, method);
}
};
usage:
str = "first1-second2-third3-last";
str.splitAndKeep(["1", "2", "3"]) == ["first", "1", "-second", "2", "-third", "3", "-last"];
str.splitAndKeep("-") == ["first1", "-", "second2", "-", "third3", "-", "last"];

An extension function splits string with substring or RegEx and the delimiter is putted according to second parameter ahead or behind.
String.prototype.splitKeep = function (splitter, ahead) {
var self = this;
var result = [];
if (splitter != '') {
var matches = [];
// Getting mached value and its index
var replaceName = splitter instanceof RegExp ? "replace" : "replaceAll";
var r = self[replaceName](splitter, function (m, i, e) {
matches.push({ value: m, index: i });
return getSubst(m);
});
// Finds split substrings
var lastIndex = 0;
for (var i = 0; i < matches.length; i++) {
var m = matches[i];
var nextIndex = ahead == true ? m.index : m.index + m.value.length;
if (nextIndex != lastIndex) {
var part = self.substring(lastIndex, nextIndex);
result.push(part);
lastIndex = nextIndex;
}
};
if (lastIndex < self.length) {
var part = self.substring(lastIndex, self.length);
result.push(part);
};
// Substitution of matched string
function getSubst(value) {
var substChar = value[0] == '0' ? '1' : '0';
var subst = '';
for (var i = 0; i < value.length; i++) {
subst += substChar;
}
return subst;
};
}
else {
result.add(self);
};
return result;
};
The test:
test('splitKeep', function () {
// String
deepEqual("1231451".splitKeep('1'), ["1", "231", "451"]);
deepEqual("123145".splitKeep('1', true), ["123", "145"]);
deepEqual("1231451".splitKeep('1', true), ["123", "145", "1"]);
deepEqual("hello man how are you!".splitKeep(' '), ["hello ", "man ", "how ", "are ", "you!"]);
deepEqual("hello man how are you!".splitKeep(' ', true), ["hello", " man", " how", " are", " you!"]);
// Regex
deepEqual("mhellommhellommmhello".splitKeep(/m+/g), ["m", "hellomm", "hellommm", "hello"]);
deepEqual("mhellommhellommmhello".splitKeep(/m+/g, true), ["mhello", "mmhello", "mmmhello"]);
});

I've been using this:
String.prototype.splitBy = function (delimiter) {
var
delimiterPATTERN = '(' + delimiter + ')',
delimiterRE = new RegExp(delimiterPATTERN, 'g');
return this.split(delimiterRE).reduce((chunks, item) => {
if (item.match(delimiterRE)){
chunks.push(item)
} else {
chunks[chunks.length - 1] += item
};
return chunks
}, [])
}
Except that you shouldn't mess with String.prototype, so here's a function version:
var splitBy = function (text, delimiter) {
var
delimiterPATTERN = '(' + delimiter + ')',
delimiterRE = new RegExp(delimiterPATTERN, 'g');
return text.split(delimiterRE).reduce(function(chunks, item){
if (item.match(delimiterRE)){
chunks.push(item)
} else {
chunks[chunks.length - 1] += item
};
return chunks
}, [])
}
So you could do:
var haystack = "aaaaaa<br />† bbbb<br />‡ cccc"
var needle = '<br \/>&#?[a-zA-Z0-9]+;';
var result = splitBy(haystack , needle)
console.log( JSON.stringify( result, null, 2) )
And you'll end up with:
[
"<br />† bbbb",
"<br />‡ cccc"
]

Most of the existing answers predate the introduction of lookbehind assertions in JavaScript in 2018. You didn't specify how you wanted the delimiters to be included in the result. One typical use case would be sentences delimited by punctuation ([.?!]), where one would want the delimiters to be included at the ends of the resulting strings. This corresponds to the fourth case in the accepted answer, but as noted there, that solution only works for single characters. Arbitrary strings with the delimiters appended at the end can be formed with a lookbehind assertion:
'It is. Is it? It is!'.split(/(?<=[.?!])/)
/* [ 'It is.', ' Is it?', ' It is!' ] */

I know that this is a bit late but you could also use lookarounds
var string = "aaaaaa<br />† bbbb<br />‡ cccc";
var array = string.split(/(?<=<br \/>)/);
console.log(array);

I've also came up with this solution. No regex needed, very readable.
const str = "hello world what a great day today balbla"
const separatorIndex = str.indexOf("great")
const parsedString = str.slice(separatorIndex)
console.log(parsedString)

Replace text but keep case

I have a set of strings that I need to replace, but I need to keep the case of letters.
Both the input words and output words are of the same length.
For example, if I need to replace "abcd" with "qwer", then the following should happen:
"AbcD" translates to "QweR"
"abCd" translates to "qwEr"
and so on.
Right now I'm using JavaScript's replace, but capital letters are lost on translation.
r = new RegExp( "(" + 'asdf' + ")" , 'gi' );
"oooAsdFoooo".replace(r, "qwer");
Any help would be appreciated.

Here’s a helper:
function matchCase(text, pattern) {
var result = '';
for(var i = 0; i < text.length; i++) {
var c = text.charAt(i);
var p = pattern.charCodeAt(i);
if(p >= 65 && p < 65 + 26) {
result += c.toUpperCase();
} else {
result += c.toLowerCase();
}
}
return result;
}
Then you can just:
"oooAsdFoooo".replace(r, function(match) {
return matchCase("qwer", match);
});

I'll leave this here for reference.
Scenario: case-insensitive search box on list of items, partial match on string should be displayed highlighted but keeping original case.
highlight() {
const re = new RegExp(this.searchValue, 'gi'); // global, insensitive
const newText = name.replace(re, `<b>$&</b>`);
return newText;
}
the $& is the matched text with case

String.prototype.translateCaseSensitive = function (fromAlphabet, toAlphabet) {
var fromAlphabet = fromAlphabet.toLowerCase(),
toAlphabet = toAlphabet.toLowerCase(),
re = new RegExp("[" + fromAlphabet + "]", "gi");
return this.replace(re, function (char) {
var charLower = char.toLowerCase(),
idx = fromAlphabet.indexOf(charLower);
if (idx > -1) {
if (char === charLower) {
return toAlphabet[idx];
} else {
return toAlphabet[idx].toUpperCase();
}
} else {
return char;
}
});
};
and
"AbcD".translateCaseSensitive("abcdefg", "qwertyu")
will return:
"QweR"

Here's a replaceCase function:
We turn the input pattern into a regular expression
We have a nested replacer function which iterates through every character
We use regular expression /[A-Z]/ to identify capital letters, otherwise we assume everything is in lowercase
function replaceCase(str, pattern, newStr) {
const rx = new RegExp(pattern, "ig")
const replacer = (c, i) => c.match(/[A-Z]/) ? newStr[i].toUpperCase() : newStr[i]
return str.replace(rx, (oldStr) => oldStr.replace(/./g, replacer) )
}
let out = replaceCase("This is my test string: AbcD", "abcd", "qwer")
console.log(out) // This is my test string: QweR
out = replaceCase("This is my test string: abCd", "abcd", "qwer")
console.log(out) // This is my test string: qwEr

You could create your own replace function such as
if(!String.prototype.myreplace){
String.prototype.myreplace = (function(obj){
return this.replace(/[a-z]{1,1}/gi,function(a,b){
var r = obj[a.toLowerCase()] || a;
return a.charCodeAt(0) > 96? r.toLowerCase() : r.toUpperCase();
});
});
}
This takes in a object that maps different letters. and it can be called such as follows
var obj = {a:'q',b:'t',c:'w'};
var s = 'AbCdea';
var n = s.myreplace(obj);
console.log(n);
This means you could potentially pass different objects in with different mappings if need be. Here's a simple fiddle showing an example (note the object is all lowercase but the function itself looks at case of the string as well)

Expanding on Ryan O'Hara's answer, the below solution avoids using charCodes and the issues that maybe encountered in using them. It also ensures the replacement is complete when the strings are of different lengths.
function enforceLength(text, pattern, result) {
if (text.length > result.length) {
result = result.concat(text.substring(result.length, text.length));
}
if (pattern.length > text.length) {
result = result.substring(0, text.length);
}
return result;
}
function matchCase(text, pattern){
var result = '';
for (var i =0; i < pattern.length; i++){
var c = text.charAt(i);
var p = pattern.charAt(i);
if(p === p.toUpperCase()) {
result += c.toUpperCase();
} else {
result += c.toLowerCase();
}
}
return enforceLength(text, pattern, result);
}

This should replace while preserving the case. Please let me know if anyone finds any flaws in this solution. I hope this helps. Thank-you!
function myReplace(str, before, after) {
var match=function(before,after){
after=after.split('');
for(var i=0;i<before.length;i++)
{
if(before.charAt(i)==before[i].toUpperCase())
{
after[i]=after[i].toUpperCase();
}
else if(before.charAt(i)==before[i].toLowerCase())
{
after[i]=after[i].toLowerCase();
}
return after.join('');
}
};
console.log(before,match(before,after));
str =str.replace(before,match(before,after));
return str;
}
myReplace("A quick brown fox jumped over the lazy dog", "jumped", "leaped");

I had a sentence where I had to replace each word with another word and that word can be longer/shorter than the word its replacing so its similar to the question but instead of a fixed length, they're dynamic.
My solution
For simplicity, I am focusing on a single word.
const oldWord = "tEsT";
const newWord = "testing";
Split both words so that I can iterate over each individual letters.
const oldWordLetters = oldWord.split("");
const newWordLetters = newWord.split("");
Now, I would iterate over the newWord letters and use its index to then get the corresponding oldWord letter in the same position. Then I would check if the old letter is capital and if it is then make the new letter in the same position capital as well.
for (const [i, letter] of newWordLetters.entries()) {
const oldLetter = oldWordLetters[i];
// stop iterating if oldWord is shorter (not enough letters to copy case).
if (!oldLetter) {
break;
}
const isCapital = oldLetter === oldLetter.toUpperCase();
// make the new letter in the same position as the old letter capital
if (isCapital) {
newWordLetters[i] = letter.toUpperCase();
}
}
The final world would be tEsTing after joining the letters again.
const finalWord = newWordLetters.join("");
console.log(finalWord); // "tEsTing"
Full code
const oldWord = "tEsT";
const newWord = "testing";
const oldWordLetters = oldWord.split("");
const newWordLetters = newWord.split("");
for (const [i, letter] of newWordLetters.entries()) {
const oldLetter = oldWordLetters[i];
// stop iterating if oldWord is shorter (not enough letters to copy case).
if (!oldLetter) {
break;
}
const isCapital = oldLetter === oldLetter.toUpperCase();
// make the new letter in the same position as the old letter capital
if (isCapital) {
newWordLetters[i] = letter.toUpperCase();
}
}
const finalWord = newWordLetters.join("");
console.log(finalWord);

I think this could work
function formatItem(text, searchText){
const search = new RegExp(escapeRegExp(searchText), 'iu')
return text?.toString().replace(search, (m) => `<b>${m}</b>`)
}
function escapeRegExp(text) {
return text?.replace(/[-[\]{}()*+?.,\\^$|#\s]/g, '\\$&') ?? '';
}

Thank you for asking this. I had the same problem when I wanted to search text and replace certain words with links, which was a slightly more specific situation because it is replacing text strings with html strings. I'll put my solution here in case anyone who finds this is doing anything similar.
let elmt = document.getElementById('the-element');
let a = document.createElement('a');
a.href = "https://www.example.com";
let re = new RegExp('the string to find', 'gi');
elmt.innerHTML = elmt.innerHTML.replaceAll(re, function (match) {
a.innerText = match;
return a.outerHTML;
});
So the regular expression ensures that it searches for case-insensitive matches, and the function as the second argument of the replaceAll function specifies that it is supposed to set the innerText of the new tag equal to the old string verbatim, before then returning the outerHTML of the whole tag.

Here is a replaceAllCaseSensitive function. If your want, you can change replaceAll by replace.
const replaceAllCaseSensitive = (
text, // Original string
pattern, // RegExp with the pattern you want match. It must include the g (global) and i (case-insensitive) flags.
replacement // string with the replacement
) => {
return text.replaceAll(pattern, (match) => {
return replacement
.split("")
.map((char, i) =>
match[i] === match[i].toUpperCase() ? char.toUpperCase() : char
)
.join("");
});
};
console.log(replaceAllCaseSensitive("AbcD abCd", /abcd/gi, "qwer"));
// outputs "QweR qwEr"
console.log(replaceAllCaseSensitive("AbcD abCd", /abcd/gi, "qwe"));
// outputs "Qwe qwE"
The function works even if replacement is shorter than match.