I have heard that JavaScript has a function called search() that can search for a string ( lets call it A ) in another string ( B ) and it will return the first position at which A was found in B.
var str = "Ana has apples!";
var n = str.search(" ");
The code should return 3 as its the first position in which the space was found in str.
And I was wondering if there is a function that can find the next spaces in my string.
For example I want to find the length of the first word in my string and I could easily do this If I knew the its starting position and its ending one.
If there is such a function, are there any better than it for such things?
You need to use String.indexOf method. It accepts the following arguments:
str.indexOf(searchValue[, fromIndex])
So you can do this:
var str = "Ana has apples!";
var pos1 = str.indexOf(" "); // 3
var pos2 = str.indexOf(" ", pos1 + 1); // 7
console.log(pos2 - pos1 - 1); // 3... length of the second word
.indexOf(…) will give you the first occurence of the " " (starting at 0):
var str = "Ana has apples!";
var n = str.indexOf(" ");
console.log(n);
If you want all occurences, this can be achieved easily using a RegExp with a while:
var str = "Ana has apples! A lot.";
var re = new RegExp(" ","ig");
var spaces = [];
while ((match = re.exec(str))) {
spaces.push(match.index);
}
// Output the whole array of results
console.log(spaces);
// You can also access the spaces position separately:
console.log('1st space:', spaces[0]);
console.log('2nd space:', spaces[1]);
⋅
⋅
⋅
Or… you can use a do {} while () loop:
var str = "Ana has apples! A lot.";
var i = 0,
n = 0;
do {
n = str.indexOf(" ");
if (n > -1) {
i += n;
console.log(i);
str = str.slice(n + 1);
i++;
}
}
while (n > -1);
Then, you can make a function of it:
var str = "Ana has apples! A lot.";
// Function
function indexsOf(str, sub) {
var arr = [],
i = 0,
n = 0;
do {
n = str.indexOf(" ");
if (n > -1) {
i += n;
arr.push(i);
str = str.slice(n + 1);
i++;
}
}
while (n > -1);
return arr;
}
var spaces = indexsOf(str, ' ')
// Output the whole array of results
console.log(spaces);
// You can also access the spaces position separately:
console.log('1st space:', spaces[0]);
console.log('2nd space:', spaces[1]);
⋅
⋅
⋅
Hope it helps.
Better for matching is to use regex. There is option like match group using group 'g' flag
var str = "Ana has apples !";
var regBuilder = new RegExp(" ","ig");
var matched = "";
while((matched = regBuilder.exec(str))){
console.log(matched + ", position : " +matched.index);
}
str = "Ana is Ana no one is better than Ana";
regBuilder = new RegExp("Ana","ig");
while((matched = regBuilder.exec(str))){
console.log(matched + ", position : " +matched.index);
}
'i' flag used to ignore case sensitive
You can check for other flags too here
Try this:
const str = "Ana has apples!";
const spaces = str.split('')
.map((c, i) => (c === ' ') ? i : -1)
.filter((c) => c !== -1);
console.log(spaces);
Then you will all the spaces positions.
I got this string:
var longText="This is a superuser test, super user is is super important!";
I want to know how many times the string "su" is in longText and the position of each "su".
I was trying with:
var nr4 = longText.replace("su", "").length;
And the difference of lenght between the main text and the nr4 divided by "su" lenght beeing 2 is resulting a number of repetitions but i bet there is a better way of doing it.
For example
var parts=longText.split("su");
alert(parts.length-1); // length will be two if there is one "su"
More details using exec
FIDDLE
var re =/su/g, pos=[];
while ((result = re.exec(longText)) !== null) {
pos.push(result.index);
}
if (pos.length>0) alert(pos.length+" found at "+pos.join(","));
Use exec. Example amended from the MDN code. len contains the number of times su appears.
var myRe = /su/g;
var str = "This is a superuser test, super user is is super important!";
var myArray, len = 0;
while ((myArray = myRe.exec(str)) !== null) {
len++;
var msg = "Found " + myArray[0] + ". ";
msg += "Next match starts at " + myRe.lastIndex;
console.log(msg, len);
}
// "Found su. Next match starts at 12" 1
// "Found su. Next match starts at 28" 2
// "Found su. Next match starts at 45" 3
DEMO
Could do :
var indexesOf = function(baseString, strToMatch){
var baseStr = new String(baseString);
var wordLen = strToMatch.length;
var listSu = [];
// Number of strToMatch occurences
var nb = baseStr.split(strToMatch).length - 1;
for (var i = 0, len = nb; i < len; i++){
var ioF = baseStr.indexOf(strToMatch);
baseStr = baseStr.slice(ioF + wordLen, baseStr.length);
if (i > 0){
ioF = ioF + listSu[i-1] + wordLen;
}
listSu.push(ioF);
}
return listSu;
}
indexesOf("This is a superuser test, super user is is super important!","su");
return [10, 26, 43]
var longText="This is a superuser test, super user is is super important!";
var count = 0;
while(longText.indexOf("su") != -1) { // NB the indexOf() method is case sensitive!
longText = longText.replace("su",""); //replace first occurence of 'su' with a void string
count++;
}
function LetterChanges(str) {
var alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
for (var i = 0; i < str.length; i++) {
var index = alphabet.indexOf(str[i])
if (/[a-zA-Z]/.test(str[i])) {
str = str.replace(str[i], alphabet.charAt(index + 1));
}
if (/[aeiou]/.test(str[i])) {
str = str.replace(str[i], alphabet.charAt(index + 26));
}
}
return str;
}
When I call LetterChanges("hello"), it returns 'Ifmmp' which is correct, but when "sent" is passed it returns 'ufOt' instead of 'tfOu'. Why is that?
str.replace() replaces the first occurrence of the match in the string with the replacement. LetterChanges("sent") does the following:
i = 0 : str.replace("s", "t"), now str = "tent"
i = 1 : str.replace("e", "f"), now str = "tfnt"
i = 2 : str.replace("n", "o"), now str = "tfot", then
str.replace("o", "O"), now str = "tfOt"
i = 3 : str.replace("t", "u"), now str = "ufOt"
return str
There are several issues. The main one is that you could inadvertently change the same letter several times.
Let's see what happens to the s in sent. You first change it to t. However, when it comes to changing the final letter, which is also t, you change the first letter again, this time from t to u.
Another, smaller, issue is the handling of the letter z.
Finally, your indexing in the second if is off by one: d becomes D and not E.
You can use String.replace to avoid that:
function LetterChanges(str) {
return str.replace(/[a-zA-Z]/g, function(c){
return String.fromCharCode(c.charCodeAt(0)+1);
}).replace(/[aeiou]/g, function(c){
return c.toUpperCase();
});
}
But there is still a bug: LetterChanges('Zebra') will return '[fcsb'. I assume that is not your intention. You will have to handle the shift.
Try this one:
function LetterChanges(str) {
var alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
var result = '';
var temp;
for (var i = 0; i < str.length; i++) {
var index = alphabet.indexOf(str[i])
if (/[a-zA-Z]/.test(str[i])) {
//str = str.replace(str[i], alphabet.charAt(index + 1));
temp= alphabet.charAt(index + 1);
index = index+1;
}
else if(str[i] == ' '){
temp = ' ';
}
if (/[aeiou]/.test(temp)) {
temp = alphabet.charAt(index + 26);
}
result += temp;
}
return result;
}
var str = 'bcd12';
str = str.replace(/[a-z]/gi, function(char) { //call replace method
char = String.fromCharCode(char.charCodeAt(0)+1);//increment ascii code of char variable by 1 .FromCharCode() method will convert Unicode values into character
if (char=='{' || char=='[') char = 'a'; //if char values goes to "[" or"{" on incrementing by one as "[ ascii value is 91 just after Z" and "{ ascii value is 123 just after "z" so assign "a" to char variable..
if (/[aeiuo]/.test(char)) char = char.toUpperCase();//convert vowels to uppercase
return char;
});
console.log(str);
Check this code sample. There is no bug in it. Not pretty straight forward but Works like a charm. Cheers!
function LetterChanges(str) {
var temp = str;
var tempArr = temp.split("");//Split the input to convert it to an Array
tempArr.forEach(changeLetter);/*Not many use this but this is the referred way of using loops in javascript*/
str = tempArr.join("");
// code goes here
return str;
}
function changeLetter(ele,index,arr) {
var lowerLetters ="abcdefghijklmnopqrstuvwxyza";
var upperLetters ="ABCDEFGHIJKLMNOPQRSTUVWXYZA";
var lowLetterArr = lowerLetters.split("");
var upLetterArr = upperLetters.split("");
var i =0;
for(i;i<lowLetterArr.length;i++){
if(arr[index] === lowLetterArr[i]){
arr[index] = lowLetterArr[i+1];
arr[index]=arr[index].replace(/[aeiou]/g,arr[index].toUpperCase());
return false;
}
if(arr[index] === upLetterArr[i]){
arr[index] = upLetterArr[i+1];
arr[index]=arr[index].replace(/[aeiou]/g,arr[index].toUpperCase());
return false;
}
}
}
// keep this function call here
// to see how to enter arguments in JavaScript scroll down
LetterChanges(readline());
I have a string, let's say Hello world and I need to replace the char at index 3. How can I replace a char by specifying a index?
var str = "hello world";
I need something like
str.replaceAt(0,"h");
In JavaScript, strings are immutable, which means the best you can do is to create a new string with the changed content and assign the variable to point to it.
You'll need to define the replaceAt() function yourself:
String.prototype.replaceAt = function(index, replacement) {
return this.substring(0, index) + replacement + this.substring(index + replacement.length);
}
And use it like this:
var hello = "Hello World";
alert(hello.replaceAt(2, "!!")); // He!!o World
There is no replaceAt function in JavaScript. You can use the following code to replace any character in any string at specified position:
function rep() {
var str = 'Hello World';
str = setCharAt(str,4,'a');
alert(str);
}
function setCharAt(str,index,chr) {
if(index > str.length-1) return str;
return str.substring(0,index) + chr + str.substring(index+1);
}
<button onclick="rep();">click</button>
You can't. Take the characters before and after the position and concat into a new string:
var s = "Hello world";
var index = 3;
s = s.substring(0, index) + 'x' + s.substring(index + 1);
str = str.split('');
str[3] = 'h';
str = str.join('');
There are lot of answers here, and all of them are based on two methods:
METHOD1: split the string using two substrings and stuff the character between them
METHOD2: convert the string to character array, replace one array member and join it
Personally, I would use these two methods in different cases. Let me explain.
#FabioPhms: Your method was the one I initially used and I was afraid that it is bad on string with lots of characters. However, question is what's a lot of characters? I tested it on 10 "lorem ipsum" paragraphs and it took a few milliseconds. Then I tested it on 10 times larger string - there was really no big difference. Hm.
#vsync, #Cory Mawhorter: Your comments are unambiguous; however, again, what is a large string? I agree that for 32...100kb performance should better and one should use substring-variant for this one operation of character replacement.
But what will happen if I have to make quite a few replacements?
I needed to perform my own tests to prove what is faster in that case. Let's say we have an algorithm that will manipulate a relatively short string that consists of 1000 characters. We expect that in average each character in that string will be replaced ~100 times. So, the code to test something like this is:
var str = "... {A LARGE STRING HERE} ...";
for(var i=0; i<100000; i++)
{
var n = '' + Math.floor(Math.random() * 10);
var p = Math.floor(Math.random() * 1000);
// replace character *n* on position *p*
}
I created a fiddle for this, and it's here.
There are two tests, TEST1 (substring) and TEST2 (array conversion).
Results:
TEST1: 195ms
TEST2: 6ms
It seems that array conversion beats substring by 2 orders of magnitude! So - what the hell happened here???
What actually happens is that all operations in TEST2 are done on array itself, using assignment expression like strarr2[p] = n. Assignment is really fast compared to substring on a large string, and its clear that it's going to win.
So, it's all about choosing the right tool for the job. Again.
Work with vectors is usually most effective to contact String.
I suggest the following function:
String.prototype.replaceAt=function(index, char) {
var a = this.split("");
a[index] = char;
return a.join("");
}
Run this snippet:
String.prototype.replaceAt=function(index, char) {
var a = this.split("");
a[index] = char;
return a.join("");
}
var str = "hello world";
str = str.replaceAt(3, "#");
document.write(str);
In Javascript strings are immutable so you have to do something like
var x = "Hello world"
x = x.substring(0, i) + 'h' + x.substring(i+1);
To replace the character in x at i with 'h'
function dothis() {
var x = document.getElementById("x").value;
var index = document.getElementById("index").value;
var text = document.getElementById("text").value;
var length = document.getElementById("length").value;
var arr = x.split("");
arr.splice(index, length, text);
var result = arr.join("");
document.getElementById('output').innerHTML = result;
console.log(result);
}
dothis();
<input id="x" type="text" value="White Dog" placeholder="Enter Text" />
<input id="index" type="number" min="0"value="6" style="width:50px" placeholder="index" />
<input id="length" type="number" min="0"value="1" style="width:50px" placeholder="length" />
<input id="text" type="text" value="F" placeholder="New character" />
<br>
<button id="submit" onclick="dothis()">Run</button>
<p id="output"></p>
This method is good for small length strings but may be slow for larger text.
var x = "White Dog";
var arr = x.split(""); // ["W", "h", "i", "t", "e", " ", "D", "o", "g"]
arr.splice(6, 1, 'F');
/*
Here 6 is starting index and 1 is no. of array elements to remove and
final argument 'F' is the new character to be inserted.
*/
var result = arr.join(""); // "White Fog"
One-liner using String.replace with callback (no emoji support):
// 0 - index to replace, 'f' - replacement string
'dog'.replace(/./g, (c, i) => i == 0? 'f': c)
// "fog"
Explained:
//String.replace will call the callback on each pattern match
//in this case - each character
'dog'.replace(/./g, function (character, index) {
if (index == 0) //we want to replace the first character
return 'f'
return character //leaving other characters the same
})
Generalizing Afanasii Kurakin's answer, we have:
function replaceAt(str, index, ch) {
return str.replace(/./g, (c, i) => i == index ? ch : c);
}
let str = 'Hello World';
str = replaceAt(str, 1, 'u');
console.log(str); // Hullo World
Let's expand and explain both the regular expression and the replacer function:
function replaceAt(str, index, newChar) {
function replacer(origChar, strIndex) {
if (strIndex === index)
return newChar;
else
return origChar;
}
return str.replace(/./g, replacer);
}
let str = 'Hello World';
str = replaceAt(str, 1, 'u');
console.log(str); // Hullo World
The regular expression . matches exactly one character. The g makes it match every character in a for loop. The replacer function is called given both the original character and the index of where that character is in the string. We make a simple if statement to determine if we're going to return either origChar or newChar.
var str = "hello world";
console.log(str);
var arr = [...str];
arr[0] = "H";
str = arr.join("");
console.log(str);
This works similar to Array.splice:
String.prototype.splice = function (i, j, str) {
return this.substr(0, i) + str + this.substr(j, this.length);
};
You could try
var strArr = str.split("");
strArr[0] = 'h';
str = strArr.join("");
this is easily achievable with RegExp!
const str = 'Hello RegEx!';
const index = 11;
const replaceWith = 'p';
//'Hello RegEx!'.replace(/^(.{11})(.)/, `$1p`);
str.replace(new RegExp(`^(.{${ index }})(.)`), `$1${ replaceWith }`);
//< "Hello RegExp"
Using the spread syntax, you may convert the string to an array, assign the character at the given position, and convert back to a string:
const str = "hello world";
function replaceAt(s, i, c) {
const arr = [...s]; // Convert string to array
arr[i] = c; // Set char c at pos i
return arr.join(''); // Back to string
}
// prints "hallo world"
console.log(replaceAt(str, 1, 'a'));
You could try
var strArr = str.split("");
strArr[0] = 'h';
str = strArr.join("");
Check out this function for printing steps
steps(3)
// '# '
// '## '
// '###'
function steps(n, i = 0, arr = Array(n).fill(' ').join('')) {
if (i === n) {
return;
}
str = arr.split('');
str[i] = '#';
str = str.join('');
console.log(str);
steps(n, (i = i + 1), str);
}
#CemKalyoncu: Thanks for the great answer!
I also adapted it slightly to make it more like the Array.splice method (and took #Ates' note into consideration):
spliceString=function(string, index, numToDelete, char) {
return string.substr(0, index) + char + string.substr(index+numToDelete);
}
var myString="hello world!";
spliceString(myString,myString.lastIndexOf('l'),2,'mhole'); // "hello wormhole!"
If you want to replace characters in string, you should create mutable strings. These are essentially character arrays. You could create a factory:
function MutableString(str) {
var result = str.split("");
result.toString = function() {
return this.join("");
}
return result;
}
Then you can access the characters and the whole array converts to string when used as string:
var x = MutableString("Hello");
x[0] = "B"; // yes, we can alter the character
x.push("!"); // good performance: no new string is created
var y = "Hi, "+x; // converted to string: "Hi, Bello!"
You can extend the string type to include the inset method:
String.prototype.append = function (index,value) {
return this.slice(0,index) + value + this.slice(index);
};
var s = "New string";
alert(s.append(4,"complete "));
Then you can call the function:
You can concatenate using sub-string function at first select text before targeted index and after targeted index then concatenate with your potential char or string. This one is better
const myString = "Hello world";
const index = 3;
const stringBeforeIndex = myString.substring(0, index);
const stringAfterIndex = myString.substring(index + 1);
const replaceChar = "X";
myString = stringBeforeIndex + replaceChar + stringAfterIndex;
console.log("New string - ", myString)
or
const myString = "Hello world";
let index = 3;
myString = myString.substring(0, index) + "X" + myString.substring(index + 1);
I did a function that does something similar to what you ask, it checks if a character in string is in an array of not allowed characters if it is it replaces it with ''
var validate = function(value){
var notAllowed = [";","_",">","<","'","%","$","&","/","|",":","=","*"];
for(var i=0; i<value.length; i++){
if(notAllowed.indexOf(value.charAt(i)) > -1){
value = value.replace(value.charAt(i), "");
value = validate(value);
}
}
return value;
}
Here is a version I came up with if you want to style words or individual characters at their index in react/javascript.
replaceAt( yourArrayOfIndexes, yourString/orArrayOfStrings )
Working example: https://codesandbox.io/s/ov7zxp9mjq
function replaceAt(indexArray, [...string]) {
const replaceValue = i => string[i] = <b>{string[i]}</b>;
indexArray.forEach(replaceValue);
return string;
}
And here is another alternate method
function replaceAt(indexArray, [...string]) {
const startTag = '<b>';
const endTag = '</b>';
const tagLetter = i => string.splice(i, 1, startTag + string[i] + endTag);
indexArray.forEach(tagLetter);
return string.join('');
}
And another...
function replaceAt(indexArray, [...string]) {
for (let i = 0; i < indexArray.length; i++) {
string = Object.assign(string, {
[indexArray[i]]: <b>{string[indexArray[i]]}</b>
});
}
return string;
}
Here is my solution using the ternary and map operator. More readable, maintainable end easier to understand if you ask me.
It is more into es6 and best practices.
function replaceAt() {
const replaceAt = document.getElementById('replaceAt').value;
const str = 'ThisIsATestStringToReplaceCharAtSomePosition';
const newStr = Array.from(str).map((character, charIndex) => charIndex === (replaceAt - 1) ? '' : character).join('');
console.log(`New string: ${newStr}`);
}
<input type="number" id="replaceAt" min="1" max="44" oninput="replaceAt()"/>
My safe approach with negative indexes
/**
* #param {string} str
* #param {number} index
* #param {string} replacement
* #returns {string}
*/
static replaceAt (str, index, replacement)
{
if (index < 0) index = str.length + index
if (index < 0 || index >= str.length) throw new Error(`Index (${index}) out of bounds "${str}"`)
return str.substring(0, index) + replacement + str.substring(index + 1)
}
Use it like that:
replaceAt('my string', -1, 'G') // 'my strinG'
replaceAt('my string', 2, 'yy') // 'myyystring'
replaceAt('my string', 22, 'yy') // Uncaught Error: Index (22) out of bounds "my string"
Lets say you want to replace Kth index (0-based index) with 'Z'.
You could use Regex to do this.
var re = var re = new RegExp("((.){" + K + "})((.){1})")
str.replace(re, "$1A$`");
You can use the following function to replace Character or String at a particular position of a String. To replace all the following match cases use String.prototype.replaceAllMatches() function.
String.prototype.replaceMatch = function(matchkey, replaceStr, matchIndex) {
var retStr = this, repeatedIndex = 0;
for (var x = 0; (matchkey != null) && (retStr.indexOf(matchkey) > -1); x++) {
if (repeatedIndex == 0 && x == 0) {
repeatedIndex = retStr.indexOf(matchkey);
} else { // matchIndex > 0
repeatedIndex = retStr.indexOf(matchkey, repeatedIndex + 1);
}
if (x == matchIndex) {
retStr = retStr.substring(0, repeatedIndex) + replaceStr + retStr.substring(repeatedIndex + (matchkey.length));
matchkey = null; // To break the loop.
}
}
return retStr;
};
Test:
var str = "yash yas $dfdas.**";
console.log('Index Matched replace : ', str.replaceMatch('as', '*', 2) );
console.log('Index Matched replace : ', str.replaceMatch('y', '~', 1) );
Output:
Index Matched replace : yash yas $dfd*.**
Index Matched replace : yash ~as $dfdas.**
I se this to make a string proper case, that is, the first letter is Upper Case and all the rest are lower case:
function toProperCase(someString){
return someString.charAt(0).toUpperCase().concat(someString.toLowerCase().substring(1,someString.length));
};
This first thing done is to ensure ALL the string is lower case - someString.toLowerCase()
then it converts the very first character to upper case -someString.charAt(0).toUpperCase()
then it takes a substring of the remaining string less the first character -someString.toLowerCase().substring(1,someString.length))
then it concatenates the two and returns the new string -someString.charAt(0).toUpperCase().concat(someString.toLowerCase().substring(1,someString.length))
New parameters could be added for the replacement character index and the replacement character, then two substrings formed and the indexed character replaced then concatenated in much the same way.
The solution does not work for negative index so I add a patch to it.
String.prototype.replaceAt=function(index, character) {
if(index>-1) return this.substr(0, index) + character + this.substr(index+character.length);
else return this.substr(0, this.length+index) + character + this.substr(index+character.length);
}
"hello world".replace(/(.{3})./, "$1h")
// 'helho world'
The methods on here are complicated.
I would do it this way:
var myString = "this is my string";
myString = myString.replace(myString.charAt(number goes here), "insert replacement here");
This is as simple as it gets.