Related
I'm working to update this function which currently takes the content and replaces any instance of the target with the substitute.
var content = textArea.value; //should be in string form
var target = targetTextArea.value;
var substitute = substituteTextArea.value;
var expression = new RegExp(target, "g"); //In order to do a global replace(replace more than once) we have to use a regex
content = content.replace(expression, substitute);
textArea.value = content.split(",");
This code somewhat works... given the input "12,34,23,13,22,1,17" and told to replace "1" with "99" the output would be "992,34,23,993,22,99,997" when it should be "12,34,23,13,22,99,17". The replace should only be performed when the substitute is equal to the number, not a substring of the number.
I dont understand the comment about the regex needed to do a global replace, I'm not sure if that's a clue?
It's also worth mentioning that I'm dealing with a string separated by either commas or spaces.
Thanks!
You could do this if regex is not a requirement
var str = "12,34,23,13,22,1,17";
var strArray = str.split(",");
for(var item in strArray)
{
if(strArray[item] === "1")
{
strArray[item] = "99"
}
}
var finalStr = strArray.join()
finalStr will be "12,34,23,13,22,99,17"
Try with this
var string1 = "12,34,23,13,22,1,17";
var pattern = /1[^\d]/g;
// or pattern = new RegExp(target+'[^\\d]', 'g');
var value = substitute+",";//Replace comma with space if u uses space in between
string1 = string1.replace(pattern, value);
console.log(string1);
Try this
target = target.replace(/,1,/g, ',99,');
Documentation
EDIT: When you say: "a string separated by either commas or spaces"
Do you mean either a string with all commas, or a string with all spaces?
Or do you have 1 string with both commas and spaces?
My answer has no regex, nothing fancy ...
But it looks like you haven't got an answer that works yet
<div id="log"></div>
<script>
var myString = "12,34,23,13,22,1,17";
var myString2 = "12 34 23 13 22 1 17";
document.getElementById('log').innerHTML += '<br/>with commas: ' + replaceItem(myString, 1, 99);
document.getElementById('log').innerHTML += '<br/>with spaces: ' + replaceItem(myString2, 1, 99);
function replaceItem(string, needle, replace_by) {
var deliminator = ',';
// split the string into an array of items
var items = string.split(',');
// >> I'm dealing with a string separated by either commas or spaces
// so if split had no effect (no commas found), we try again with spaces
if(! (items.length > 1)) {
deliminator = ' ';
items = string.split(' ');
}
for(var i=0; i<items.length; i++) {
if(items[i] == needle) {
items[i] = replace_by;
}
}
return items.join(deliminator);
}
</script>
I'm looking for a regex that will remove all characters that have been repeated in a string. I already solved this using a loop. Just wondering if there is a regex that can do the same.
this is what i have so far:
function onlyUnique(str) {
var re = /(.)(?=.*\1)/g
return str.replace(re, '');
}
This string:
"rc iauauc!gcusa_usdiscgaesracg"
should end up as this:
" !_de"
You can use Array#filter with Array#indexOf and Array#lastIndexOf to check if the element is repeated.
var str = "rc iauauc!gcusa_usdiscgaesracg";
// Split to get array
var arr = str.split('');
// Filter splitted array
str = arr.filter(function (e) {
// If index and lastIndex are equal, the element is not repeated
return arr.indexOf(e) === arr.lastIndexOf(e);
}).join(''); // Join to get string from array
console.log(str);
document.write(str);
well, no idea if regex can do that, but you could work it out using for loop, like:
function unikChars(str) {
store = [];
for (var a = 0, len = str.length; a < len; a++) {
var ch = str.charAt(a);
if (str.indexOf(ch) == a && str.indexOf(ch, a + 1) == -1) {
store.push(ch);
}
}
return store.join("");
}
var str = 'rc iauauc!gcusa_usdiscgaesracg';
console.log(unikChars(str)); //gives !_de
Demo:: jsFiddle
Your regex searches pairs of duplicated characters and only removes the first one. Therefore, the latest duplicate won't be removed.
To address this problem, you should remove all duplicates simultaneously, but I don't think you can do this with a single replace.
Instead, I would build a map which counts the occurrences of each character, and then iterate the string again, pushing the characters that appeared only once to a new string:
function onlyUnique(str) {
var map = Object.create(null);
for(var i=0; i<str.length; ++i)
map[str[i]] = (map[str[i]] || 0) + 1;
var chars = [];
for(var i=0; i<str.length; ++i)
if(map[str[i]] === 1)
chars.push(str[i]);
return chars.join('');
}
Unlike indexOf, searches in the hash map are constant on average. So the cost of a call with a string of n characters will be n.
If you want to do it with a regex, you can use your own regex with a callback function inside a replace.
var re = /(.)(?=.*\1)/g;
var str = 'rc iauauc!gcusa_usdiscgaesracg';
var result = str;
str.replace(re, function(m, g1) {
result = result.replace(RegExp(g1.replace(/[.*+?^${}()|[\]\\]/g, "\\$&"), "g"), '');
});
document.getElementById("r").innerHTML = "'" + result + "'";
<div id="r"/>
The idea is: get the duplicated character, and remove it from the input string. Note that escaping is necessary if the character might be a special regex metacharacter (thus, g1.replace(/[.*+?^${}()|[\]\\]/g, "\\$&") is used).
Another idea belongs to Washington Guedes in his deleted answer, I just add my own implementation here (with removing duplicate symbols from the character class and escaping special regex chars):
var s = "rc iauauc!gcusa_u]sdiscgaesracg]";
var delimiters= '[' + s.match(/(.)(?=.*\1)/g).filter(function(value, index, self) { // find all repeating chars
return self.indexOf(value) === index; // get unique values only
}).join('').replace(/[.*+?^${}()|[\]\\]/g, "\\$&") + ']'; // escape special chars
var regex = new RegExp(delimiters, 'g'); // build the global regex from the delimiters
var result = s.replace(regex, ''); // obtain the result
document.getElementById("r2").innerHTML = "'" + result + "'";
<div id="r2"/>
NOTE: if you want to support newline symbols as well, replace . with [^] or [\s\S] inside the regex pattern.
function onlyUnique(str) {
// match the characters you want to remove
var match = str.match(/(.)(?=.*\1)/g);
if (match) {
// build your regex pattern
match = '[' + match.join('') + ']';
}
// if string is already unique return the string
else {
return str
}
// create a regex with the characters you want to remove
var re = new RegExp(match, 'g');
return str.replace(re, '');
}
So using jquery and I have a string of coordinates like so:
38.313072,-89.863845,38.312675,-89.863586,38.310405,-89.862091,38.310405,-89.862091,38.309913,-89.861976,38.309768,-89.861976,38.309768,-89.861976
I need them to look like this:
[38.313072, -89.863845] [38.312675, -89.863586] [38.310405, -89.862091] [38.310405,-89.862091] [38.309913, -89.861976] [38.309768, -89.861976] [38.309768, -89.861976] [38.30965, -89.861991]
So I need to figure out how to replace every other comma with a space and bracket sets of coordinates.
Ideas?
You can use regex.
([-\d.]+),([-\d.]+),?
Regex Explanation and Live Demo
[-\d.]: Character class, - will match literal - hyphen, \d will match a single digit, . will match . literally. When mentioned inside class sequence doesn't matter.
+: Matches one or more occurrences of the previous matches
(...): Capturing Group. The matches inside the braces are captured and returned in $1, $2, ...
,?: To not match the every other comma
g: Global Match. To match all possible occurrences.
jsFiddle Demo
var str = '38.313072,-89.863845,38.312675,-89.863586,38.310405,-89.862091,38.310405,-89.862091,38.309913,-89.861976,38.309768,-89.861976,38.309768,-89.861976';
var result = str.replace(/([-\d.]+),([-\d.]+),?/g, '[$1, $2] ').trim();
document.getElementById('output').innerHTML = JSON.stringify(result, 0, 2);
<pre id="output"></pre>
Use Regex !
It's perfect for this !
var str = "38.313072,-89.863845,38.312675,-89.863586,38.310405,-89.862091,38.310405,-89.862091,38.309913,-89.861976,38.309768,-89.861976,38.309768,-89.861976,38.30965, -89.861991";
var expected = "[38.313072, -89.863845] [38.312675, -89.863586] [38.310405, -89.862091] [38.310405,-89.862091] [38.309913, -89.861976] [38.309768, -89.861976] [38.309768, -89.861976] [38.30965, -89.861991]"
var computed = str.replace(/(([0-9\-. ]*)(,)([0-9\-. ]*)),?/g, '[$2 $3 $4] ');
document.write('computed<br>')
document.write(str.replace(/(([0-9\-. ]*)(,)([0-9\-. ]*)),?/g, '[$2 $3 $4] '))
document.write( "<hr>expected<br>" + expected )
Not pretty, but it works:
var coordinateString = "38.313072,-89.863845,38.312675,-89.863586,38.310405,-89.862091,38.310405,-89.862091,38.309913,-89.861976,38.309768,-89.861976,38.309768,-89.861976";
var coordinateArray = coordinateString.split(",");
var coordinateResult = "";
var i = 0;
while(i<coordinateArray.length){
coordinateResult +="[" + coordinateArray[i] + ", " + coordinateArray[i+1] + "]";
i += 2;
}
This should do the work.
function toPairs(src){
// Split the string into values.
var values = src.split(',');
// Group these values 2 by 2.
var pairs = [];
for(var i = 0; i < values.length; i += 2){
pairs.push("[" + values[i] + ", " + values[i + 1] + "]");
}
// Join with a whitespace.
return pairs.join(" ");
}
document.body.innerHTML = toPairs("38.313072,-89.863845,38.312675,-89.863586,38.310405,-89.862091,38.310405,-89.862091,38.309913,-89.861976,38.309768,-89.861976,38.309768,-89.861976");
Try utilizing while loop , Array.prototype.splice() , returning results as array of arrays
var str = "38.313072,-89.863845,38.312675,-89.863586,38.310405,-89.862091,38.310405,-89.862091,38.309913,-89.861976,38.309768,-89.861976,38.309768,-89.861976"
, res = []
, arr = str.split(",");
while (arr.length) res.push(arr.splice(0, 2))
console.log(res, JSON.stringify(res, null, 2))
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)
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.