I have this JavaScript:
var str = "abcdefoihewfojias".split('');
for (var i = 0; i < str.length; i++) {
var xp = str[i] = "|";
}
alert( str.join("") );
I aim to replace every fourth letter in the string abcdefoihewfojias
with |
, so it becomes abc|efo|....etc
,but I do not have a clue how to do this.
I have this JavaScript:
var str = "abcdefoihewfojias".split('');
for (var i = 0; i < str.length; i++) {
var xp = str[i] = "|";
}
alert( str.join("") );
I aim to replace every fourth letter in the string abcdefoihewfojias
with |
, so it becomes abc|efo|....etc
,but I do not have a clue how to do this.
7 Answers
Reset to default 8You could just do it with a regex replace:
var str = "abcdefoihewfojias";
var result = str.replace(/(...)./g, "$1|");
console.log(result);
To support re-usability and the option to wrap this in an object/function let's parameterise it:
var str = "abcdefoihewfojias".split('');
var nth = 4; // the nth character you want to replace
var replaceWith = "|" // the character you want to replace the nth value
for (var i = nth-1; i < str.length-1; i+=nth) {
str[i] = replaceWith;
}
alert( str.join("") );
This might help you solve your problem
var str = "abcdefoihewfojias".split("");
for (var i = 3; i < str.length - 1; i+=4) {
str[i] = "|";
}
alert( str.join("") );
You go with for loop from the the first char that you want to replace (the 3 char) until the one digit before the end and replace every 4 places.
If the for loop will go from the str.length and not to str.length-1 sometimes at the last char will be |.
.map one-liner
You can use this one-liner:
var str = "abcdefoihewfojias";
str.split('').map(function(l,i) {
return (i + 1) % 4 ? l : '|';
}).join('');
%
returns the remainder. So:
# | Result (# + 1) % 4
---|-------
0 | 1
1 | 2
2 | 3
4 | 0 // Bingo!
ES6 alternative
With ES6, you can do:
[...str].map((l,i) => (i + 1) % 4 ? l : '|')
Simple just use modulus
https://jsfiddle.net/ctfsorwg/
var str = "abcdefoihewfojias";
var outputStr = str.split("");
for (var i = 0; i < outputStr.length; i++) {
if(!((i+1)%4))outputStr[i] = '|';
}
alert( "Before: " + str + "\nAfter: " + outputStr.join(""));
While there are several answers already, I thought I'd offer a slightly alternative approach, using Array.prototype.map()
, wrapped in a function that can be adapted by the user (to update the value of n
in the nth
character, and change the replacement character used):
// defining the named function, with an 'opts' argument:
function replaceNthWith(opts) {
// setting the default options:
var defaults = {
// defining the nth character, in this case
// every fourth:
'nth': 4,
// defining the character to replace that
// nth character with:
'char': '|'
};
// Note that there's no default string argument,
// so that one argument must be provided in the
// opts object.
// iterating over each property in the
// opts Object:
for (var property in opts) {
// if the current property is a property of
// this Object, not inherited from the Object
// prototype:
if (opts.hasOwnProperty(property)) {
// we set that property of the defaults
// Object to be equal to that property
// as set in the opts Object:
defaults[property] = opts[property];
}
}
// if there is a defaults.string property
// (inherited from the opts.string property)
// then we go ahead; otherwise nothing happens
// note: this property must be set for the
// function to do anything useful:
if (defaults.string) {
// here we split the string supplied from the user,
// via opts.string, in defaults.string to form an
// Array of characters; we iterate over that Array
// with Array.prototype.map(), which process one
// Array and returns a new Array according to the
// anonymous function supplied:
return haystack = defaults.string.split('').map(function(character, index) {
// here, when the index of the current letter in the
// Array formed by Array.prototype.split() plus 1
// (JavaScript is zero-based) divided by the number
// held in defaults.nth is equal to zero - ensuring
// that the current letter is the 'nth' index we return
// the defaults.char character; otherwise we return
// the original character from the Array over which
// we're iterating:
return (index + 1) % parseInt(defaults.nth) === 0 ? defaults.char : character;
// here we join the Array back into a String, using
// Array.prototype.join() with an empty string:
}).join('');
}
}
// 'snippet.log()' is used only in this demonstration, in real life use
// 'console.log()', or print to screen or display in whatever other
// method you prefer:
snippet.log( replaceNthWith({ 'string': "abcdefoihewfojias" }) );
function replaceNthWith(opts) {
var defaults = {
'nth': 4,
'char': '|'
};
for (var property in opts) {
if (opts.hasOwnProperty(property)) {
defaults[property] = opts[property];
}
}
if (defaults.string) {
return haystack = defaults.string.split('').map(function(character, index) {
return (index + 1) % parseInt(defaults.nth) === 0 ? defaults.char : character;
}).join('');
}
}
// 'snippet.log()' is used only in this demonstration, in real life use
// 'console.log()', or print to screen or display in whatever other
// method you prefer.
// calling the function, passing in the supplied 'string'
// property value:
snippet.log( replaceNthWith({
'string': "abcdefoihewfojias"
}) );
// outputs: abc|efo|hew|oji|s
// calling the function with the same string, but to replace
// every second character ( 'nth' : 2 ):
snippet.log( replaceNthWith({
'string': "abcdefoihewfojias",
'nth': 2
}) );
// outputs: a|c|e|o|h|w|o|i|s
// passing in the same string once again, working on every
// third character, and replacing with a caret ('^'):
snippet.log( replaceNthWith({
'string': "abcdefoihewfojias",
'nth': 3,
'char' : '^'
}) );
// outputs: ab^de^oi^ew^oj^as
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<script src="http://tjcrowder.github.io/simple-snippets-console/snippet.js"></script>
References:
Array.prototype.map()
.Array.prototype.join()
.for...in
loop.- JavaScript Remainder (
%
) operator. Object.prototype.hasOwnProperty()
.String.prototype.split()
.
function replaceWith(word,nth, replaceWithCh) {
//match nth position globally
//'\S' is for non-whitespace
let regex = new RegExp("(\\S{" + (nth - 1) + "})\\S", "g");
// '$1' means single group
// after each group position replaceWithCharecter
let _word = word.replace(regex, "$1" + replaceWithCh);
return _word;
}
const str = "abcdefoihewfojias";
const result = replaceWith(str, 3, "X");
console.log(result);