Replace Every Nth Character From A String

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 ev

Solution 1:

You could just do it with a regex replace:

var str = "abcdefoihewfojias";
    
var result = str.replace(/(...)./g, "$1|");

console.log(result);

Solution 2:

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("") );

Solution 3:

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 |.

Solution 4:

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(""));

Solution 5:

.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 : '|')