JavaScript: Replace all but the last number of characters with the specified mask character
JavaScript fundamental (ES6 Syntax): Exercise-85 with Solution
Write a JavaScript program to replace all but the last number of characters with the specified mask character.
- Use String.prototype.slice() to grab the portion of the characters that will remain unmasked.
- Use String.padStart() to fill the beginning of the string with the mask character up to the original length.
- If num is negative, the unmasked characters will be at the start of the string.
- Omit the second argument, num, to keep a default of 4 characters unmasked.
- Omit the third argument, mask, to use a default character of '*' for the mask.
Sample Solution:
JavaScript Code:
//#Source https://bit.ly/2neWfJ2
const mask = (cc, num = 4, mask = '*') =>
('' + cc).slice(0, -num).replace(/./g, mask) + ('' + cc).slice(-num);
console.log(mask(1234567890));
console.log(mask(1234567890, 3));
console.log(mask(1234567890, -4, '$'));
Sample Output:
******7890 *******890 $$$$567890
Pictorial Presentation:
Flowchart:

Live Demo:
See the Pen javascript-basic-exercise-85-1 by w3resource (@w3resource) on CodePen.
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JavaScript: Tips of the Day
Chunks an array into n smaller arrays
Example:
const tips_chunkIntoN = (arr, n) => { const size = Math.ceil(arr.length / n); return Array.from({ length: n }, (v, i) => arr.slice(i * size, i * size + size) ); } console.log(tips_chunkIntoN([1, 2, 3, 4, 5, 6, 7,8], 4));
Output:
[[1,2],[3,4],[5,6],[7,8]]
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