Generate 10 unique random integers

Question

I'm working on a loop that adds 10 unique random integers from 1-10 to an array, so the expected result would be 2,1,4,6,10,7,9,8,5,3 I manage to generate random numbers and only add the unique ones, my problem is, its not complete 10 items, I think the problem is upon checking I it doesn't generate again.

Hope you help me.

Thanks.

 for (let i = 0; i < 10; i++) { var arrNum = []; setTimeout(function() { var r = Math.floor(Math.random() * 10) + 1; while (arrNum.includes(r) == false) { arrNum.push(r); } if (i == 9) { $('ul').append('<li>' + arrNum + '</li>'); } }, 500); }

 ul li{ list-style-type: none; display: inline-block; }

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <ul></ul>

Answer 1

  while (arrNum.includes(r) == false) {
   arrNum.push(r);
 }

That adds the number as long as it doesnt exist, so it will only run once, and if the number exists already it doesnt generate a new one, you want:

   const result = [];
   for(let i = 1; i <= 10; i++) {
       let random;
       do {
         random = 1 + Math.floor(Math.random() * 10);
      } while(result.includes(random));
      result.push(random);
  }

But its probably way more easy to just shuffle an array with the numbers 1 to 10.

Answer 2

Your code has a mistake. The i is not the ideal way to check for 10 occurrences as it is incremented even if variable r does not satisfy your condition - arrNum.includes(r) == false .

The check should be made by array length.

Replace if (i == 9) with if(arrNum.length = 10) in your code.

Answer 3

Create an array with numbers from 1 to N (first line). Replace N with 10 in your case. Then sort it randomly.

const numbers = [ ...Array(N).keys() ].map(o => o + 1);
numbers.sort(() => Math.random() - 0.5);
console.log(numbers); // [ 6, 4, 5, 2, 1, 9, 3, 7, 8, 10 ]