Console log shows undefined for a specific value of an array

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I am trying to store the values that correspond to Sunday and Saturday in the table arr2 and display it in the console, the variable a is the number of days in the current month, I calculated it before, when I display the whole table console.log(arr2 ), it displays the values correctly, but when I try to display a specific value eg console.log(arr2[1]) it shows an undefined message in the console.

What I have tried:

<pre lang="Javascript">let arr2 = []; //catch index of samedi's and dimanche's
  let j=0;//initialize index of arr2
  console.log(arr2)
  console.log(arr2[0]);

JavaScript

const renderDay = () => {
         let td = [];
         for (let i = 1; i <= a; i++) {
          if(days[i].toLocaleDateString("fr-FR", options).slice(0,3)==='sam' ||
          days[i].toLocaleDateString("fr-FR", options).slice(0,3)==='dim'){
             td.push(<td className="test" key={i}>{days[i].toLocaleDateString("fr-FR", options).slice(0,3)}</td>);
             arr2[j]=i;
             j++;
          }else
             td.push(<td key={i}>{days[i].toLocaleDateString("fr-FR", options).slice(0,3)}</td>);
            }
      return td;
       };

the output of console.log(arr2) is an array contains

index of sunday's and saturday's of the current month and that's is what I want, but the problem is when I try to console log just specific value example : (arr2[2]) it shows in console undefined

Posted 4-Nov-21 7:21am

Yassine El Aissati

Updated 12-Nov-21 6:07am

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Comments

Richard MacCutchan 4-Nov-21 13:31pm

You never store any items in arr2.

Yassine El Aissati 4-Nov-21 13:37pm

I store items in arr2 in renderDay function ,when I use console.log(arr2) it shows me all the correct values who i wanted in the console,but i can't access to every specific value , when i type console.log(arr2[1]) it shows undefined in the console

Richard MacCutchan 4-Nov-21 13:43pm

You have two separate blocks of code so it is difficult to see which comes where.

Richard Deeming 4-Nov-21 13:38pm

You can replace arr2[j] = i; j++; with arr2.push(i);, which will do the same thing.

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ahilang37 · Answer 1 · 2021-11-12T06:07:00

// This is an empty Array without any objects/ values in it.
let arr2 = [];

console.log(arr2); // []
console.log(arr2[0]); // undefined

// Solution1: pushing the value to the array one by one
arr2.push(1);

console.log(arr2); // [ 1 ]
console.log(arr2[0]); // 1

// Solution2: Mapping default values to the array of size 4
let arr3 = [...Array(4)].map(x => 0);
console.log(arr3);
console.log(arr3[0]);
console.log(arr3[1]);
console.log(arr3[5]);

3xh89vqvr - JavaScript - OneCompiler[^]