Javascript join not working on created array

Question

Strange issue on join. It works on initial array, but not with filtered array. Tested also functions for string, but resultA is not string also.

What is wrong with the code:

const constants = [2, 2, 4, 2, 6, 4, 7, 8];
const stdFound = [];

// Makes list of indexes - positions from matching value 2 in constants
constants.filter(function(elem, index, array) {
  if (elem == (2)) {
    stdFound.push(index);
  }
});

//Makes new array from results by indexes found
const resultA = [constants.filter((x, i) => stdFound.includes(i))];
document.getElementById("resultA").innerHTML = resultA;
//Makes string by join
document.getElementById("resultJ").innerHTML = resultA.join(" x ");

<p> ARRAY : <span id="resultA"></span>
  <p> JOIN x : <span id="resultJ"></span></p>

Right now, this code results: ARRAY : 2,2,2 JOIN x : 2,2,2

so clearly, join is not working.

resultA is an array containing one array. If you console.log it, you will see that resultA === [[2,2,2]]. Therefore, when you joint it, it just gives [2,2,2]. "join is not working", it is. Your dataset is wrong, is all. constants.filter returns an array, and you are encapsulating it inside [ ], creating an array of one array. — Jeremy Thille
– Jeremy Thille, Commented Oct 8, 2021 at 9:03

Tushar Shahi · Accepted Answer · 2021-10-08 09:03:51Z

2

filter() already returns an array. No need to wrap in this [].

console.log(resultA) will show that.

const constants = [2, 2, 4, 2, 6, 4, 7, 8];
const stdFound = [];

// Makes list of indexes - positions from matching value 2 in constants
constants.filter(function(elem, index, array) {
  if (elem == (2)) {
    stdFound.push(index);
  }
});

//Makes new array from results by indexes found
const resultA = constants.filter((x, i) => stdFound.includes(i));
document.getElementById("resultA").innerHTML = resultA;

//Makes string by join
document.getElementById("resultJ").innerHTML = resultA.join(" x ");

<p> ARRAY : <span id="resultA"></span>
  <p> JOIN x : <span id="resultJ"></span></p>

answered Oct 8, 2021 at 9:03

Tushar Shahi

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2 Comments

Mait N Over a year ago

Thank you Jeremy Thille and Tushar Shahi! Issue resolved, it was double [[]].

Tushar Shahi Over a year ago

Sure. Use the greentick to mark this as accepted to complete the question

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