Solve - findIndex is not a function Error in JavaScript

Solve - findIndex is not a function Error #

The "findIndex is not a function" error occurs for multiple reasons:

• Using the findIndex method in a browser that doesn't support it.
• Using the findIndex method on a value that is not an array.

Here is an example of how the error occurs.

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const arr = {};

// ⛔️ TypeError: findIndex is not a function
const result = arr.findIndex(element => element % 2 === 0);

We called the Array.findIndex() method on an object and got the error back.

To solve the "findIndex is not a function" error, make sure to only call the findIndex() method on arrays and in browsers that support it. The findIndex method can only be called on arrays and returns the index of the first element that passes the test.

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const arr = [3, 4, 9, 12];

const result = arr.findIndex(element => element % 2 === 0);

console.log(result); // 👉️ 1

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// ✅ Supported in Internet Explorer
const arr = [3, 4, 9, 12];

let index = -1;

const result = arr.some((element, idx) => {
  if (element % 2 === 0) {
    index = idx;
    return true;
  }

  return false;
});

console.log(result); // 👉️ true

console.log(index); // 👉️ 1

We used the some() method to implement something similar to the findIndex() method.

Our solution works in Internet Explorer, but is a bit more verbose.

You can conditionally check if the value is an array by using the Array.isArray method.

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const arr = null;

const result = Array.isArray(arr) ? arr.findIndex(element => element % 2 === 0) : -1;

console.log(result); // 👉️ -1

We used a ternary operator, which is very similar to an if/else statement.

If the value is an array, we return the result of calling the findIndex method on it, otherwise we return -1. This way, you won't get an error, even if the value is not an array.

If you have an array-like object which you're trying to convert to an array before calling the findIndex method, use the Array.from method.

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const set = new Set([3, 5, 9, 12]);

const result = Array.from(set).findIndex(e => e % 2 === 0);
console.log(result); // 👉️ 3

Before calling the findIndex method, we convert the value to an array.

You could also use the spread syntax (...) to achieve the same result.

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const set = new Set([3, 5, 9, 12]);

const result = [...set].findIndex(e => e % 2 === 0);
console.log(result); // 👉️ 3

If the error persists, console.log the value you're calling the findIndex method on and make sure it's an array.