Trying to Change The Condition of without using null-CodePudding

I am trying to achieve a check for duplication and push items to array in the below code the functionality works well but instead of using a null is there any other way of doing it

array2.forEach((item) =>
   array1.includes(item)
 ? null
 : array1.push(item),
);

CodePudding user response：

You can use && to short circuit the operation

const array1 = [1, 2, 3, 4, 5, 6],
  array2 = [4, 5, 6, 7, 8, 9, 10];
array2.forEach((item) => !array1.includes(item) && array1.push(item));
console.log(array1)

CodePudding user response：

You can use a set

const set = new Set();
const brr = ["one","one"].forEach(item=> set.add(item));
Array.from(set) 
// --> ["one"]

CodePudding user response：

this should works:

array1.push( ... array2.filter(item => !array1.includes(item) ) )

or you can switch arrays to Set()'s and use:

array2.forEach( item => array1.add(item) )

CodePudding user response：

You can use ES6 features like spread operator(...) and Set

let array1 = [1,2,3,4,5,6];
let array2 = [4,5,6,7,8,9,10];
const mergedArrays = [...array1, ...array2];
array1 = [...new Set(mergedArrays)];    // make the array elements unique
console.log(array1);

CodePudding user response：

you can use "undefined" instead of "null", undefined means do nothing, and null is by itself a value.

ray2.forEach((item) =>
   array1.includes(item)
 ? undefined
 : array1.push(item),
);