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Intentando cambiar la condición de sin usar nulo

Estoy tratando de lograr una verificación de duplicación y empujar elementos a la matriz en el siguiente código, la funcionalidad funciona bien, pero en lugar de usar un valor nulo, ¿hay alguna otra forma de hacerlo?

 array2.forEach((item) => array1.includes(item) ? null : array1.push(item), );
about 4 years ago · Juan Pablo Isaza
3 answers
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0

Puede usar && para cortocircuitar la operación

 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)

about 4 years ago · Juan Pablo Isaza Report

0

Puedes usar un conjunto

 const set = new Set(); const brr = ["one","one"].forEach(item=> set.add(item)); Array.from(set) // --> ["one"]
about 4 years ago · Juan Pablo Isaza Report

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Puede usar funciones de ES6 como operador de propagación (...) y 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);

about 4 years ago · Juan Pablo Isaza Report
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