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Javascript - array includes() method opposite
1
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What is the correct way of removing objects that have a certain property?
I can use includes() in the followind way:
const array = [
{ id: 1, name: 'Tom1' },
{ id: 2, name: 'Ann' },
{ id: 3, name: 'Tom2' },
{ id: 4, name: 'Kate' },
];
const result = array.filter((item) => item.name.includes('Tom'));
To get the result:
[ { id: 1, name: 'Tom1' }, { id: 3, name: 'Tom2' } ]
But how can I do the opposite to remove the objects above, keeping everything but them?
1 answer
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Just negate the result of the function.
// ONLINE-RUNNER:browser;
const array = [
{ id: 1, name: 'Tom1' },
{ id: 2, name: 'Ann' },
{ id: 3, name: 'Tom2' },
{ id: 4, name: 'Kate' },
];
const result = array.filter((item) => !item.name.includes('Tom'));
console.log(JSON.stringify(result, null, 4)); // [ { id: 2, name: 'Ann' }, { id: 4, name: 'Kate' } ]
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