My goal is to replace the value "3" with "4" in the array. After this operation, only "1":"2" will remain in the array.
const array = [];
array.push({ "1": "2" })
array.push({ "3": "4" })
const index = array.indexOf(3);
if (index > -1) {
array.splice(index, 1);
}
console.log(array)
When dealing with OBJECT values
If you are using indexOf with objects, it is important to compare them with the exact same object reference. You must have the identical reference to successfully remove it from the array where it was previously added.
Objects are compared based on references, while primitives are compared based on values. For further details, refer to How to determine equality for two JavaScript objects?
const array = [];
const obj = { '3': '4' };
array.push({ '1': '2' });
array.push(obj);
const index = array.indexOf(obj);
console.log(index);When working with PRIMITIVE values
If you are dealing with an array of primitive values, you can safely use indexOf because primitives are compared based on their values as shown below:
const array = ['1', '2', '3'];
console.log(array.indexOf('3'));1) An alternative approach is to use findIndex with hasOwnProperty to achieve a similar result. Credits to pilchard for this suggestion.
const array = [];
array.push({ '1': '2' });
array.push({ '3': '4' });
const index = array.findIndex((o) => o.hasOwnProperty('3'));
if (index > -1) {
array.splice(index, 1);
}
console.log(array);2) Another option is to use findIndex to locate the index of an object that contains 3 as a key.
const array = [];
array.push({ '1': '2' });
array.push({ '3': '4' });
const index = array.findIndex((o) => Object.keys(o).some((k) => k === '3'));
if (index > -1) {
array.splice(index, 1);
}
console.log(array);To implement the Array.pop() method effectively, you can follow this example:
let myArray = [];
myArray.push({ a: 'b' });
myArray.push({ x: 'y' });
myArray.pop();
console.log(myArray);
After adding both elements to the array, the structure would be as follows:
[ {"1":"2"}, {"3":"4"} ]As a result, when using the indexOf method, it will not return anything.
To achieve the desired outcome, consider utilizing an object and using delete to eliminate any unwanted properties. For example:
const obj = {}
obj["1"] = "2";
obj["3"] = "4"
console.log("original obj : ", obj)
delete obj["3"];
console.log("filtered obj : ", obj)filter the elements. When an element's key matches your specified key, it will not be included in the filtered array.
const arr = [];
arr.push({ 1: '2' })
arr.push({ 3: '4' })
function excludeElement(arr, k) {
return arr.filter(obj => {
return Object.keys(obj)[0] !== k;
});
}
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