What is the best way to acquire the opposing attribute?

There is a specific object I am working with that will always have 2 properties, never more. The object looks like this:

var people = {
   'John': { ... },
   'Peter': { ... }
}

In addition, there is a variable var name = 'John'.

I am looking for an efficient way to retrieve the value of the property 'Peter' when the value of name is 'John'. This should be done without explicitly coding the names John and Peter in the function. The goal is to dynamically fetch the opposite property based on the value stored in the name variable.

Answer №1

const username = 'Alice'; // or any name
const allNames = Object.keys(users);

const secondUser = allNames.find(user => user !== username);
users[secondUser] // this will display the value of the second user's property

Answer №2

Object.keys function will provide an array containing the property names.

The filter method allows you to further refine that array.

For instance:

const name = "Sara";
const countries = {
  'Sara': 1,
  'Dylan': 1
};
const [result] = Object.keys(countries).filter(country => country !== name);

console.log({
  result
});

Answer №3

I recently created a function that accomplishes this task effortlessly. It requires a key and an object as parameters, and it will retrieve the value of the element in the specified object using the inverse of the provided key. Please note that this function is designed to work only with objects containing two keys.

var animals = {
    'Lion': { 'sound':['roar','growl'] },
    'Elephant': { 'feature':['trunk','tusk'] }
}
getInverseObjectValue = (key,obj) => { // Function declaration for 'getInverseObjectValue'
    let objKeys = Object.keys(obj); // Create an array of keys from the object
    let invertedKey = objKeys.filter(k => k != key)[0]; // Filter through the keys to find the non-matching key
    return obj[invertedKey]; // Retrieve the value based on the inverted key
}
console.log(getInverseObjectValue('Elephant',animals));

Answer №4

Here is a possible solution:

for (let index in array){
    if (index !== target) return array[index];
}

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