How to get value in an object array by key

Question

how can I get value in an object array by a key, which is also in this object array.

the object array looks like this:

const objectArray = [
    {key: "1", value: "12321"},
    {key: "2", value: "asdfas"}
]

I have now the value of key, e.g. key = 1, but I want to get 12321 as result.

any solution?

Answer 1

You can use .find() to achieve it.

Try this:

Working Demo

this.objectArray.find(x => x.key == "1").value

To handle exception, if the item doesn't exists in the array, do this:

let item = this.objectArray.find(x => x.key == "1")
this.value = item ? item.value : null

Answer 2

You can do this using filter() and use the value of the key you already have.

const objectArray = [
    {key: "1", value: "12321"},
    {key: "2", value: "asdfas"}
]

const el = objectArray.filter(item => item.key == 1)[0];

el
  ? console.log(el.value) // gives 12321
  : console.log('none listed')

Answer 3

objectArray.find(e => e.key == "1")

ref https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/find

Answer 4

objectArray.forEach(function(item) {
    Object.keys(item).forEach(function(key) {
        console.log("key:" + key + "value:" + item[key]);
    });
});

Answer 5

This question is dated, but I'm posting my response hoping it will help others.

This is what you need:

objectArray
    .filter((obj) => obj.key === key)
    .map((item) => {
        return item.value;
    }
);

Below is the fully functional code:

const objectArray = [
{ key: '1', value: '12321' },
{ key: '2', value: 'asdfas' },];



/* This function will return the value you're looking for. */

function getObjectValueById(key: string) {
    return objectArray
        .filter((obj) => obj.key === key)
        .map((item) => {
            return item.value;
        }
    );
}

Because your key is a string, I am using key: string in the function's argument. If your key is a number, then you can modify the function argument to key: number, and rest of the code will still work.