How to find a value from array object in JavaScript?

Question

I have a array object I want to return and if it will find in this array otherwise it will return or I want to just return value not key.

Currently it returns an object and I need only a value.

const arrObj = [
 {
 "relation_type": "undefined"
 },
 {
 "relation_type": "or"
 },
 {
 "relation_type": "and"
 },
 {
 "relation_type": "or"
 },
 {
 "relation_type": "or"
 }
]
let obj = arrObj.find((o) => {
 if (o.relation_type === "and") {
 return true;
 }
});
console.log(obj);

• Joseph: could you write what you expect exactlly?
• Felix Kling: If I understand that correctly you want something like const result = arrObj.some(o => o.relation_type === 'and') ? 'and' : 'or'.
• Rohit Verma: Thanks you @Felix Kling it working fine as I want.

Answer 1

You could do something like that :

let obj = arrObj.find(o => o.relation_type === "and") ? "and" : "or"

• Felix Kling: "One way to achieve this would be to return o.relation_type instead of true inside your obj variable." I assume you are referring to the call to .find but no, that's not how .find works. It doesn't return the value returned from the callback.
• Felix Kling: arrObj.find(o => o.relation_type === "and").relation_type would always return "and" or throw an error.
• Rohit Verma: @rmfy what about if there is "and" not available in array?
• rmfy: @FelixKling you're right, updated my OP !

Answer 2

Maybe we can simply use destruction:

let {relation_type} = arrObj.find((o) => {
 if (o.relation_type === "and") {
 return true;
 }})

Answer 3

You can use the .map method and loop over and access the relation type as a property here is a working example

const arrObj = [
 {
 "relationtype": "undefined"
 },
 {
 "relationtype": "or"
 },
 {
 "relationtype": "and"
 },
 {
 "relationtype": "or"
 },
 {
 "relationtype": "or"
 }
]
{arrObj.map((object,index)=> 
console.log(object.relationtype)
 )}

.map() MDN Docs