javascript findindex

JavaScript
//The findIndex() method returns the index of the first element 
//in the array that satisfies the provided testing function.
//Otherwise, it returns -1, indicating that no element passed the test.
const array1 = [5, 12, 8, 130, 44];

const isLargeNumber = (element) => element > 13;

console.log(array1.findIndex(isLargeNumber));
// expected output: 3
a = [
  {prop1:"abc",prop2:"qwe"},
  {prop1:"bnmb",prop2:"yutu"},
  {prop1:"zxvz",prop2:"qwrq"}];
    
index = a.findIndex(x => x.prop2 ==="yutu");

console.log(index);const array1 = [5, 12, 8, 130, 44];
const search = element => element > 13;
console.log(array1.findIndex(search));
// expected output: 3

const array2 = [
  { id: 1, dev: false },
  { id: 2, dev: false },
  { id: 3, dev: true }
];
const search = obj => obj.dev === true;
console.log(array2.findIndex(search));
// expected output: 2let myList = ['foo', 'bar', 'qux'];

myList.indexOf('bar');	// 1// 	findIndex(callback fn)  

//	.... return index (when condition meets)
//  .... return -1 (if condition not meets)

const array = [5, 12, 8, 130, 44];

/// it returns the index of number which satisfy the condition true
const index = array.findIndex((item)=> item>10);   //1

/// now we can check what element at that index...
console.log(array[index]); // array[1]var fruits = ["Banana", "Orange", "Apple", "Mango"];
return fruits.indexOf("Apple"); // Returns 2
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