I have an array that looks like this:
var array = [[10, 8, 2], [5, 7, 1], [3, 9, 4]];
I need something that tells me if the sub object exists, without throwing an error. Here's an example:
elementExsits(array[0][4]);
//false
elementExsists(array[1][2]);
//true
The function elementExsists would verify it the path exists. I've tried:
if (typeof array[3][2] !== 'undefined') {};
But it just says
Cannot read property '2' of undefined
You check one property at a time. Stop if you encounter undefined:
if (typeof array[3] !== "undefined" && typeof array[3][2] !== "undefined") {
}
Or better:
if (3 in array && 2 in array[3]) {
}
A simple if statement can decide if an element exists:
var array = [[10, 8, 2], [5, 7, 1], [3, 9, 4]];
if (array[1] && array[1][2]) {
console.log('exists');
} else {
console.log('doesn\'t exist');
}
if (array[10] && array[10][2]) {
console.log('exists');
} else {
console.log('doesn\'t exist');
}Trying to access an array value that doesn't exist returns undefined, so if statements will switch based on the result.
array[10][1].
Salman's answer works for your scenario. Or, you can also create a dynamic function to return the value at the index, or undefined if not found.
var array = [[10, 8, 2], [5, 7, 1], [3, 9, 4]];
function elementExists(array, ...indices){
return indices.reduce((el, i) => el && el[i], array)
}
console.log(elementExists(array, 0, 4))
console.log(elementExists(array, 1, 2))
console.log(elementExists(array, 3, 2))
array = [[10, [1,[11,12]],8, 2]] and elementExists(array,0,1,1,0) gives you 11. (Four levels deep).You could use in operator and check if the index exists. Then proceed with the next nested array.
This solution works with the array and an array of indices and uses a short circuit with Array#every, if the index is not in the given array.
function elementExsists(array, indices) {
return indices.every(i => Array.isArray(array) && i in array && (array = array[i], true))
}
var array = [[10, 8, 2], [5, 7, 1], [3, 9, 4]];
console.log(elementExsists(array, [0, 4])); // false
console.log(elementExsists(array, [1, 2])); // trueYou can take an array, and a list of indexes, and using Array.every() iterate the indexes. If all indexes are not undefined the element exists. If event one is undefined, the element doesn't exist.
const elementExists = (array, ...indexes) => {
let current = array;
return indexes.every(index => {
current = array[index];
return current !== undefined;
});
};
const array = [[10, 8, 2], [5, 7, 1], [3, 9, 4]];
console.log(elementExists(array, 0, 4)); // false
console.log(elementExists(array, 1, 2)); // true
console.log(elementExists(array, 3, 2)); // false
function elementExists(a, i, j) {
return Array.isArray(a) && Array.isArray(a[i]) && a[i].length > j;
}
console.log(elementExists(null, 0, 1));
console.log(elementExists(undefined, 0, 1));
console.log(elementExists(0, 0, 1));
console.log(elementExists('foo', 0, 1));
console.log(elementExists([], 0, 1));
console.log(elementExists([null], 0, 1));
console.log(elementExists([0], 0, 1));
console.log(elementExists([[]], 0, 1));
console.log(elementExists([[],[]], 0, 1));
console.log(elementExists([[3],[9]], 0, 1));
console.log(elementExists([[3,2],[9]], 0, 1));