It is possible to validate an array like this:
const Array = [[1, 2], ['hello', 'hi'], []]
I need to kwon when one array is empty or is ".length === 0" but only if one of the array is empty it say an error and if all are empty is the same error.
for example this is an error
const Array = [[1, 2], ['hello', 'hi'], []]
const Array = [[1, 2], [], []]
const Array = [[], ['hello', 'hi'], []]
when all array has values then do something else like show an alert
You can use the .some() function to check if one or more items of an Array matches a certain condition. Here is what it would look like applied to your specific case:
const array_of_arrays = [[1, 2], ['hello', 'hi'], []]
let contains_empty_arrays = array_of_arrays.some((sub_array) => sub_array.length === 0)
console.log(contains_empty_arrays )Using Array.prototype.every()
const data = [[], ["hello", "hi"], []],
check = data.every(item => item.length);
if (!check) {
alert("Empty Values");
}You can do something like this.
const myArray = [[1, 2], ['hello', 'hi'], []];
if (myArray.find((innerArray) => innerArray.length === 0)) {
console.log("Some error")
} else {
console.log("Do something else");
}Can you please try running the snippet below:
const myArray = [[1, 2], ['hello', 'hi'], []];
var with_empty = false;
for (i = 0; i < myArray.length; i++) {
if(myArray[i].length == 0){
with_empty = true;
break;
}
}
if(with_empty)
alert('There is an empty value!');Maybe you want to do something like:
function fullNest(arrayOfArrays){
for(let a of arrayOfArrays){
if(!a.length){
return false;
}
}
return true;
}
const nest = [[1, 2], ['hello', 'hi'], []];
console.log(nest); console.log(fullNest(nest)); console.log('-'.repeat(50));
nest[2].push('full now'); console.log(nest); console.log(fullNest(nest));
const, which shouldn't beArrayanyways, sinceArrayis a predefined constructor.