the title is bit confusing so sorry about that, so what i have is a too arrays one of the array contain more then the second array
x = [1,2,3,4,5];
y = [{
x: "test1",
y: "test2",
z: "test3",
w: `test4`
}]
so what I want to do is for example
for (let i = 0; i < x.length; i++) {
console.log(y[i])
}
which would only log out the first one one time but what i want is to log y as many as x length hope that was clear enough
You can use the remainder operator to get the corresponding item from the shorter array:
const x = [1, 2, 3, 4, 5];
const y = [{"x":"test1","y":"test2","z":"test3","w":"test4"},{"x":"test21","y":"test22","z":"test23","w":"test24"}]
for (let i = 0; i < x.length; i++) {
const idx = i % y.length;
console.log(y[idx])
}Or you can use the last index of the y array, if the current i value is over the length of the last item index in the y array:
const x = [1, 2, 3, 4, 5];
const y = [{"x":"test1","y":"test2","z":"test3","w":"test4"},{"x":"test21","y":"test22","z":"test23","w":"test24"}]
for (let i = 0; i < x.length; i++) {
const idx = Math.min(i, y.length - 1)
console.log(y[idx])
}