Please help here I wanna print in the console all array elements in condition that they are not numbers and they do not start with letter "A"
I think the problem is where to put the i += 1; but actually when I am changing its position the output is not what like I need.
I tried to do it with for and every thing turned out okay, but I don't know what is the problem with while.
Here is the code:
let friends = ["Ahmed", "Sayed", "Ali", 1, 2, "Mahmoud", "Amany"];
let i = 0;
while (i < friends.length) {
if (friends[i][0] === "A" || typeof friends[i] === "number") {
continue;
}
console.log(friends[i]);
i += 1;
}
I tried to use while to do what I've said previously
Since you already have an array, a better approach for this would be to just loop through it.
That way you could avoid while loop entirely.
let friends = ["Ahmed", "Sayed", "Ali", 1, 2, "Mahmoud", "Amany"];
const filteredFriends = friends.filter(friend => friend[0] !== 'A' && typeof friend !== 'number');
console.log(filteredFriends);documentation for filter()
The reason you were getting an infinite loop is because of the continue statement inside if. That skips the rest of the code for that iteration and i += 1 doesn't get executed.
When you continue, you do not increment i, so in next iteration i stays the same.
You can always use for ... of to drop need for manually incrementing i
let friends = ["Ahmed", "Sayed", "Ali", 1, 2, "Mahmoud", "Amany"];
for (const friend of friends) {
if (typeof friend === "number" || friend[0] === "A") {
continue;
}
console.log(friend);
}You need to increase i every time you loop or your while condition will never resolve (unless all meet your condition).
while (i < friends.length) {
if (friends[i][0] !== "A" && typeof friends[i] !== "number") {
console.log(friends[i]);
}
i += 1;
}
*I changed the condition to look for the desired result rather than the negative (just a personal preference).