javascript;storing values in array

Question

I am trying to store values in array using javascript..but i get strange error in javascript.below is my code

var a = 1;
for(i=0;i<4;i++)
{

var all = new Array();
all[i]=a;
a++;
}

alert(all[1]);
alert(all[2]);
alert(all[3]);

please check the code here: http://jsfiddle.net/D8Suq/

for all[1] and all[2] i am getting undefined error..but all[3] is working fine,,,am confused.some one please help me

Answer 1

You are reassigning your array in every loop iteration (which deletes everything in it) instead of only before the whole loop.

This should work as expected:

var a = 1;
var all = new Array();
for(i=0;i<4;i++)
{
    all[i]=a;
    a++;
}

alert(all[1]);
alert(all[2]);
alert(all[3]);

Answer 2

You are re-initializing the array inside of the for-loop, overwriting any data you had previously written. Move new Array() (or better [] , the array literal) outside the loop

Answer 3

 var all = [];
    for (i = 0; i <= 4; i++) {
      let element = i;
      all.push(element);
    }
    alert(all[1]);
    alert(all[2]);
    alert(all[3]);
    alert(all[4]);

Answer 4

Try some thing like:

const Coin = [
  "Heads",
  "Tails"];

Answer 5

You are recreating the array on each iteration. Try this:

var all = []; // Moved up, and replaced with bracket notation.
var a = 1;
for(i=0;i<4;i++)
{
all[i]=a;
a++;
}

alert(all[1]);
alert(all[2]);
alert(all[3]);

Answer 6

Your issue here is that you're reinstantiating a new Array on each iteration of the loop. So, the first time around, you set a value in that array. The second time around, you redefine the all variable to be a completely new array, which undoes the work you did in the last iteration.

The easiest thing to do is just move var all = new Array() and put it before your loop.

Answer 7

You are redefining your array inside the for loop. You need to define it outside.

var a = 1;
var all = new Array();
for(i=0;i<4;i++)
{
  all[i]=a;
  a++;
}

alert(all[1]);
alert(all[2]);
alert(all[3]);

Answer 8

var a = 1;
var all = new Array();
for(i=0;i<4;i++)
{


all[i]=a;
a++;
}

alert(all[0]);
alert(all[1]);
alert(all[2]

Answer 9

You need to put var all = new Array() outside your loop. You're creating a new all[] four times.

var a = 1;
var all = new Array();
for(i=0;i<4;i++)
{
   all[i]=a;
   a++;
}

alert(all[1]);
alert(all[2]);
alert(all[3]);