How to get values of array into global variable

Question

How do I get values of array into global variable?

var p = new Array();
var all_Y_values = 0;
for (var r = 0; r < embeddedCells.length; r++)
{
    p[r] = embeddedCells[r].attributes.position.y;
    all_Y_values = p[r], all_Y_values; 
    console.log("all y values: " + all_Y_values); //prints all values
}
console.log("all y values: " + all_Y_values); //prints only last value

Right now inside the loop I am able to print all values inside loop but when I print the same outside loop its printing only last value.

all_Y_values = p[r], all_Y_values; is parsed as (all_Y_values = p[r]), all_Y_values;, which is equivalent to all_Y_values = p[r]; all_Y_values; which is equivalent to all_Y_values = p[r];. — melpomene
– melpomene, Commented Sep 18, 2015 at 11:47
if you want all the values into all_Y_values, make it a string and concatenate them — Vamsi
– Vamsi, Commented Sep 18, 2015 at 11:47
Or just use p, which already is an array containing all of your values. — melpomene
– melpomene, Commented Sep 18, 2015 at 11:50
@Vamsi Its printing "undefined" when I make all_Y_values as string — kittu
– kittu, Commented Sep 18, 2015 at 11:50

Kustolovic · Accepted Answer · 2015-09-18 11:52:36Z

3

Your collection of values is already inside "p":

var p = new Array();
var all_Y_values = 0;
for (var r = 0; r < embeddedCells.length; r++) {
  p[r] = embeddedCells[r].attributes.position.y; 
  console.log("current y value: " + p[r]); //prints current value
}
console.log("all y values: " + p.join(','));

p.s. : p and all_Y_values are not global, but local variables. In javascript only, function create a new context. Loops are not.

answered Sep 18, 2015 at 11:52

Kustolovic

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1 Comment

kittu Over a year ago

Thanks. I am able to print the values now. I want to covert these values to array. How do I do that?

Daniel Rosano · Accepted Answer · 2015-09-18 13:30:25Z

2

This should print all y values at the end (ps : new version using forEach)

var p = new Array();
embeddedCells.forEach (function (e, i) {
    p[i] = e.attributes.position.y;
    console.log("current y value: " + p[i]); //prints current value
});
console.log("all y values: " + p.join(", "));

Hope it works

edited Sep 18, 2015 at 13:30

answered Sep 18, 2015 at 11:49

Daniel Rosano

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