Question

I have a problem with my arrays in JavaScript. I can’t seem to get the value correctly.

I create my array in PHP like this:

$data = Array();        
$get = mysql_query("SELECT x,y,sid FROM table WHERE uid='1'");
    while($row = mysql_fetch_assoc($get)){
        $data[$row['x']] = Array();
        $data[$row['x']][$row['y']] = $row['sid'];
        }
$data = json_encode($row)

EDIT The json_encode comes out as “false” /EDIT

I then assigned this $data to a JS variable as sdata.

So then i try to get the value in JS but its not working. I get an undefined error.

This is my Javascript:

var i = 1;
var j = 5;
    if(sdata["x"] == i && sdata["y"] == j){
    alert(sdata["x"]["y"]["sid"]);
            }

Its meant to alert me the value of “sid” but i get:
Undefined

Any ideas where my mistake is?

Answer 1

Given your json

alert(sdata["x"]["y"]["sid"]);

should be

alert(sdata["sid"]);

as the json encoded data you show is only a one dimensional array

EDIT

If your json comes out as false, that’s a different story:

$data = json_encode($row);

Should be inside your loop, for good practice, and if you’re just encoding the $row why even bother with

    $data[$row['x']] = Array();
    $data[$row['x']][$row['y']] = $row['sid'];

Otherwise try

print_r($row);

within your while statement and make sure your expected result is coming through from mysql in the first place.

EDIT

If you are trying to get the $data variable in json then use

$data = json_encode($data);

outside of your loop.

EDIT

JSFiddle to return sid for given x and y – I’m sure it can be done in a cleaner way, but it’s a start
http://jsfiddle.net/HWByj/

var sdata = {"4":{"4":"1"},"7":{"1":"0"}};
var i = 4;
var j = 4;

if(typeof sdata[i] != 'undefined')
{
    if(typeof sdata[i][j] != 'undefined')
    {
        alert(sdata[i][j]); //gives you the value of sid
    }
}
        ​