PHP data passed as string instead of integer

Question

I have a slight issue with parsing data in an array from PHP to JavaScript.

My JavaScript is thinking the data is a string and not an integer. Resulting in the wrong answer with the maths.

My JS maths is this:

abposx = data[d]["x"]+offset_x;

If data[d]["x"] was 20 and offset_x was 0

The answer becomes 200 instead of 20.

I'm wondering how I can get JavaScript to use data[d]["x"] as an number and not a string?

This jist of my php:

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

The json Encode looks like this:

[{"x":"283","y":"99","sid":"1"}]

Hope you can help!

Can't you just cast it as an int? (int)data[d]["x"]+offset_x; — Sandeep Bansal
– Sandeep Bansal, Commented Mar 6, 2012 at 21:17
Could you post the PHP where you set the value? Are you using ajax? — bfavaretto
– bfavaretto, Commented Mar 6, 2012 at 21:18
No, i tried that but i got Unexpected identifier as an error. — Sir
– Sir, Commented Mar 6, 2012 at 21:18

Alexander Pavlov · Accepted Answer · 2012-03-06 21:18:41Z

3

abposx = parseInt(data[d]["x"], 10) + offset_x;

answered Mar 6, 2012 at 21:18

Alexander Pavlov

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2 Comments

Alexander Pavlov Over a year ago

can you inspect the value of data[d]["x"]? What is it? Unless it is not a number, this code should work.

Alexander Pavlov Over a year ago

Glad to know there is no miracle in this case :)

Tufan Barış Yıldırım · Accepted Answer · 2012-03-06 21:22:25Z

1

parseInt can be used javascriptside, if you want do this phpside use as;

while($row = mysql_fetch_assoc($get)) {
    $data[] = array_map('intval',$row);
}

answered Mar 6, 2012 at 21:22

Tufan Barış Yıldırım

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Comments

Derek Harrington · Accepted Answer · 2012-03-06 21:18:47Z

0

use parseInt(string), which returns the string as an integer

answered Mar 6, 2012 at 21:18

Derek Harrington

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1 Comment

gen_Eric Over a year ago

Don't forget the radix, parseInt(string, 10)

Keith Frey · Accepted Answer · 2012-03-06 21:25:09Z

0

Try using type conversion:

var absData = data[d]["x"] - 0;
abposx = absData+offset_x;

edited Mar 6, 2012 at 21:25

answered Mar 6, 2012 at 21:18

Keith Frey

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1 Comment

Alexander Pavlov Over a year ago

This should be JavaScript code, there is no typecasting in JavaScript, nor int type...

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