PHP if condition not working [duplicate]

Question

I am having some problems with setting the value of "swimsuit" variable as defined below, however when I set the value to "1" in my MYSQL database, it sets $real=true but elsewise it says "real is undefined".

$query4 = "SELECT swimsuit FROM player_equipment WHERE player_name='xhyderz'";
$result = mysql_query($query4,$link);
$row = mysql_fetch_assoc($result);
$swim = $row['swimsuit'];
$real=null;
if($swim==0){
    $real=false;
}else{
    $real=true;
}
echo "<script type='text/javascript'> var swimsuit=".$real."; </script>";

If you echo a boolean false it prints nothing so your javascript won't work as is — Adam
– Adam, Commented Apr 15, 2014 at 16:27
You just can write: $real='true'; if($swim==0)$real='false'; — Milaci
– Milaci, Commented Apr 15, 2014 at 16:28
Note, you are using deprecated functions, consider using PDO or mysqli. — user1180790
– user1180790, Commented Apr 15, 2014 at 16:29

cmbuckley · Accepted Answer · 2014-04-15 16:28:36Z

4

When you concatenate a boolean variable with a string like that, you get the following:

var_dump(true  . ""); // "1"
var_dump(false . ""); // ""

So if $real is false, your JavaScript looks like:

var swimsuit=;

You can either use the strings "true" or "false" for $real, or you can use json_encode:

echo "var swimsuit=" . json_encode($real) . ";";

answered Apr 15, 2014 at 16:28

cmbuckley

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Comments

iizno · Accepted Answer · 2014-04-16 11:13:43Z

1

Try :

if($swim==0){
    $real="false";
}else{
    $real="true";
}

Edit:

Look at cbuckley's explaination, way more clear and complete than mine.

edited Apr 16, 2014 at 11:13

answered Apr 15, 2014 at 16:27

iizno

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