Variable variable not working

Question

I have some code that doesn't seem to work. I'm using a form to post a value (green, blue, red, etc) and make it look up something in the database using that post value as part of a variable name.

$mine = 'green';

$querymine = $dbcon->prepare('SELECT * FROM mine WHERE playerid = :playerid');
$querymine->execute(array(
    'playerid' => $playerid
));
$rowmine = $querymine->fetch(PDO::FETCH_ASSOC);

if (${'rowmine[\'has'.$mine.'\']'} == 0) {
    header("location:mine.php");
    die();
}

It is looking up the column "hasgreen", which is INT 4 in this case. The problem is, it tells me that $rowmine['hasgreen'] is NULL, thus I'm getting kicked back to "mine.php". I'm not sure what the problem is, maybe the use of \' in the variablename?

ThiefMaster · Accepted Answer · 2014-04-08 22:38:53Z

3

You have an array. There is no reason to use a variable variable here.
You can simply use $rowmine['has'.$mine] instead!

Since you are probably a beginner: If you use variable variables you should always consider using an array instead. Variable variables are a great way to get unreadable spaghetti code.

answered Apr 8, 2014 at 22:38

ThiefMaster

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3 Comments

s1h4d0w Over a year ago

Thanks, it works! I'm indeed a beginner, trying to learn PHP by Googling and experimenting.

ThiefMaster Over a year ago

Please be very careful. There is an incredible amount of CRAP (yes, bold uppercase crap) out there when you search for PHP tutorials and stuff. Especially if it's not very recent. To be honest, this is one of the main reasons (besides the language being much nicer) why I'd always recommend someone to learn Python instead of PHP. Anyway, virtual +1 for doing SQL the right way with prepared statements!

s1h4d0w Over a year ago

Thanks! I try to be as safe as possible and immediately use the right protection against SQL injection. I spend a lot of time researching the correct ways of implementing security and making sure no-one can mess with my code or database. Only thing I need to start doing is commenting my code :)

CMPS · Accepted Answer · 2014-04-08 22:38:30Z

1

No need for the {}, try this:

 if ($rowmine['has'.$mine] == 0)

answered Apr 8, 2014 at 22:38

CMPS

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