PHP: variable based on array contents

Question

Imagine this:

$array(type=>$user_input);

$level1 = "arbitrary 1";
$level2 = "arbitarty 2";

if( $type && $type != '' ){
        switch ($type){
            case 'level1':
                $levels = $level1;
                break;
            case 'level2':
                $levels = $level2;
                break;
            }
    }else{
        $levels = $level1 . $level2;
    }

This works, but seems repetitive -- especially with 10, 20 levels...

How would I do this:

$array(type=>$user_input);

$level1 = "arbitrary 1";
$level2 = "arbitarty 2";

if( $type && $type != '' ){
        $levels = (use the type contained in the variable named by the user_input)
    }else{
        $levels = $level1 . $level2;
    }

Since I have lost my ability to speak in proper English, I hope my code explanation is self-explanatory.

$array(type=>$user_input); makes absolutely no sense to me... What is this suposed to do? — Pevara
– Pevara, Commented Oct 8, 2013 at 23:00
The array is built by other code that contains a "user input" of level1, level2, level3 etc... — OliverSamson43
– OliverSamson43, Commented Oct 8, 2013 at 23:05

Community · Accepted Answer · 2023-11-17 19:30:35Z

1

You could use variable variables:

$level = $$type;

edited Nov 17, 2023 at 19:30

CommunityBot

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answered Oct 8, 2013 at 22:59

TheWolf

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2 Comments

OliverSamson43 Over a year ago

Awesome. I was just looking at variable variables... what a rabbit hole ;)

TheWolf Over a year ago

Indeed, it is. You really need to be careful when using it in order to make sure there's no unexpected value in $type which is why using an array as other have suggested may be the better approach.

Pevara · Accepted Answer · 2013-10-08 23:07:14Z

0

Not sure what you are trying to achieve here, but if I understand you correctly I would probably do something like this:

$levelList = array(
 'level1' => "arbitrary 1",
 'level2' => "arbitarty 2"
);

if ($type){
   $levels = isset($levelList[$type]) ? $levelList[$type] : implode('', $levelList);
}

using the $$variable syntax is valid indeed, but it seems a bit dirty to me, and an array seems just more appropriate here anayway

answered Oct 8, 2013 at 23:07

Pevara

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Comments

AlexP · Accepted Answer · 2013-10-08 23:09:10Z

0

Wouldn't it make more sense to use an array?

$levels = array('Level 1', 'Level 2', 'Level 3'); 
$key    = array_search($type, $levels);
$result = $levels[$key];

answered Oct 8, 2013 at 23:09

AlexP

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