php 4 variable variables

Question

$str='input_arr["username"]';

$input_arr=array();

$$str='abcd';

print_r($input_arr);

When I run the above code it only prints Array().

I expected it to print Array([username]=>'abcd')

What am I doing wrong?

This is in php 4 by the way.

Thanks a lot.

Edit:What am I trying to do?

$input_arr is supposed to be a static variable to hold validated user input.However, I only recently realised that php4.3 doesnt support self::$input_arr so I had to edit my script to bar($input_arr['name'],$value); so that I can save the value to a static variable in bar();since $input_arr['name'] does not exists in the current scope, I had to make it a string.

You are using arrays wrong. Variable variables are a pain to debug. What exactly are you trying to do? I am sure there's a better way to do it. — quantumSoup
– quantumSoup, Commented Aug 17, 2010 at 13:21
I did some basic tests with this and you're right, it doesn't work. On my end it stops working when I try fiddling with the elements of the array through variable variables. If I mess around with the array as a whole, everything I try works just fine. But it sure looks like you're crossing the lake to get the water here, what exactly are you trying to accomplish? — Teekin
– Teekin, Commented Aug 17, 2010 at 13:46

janmoesen · Accepted Answer · 2010-08-17 13:37:43Z

3

I would strive to avoid eval at all costs. Use PHP's built-in tokenizer.

<?php
error_reporting(-1);
$input = array(3 => array(5 => 'some value'));

echo '$input: '; var_dump($input);

echo '$input[3][5] (directly): '; var_dump($input[3][5]);

$str = '$input[3][5]';
echo "$str (as a variable variable): "; var_dump($$str);

echo "$str (using eval - don't use this!): "; var_dump(eval("return $str;"));

$var = null;
foreach (token_get_all("<?php $str") as $token) {
    if (isset($token[1]) && $token[1] === '<?php') {
        continue;
    }
    if (isset($token[0]) && $token[0] === T_VARIABLE) {
        $varName = substr($token[1], 1);
        $var = $$varName;
    } else if ($token === '[') {
        $currentIndex = null;
    } else if (isset($token[0]) && $token[0] === T_LNUMBER) {
        $currentIndex = $token[1];
    } else if ($token === ']') {
        $var = $var[$currentIndex];
    } else {
        // Handle/complain about unrecognized input.
    }
}
echo "$str (using tokenizer): "; var_dump($var);

The output:

$input: array(1) {
  [3]=>
  array(1) {
    [5]=>
    string(10) "some value"
  }
}
$input[3][5] (directly): string(10) "some value"
$input[3][5] (as a variable variable): 
Notice: Undefined variable: $input[3][5] in - on line 10
NULL
$input[3][5] (using eval - don't use this!): string(10) "some value"
$input[3][5] (using tokenizer): string(10) "some value"

edited Aug 17, 2010 at 13:37

answered Aug 17, 2010 at 13:31

janmoesen

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2 Comments

janmoesen Over a year ago

@Mystery person: care to explain the downvote? I think this is the safest approach (keeping in mind that variable variables still smell like bad code design) and I made it clear that eval is evil. Are you objecting to the use of the tokenizer instead of eval, perhaps?

danp Over a year ago

someone going nuts on the downvotes :p code looked clean to me, upvoted.

grossvogel · Accepted Answer · 2010-08-17 14:44:31Z

0

It's still difficult to tell what you're trying to do, but it sounds like you want $str to determine where inside $input_arr a piece of data lives. If so, you should store only the array key(s) in $str, not a string representation of the code.

In your last example, it's as simple as setting $str = 'name' and then using $input_arr[$str] to access $input['name']. In the first case, you could use an array $keys = array(3,5) instead of $str, and then $input_arr[$keys[0]][$keys[1]] would be equivalent to $input_arr[3][5].

answered Aug 17, 2010 at 14:44

grossvogel

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Comments

Julien Roncaglia · Accepted Answer · 2010-08-17 15:16:58Z

0

It could work with 2 variables if you really want this. Even better if you use a reference to the array instead of a variable variable.

$input_arr = Array();

function somefunction($array, $key)
{
    ${$array}[$key] = 'abcd';
}

function betterfunction(&$array, $key)
{
    $array[$key] = 'abcd';
}

somefunction('input_arr', 'username');
betterfunction($input_arr, 'username');

answered Aug 17, 2010 at 15:16

Julien Roncaglia

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