Converting eval code into ordinary php code

Question

Is it possible to have just normal php code and not use eval() function in such case?

My eval code is following:

eval('$'.$var.'["'.$key.'"]="'.$value.'";');

And what I am doing is following:

function config_update($array, $var="result")
{ 
    global $result, $dbi;

    while (list($key,$value)=each($array))
    {
        if (is_array($value))
        { 
            config_update($value, $var.'["'.$key.'"]'); 
        } 
        else
        { 
            for ($i=0; $i<count($value);$i++)
            {
                if ($value == "{FROM_DB}")
                {
                    $query = $dbi->prepare("SELECT `value` FROM `".PRE."config` WHERE `key`=?;")->execute($key)->results();

                    if (!$query) 
                    {
                        trigger_error("There is no such key as `{$key}` in config database",E_USER_ERROR);
                        exit;
                    }
                    else
                        $value = str_replace("{FROM_DB}",$query[0]['value'],$value);
                }
                eval('$'.$var.'["'.$key.'"]="'.$value.'";');
                //$$var[$key] = $value; #smth like that..
            }
        }
    }
}

I basically need to update variable values with {FROM_DB} with values coming from DB.

Warning: Illegal string offset 'doc_root' in H:\xampp\htdocs\app_portable\boot.php on line 80 and not only doc_root such message loops for different configuration key names... — Davit
– Davit, Commented Jan 20, 2013 at 23:21

WhyteWolf · Accepted Answer · 2013-01-20 23:41:24Z

2

from the php manual on variable variables you should use {} when working with array based variable variables to avoid unknown behavior.

${$var}[$key] = $value

see Variable variables

answered Jan 20, 2013 at 23:41

WhyteWolf

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2 Comments

WhyteWolf Over a year ago

no problem, anything to reduce the use of eval out in the wild.

user166390 Over a year ago

This is still ugly as it still uses "variable variables". I would recommend switching to the proper root ADT. (Very few languages support "variable variables"; and yet almost all computer languages support this same goal/operation.)

hohner · Accepted Answer · 2013-01-20 23:37:12Z

1

If you were to use:

$$var = array();
$$var[$key] = $value;

It would evaluate as:

$result[$key] = $value;

Assuming that $var = 'result'.

answered Jan 20, 2013 at 23:37

hohner

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2 Comments

Davit Over a year ago

I know, but it does not work in my case for unknown reason and that is what I try to ask

hohner Over a year ago

Have you defined it as an array beforehand?

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