php function does not work when passing variable as argument

Question

Pls see my code below, which is meant to validate whether a variable has on of the specific value (and based on solution example of Is it possible to add to filter_var() function user-defined parameters?)

I tested two cases (A, B) as descibed in the comments in the code below. I dont understand why the function is not working properly in case B? (which I tested by means of ideone.com)

function validate_select($val, $myoptions)
{
//print_r($myoptions);
for($i=0;$i<count($myoptions);$i++){
    if($val==$myoptions[$i]){
        return $val;
    }
}
    return false;
}

$testVar = 'apple';
$myoptions = array('banana','pear','apple');
$result = filter_var($testVar, FILTER_CALLBACK, array('options' => function($var) {
  //return validate_select($var, array('banana','pear','apple')); //case A: returns correct value 'apple'
  return validate_select($var, $myoptions); //case B: returns unexpected value false
}));
echo($result);

Thiago França · Accepted Answer · 2014-04-16 17:08:08Z

2

$myoptions is outside of your function... try add function ($var) use ($myoptions) like

$result = filter_var($testVar, FILTER_CALLBACK, array('options' => function ($var) use ($myoptions) {
    return validate_select($var, $myoptions);
}));

edited Apr 16, 2014 at 17:08

answered Apr 16, 2014 at 16:57

Thiago França

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1 Comment

Joppo Over a year ago

thnx. However, that still does not seem to work (returns false in described example) (How does your line of code 'know' to call function validate_select()?

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