php variable scope in functions

Question

I'm having difficulty with variable scope. I have a function that looks up a lab value from a multidimensional array. I want to pass the function the name of the lab and have it look it up. The $lab I pass to the get_lab function however is not accessible to the second function used in the array_filter. Where am I going wrong with scope?

function get_lab($lab){
    $result = array_filter($labs_array, function($v) { 
        return $v[1] == $lab; 
    });
    print_r($result);
}

Victor Smirnov · Accepted Answer · 2016-04-25 15:07:56Z

3

You should declare variable in use clause like this

function get_lab($lab){
    $result = array_filter($labs_array, function($v) use ($lab) { 
        return $v[1] == $lab; 
    });
    print_r($result);
}

Check the manual for anonymous functions

answered Apr 25, 2016 at 15:07

Victor Smirnov

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Dale · Accepted Answer · 2016-04-25 15:07:19Z

2

You need to pass $lab to inner function..

One way is to use the use keyword

function get_lab($lab){
    $lab = $lab;
    $result = array_filter($labs_array, function($v) use ($lab) { 
        return $v[1] == $lab; 
    });
    print_r($result);
}

answered Apr 25, 2016 at 15:07

Dale

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