How can I use argument of outer function in an inner function in php?

Question

I am trying to build a customized filter function using array_filter() in php



    $lst = [["indexNo" => '123', "name" =>"aaa", grade => 8],
    ["indexNo" => '124', "name" =>"abc", grade => 8],
    ["indexNo" => '125', "name" =>"avb", grade => 9]
    ];
    
    function filterByIndex($grade){
            global $lst;
            return array_filter($lst, function($record){
                
                $record['grade'] == $grade;
            });
    
            
        }

The compiler says that the $garde in line($record['grade'] == $grade) is not declared.

What can I do for it?

AbraCadaver · Accepted Answer · 2021-02-25 03:48:58Z

2

Three issues. First, $grade is not the same as $garde in the function definition. Second, use the use() construct to get the variable into the anonymous function. Third, you need to return true or false for the filter to work:

function filterByIndex($grade){
        global $lst;
        return array_filter($lst, function($record) use($grade){
            return ($record['grade'] == $grade);
        });
}

Also, in your $lst definition you are using the index garde but in the function you use index grade.

Working Demo.

edited Feb 25, 2021 at 3:48

answered Feb 25, 2021 at 3:34

AbraCadaver

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1 Comment

Mohamed Ishad Over a year ago

It works for me. actually I didn't know about use() as I am beginner for php. Thank you

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