Undefined variable: value in PHP

Question

I was trying to print the value of a variable from a function, but it is showing undefined variable value. Please help me and tell me how to solve this problem; I'm pretty new to PHP coding. I tried the function as public, but it shows some other error.

<?php

    if(isset($_POST['btn_submit']))
    {
        $num1 = $_POST['cards1'];
        $num2 = $_POST['cards2'];
        $num1=substr($num1,0,(strlen($num1)-1));
        $num2=substr($num2,0,(strlen($num2)-1));
    }

    function get_value() {

        switch ($num2) {

            case 'A' :
                $value = 11;
                break;
            case 'J':
            case 'Q':
            case 'K':
                $value = 10;
                break;
            default:
                $value = $num2;
                break;

        }

            //return $value;
    }   
    echo '<script type="text/javascript">alert("Out Put is'.$value.' ");</script>';

        //$total_value =  $num1+$num2;      


  ?>

Here `$value` is local to the function. Make it global and you can use then — prava, Jun 12 '14 at 13:04
@prava You should avoid using the `global` keyword. There's almost always a much better way to do it. My comment above being a great example. — John Conde, Jun 12 '14 at 13:05
and here OP has to return the value also :) and then can use it :) — prava, Jun 12 '14 at 13:06

score 0 · Answer 1 · answered Jun 12 '14 at 13:05

0

$num2 is local to your function. You need to pass it in as a parameter:

function get_value($num2) {

*** YOUR CODE

}

answered Jun 12 '14 at 13:05

GGio

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still showing undefined variable error for value – YMO Jun 12 '14 at 13:10
did you pass in $num2 into the function? – GGio Jun 12 '14 at 13:13
yes i did function get_value($num2) { switch ($num2) { case 'A' : $value = 11; break; case 'J': case 'Q': case 'K': $value = 10; break; default: $value = $num2; break; } //return $value; } – YMO Jun 12 '14 at 13:14
remove comment from here - `//return $value;` You should return the value and then use it here `echo '';` – prava Jun 12 '14 at 13:16

Undefined variable: value in PHP

1 Answers1