在函数内部分配变量时出错

Question

Can you please take a look at this script and let me know why I am getting the

Undefined variable: step1

error?

<?php

$step1;

function getNums($num1, $num2){
 $diff = $num2 - $num1;
  $steps =[
        round($num1 + $diff/4), 
        round($num1 + $diff/2), 
        round($num1 + $diff*.75), 
        $num2
    ];
 $step1 = $steps[1];
}

getNums(50, 400);
echo $step1;

?>

Answer 1

The code inside your function is in a different scope than the code running outside of it, which is why you are getting an error about $step1 being undefined - it is being defined outside the function. If you want to be able to refer to it inside your function you will either need to pass it in as an argument to your function by reference or make the variable global.

Pass by reference

function getNums( $num1, $num2, &$step1 ){
    // ... your code
}
// pass the variable by reference
getNums( 50, 400, $step1 );
echo $step1;

Using global

// accessible globally
global $step1;

function getNums( $num1, $num2 ){
    global $step1;
    // ... your code, with $step1 accessible
}

getNums( 50, 400 );
echo $step1;