Defining a variable as part of a function

Question

Im having trouble setting a variable as part of a function, im sure its down the syntax more than anything, what im trying to do is use a if/else statement and as part of it, to return a variable $Bname.

But im not sure how to set it.

My function is as follows

 function bestPriceFunc($var1, $var2) {
      if ($var1 < $var2) {
          return $var1;
          $Bname = 'will';
          return $Bname;

      } else {
          return $var2;
          $Bname = 'Lad';
          return $Bname;
      }
  }

is this the correct way to do it or could i just write return $Bname = 'Lad'; . Ive tried both but i getting an undefined variable error when i load the page

Answer 1

You can only use return once per function as it terminates the function the first time it is executed.

Try the following or if you don't need to return $var1 and $var2, simply remove those lines.

<?php
function bestPriceFunc($var1, $var2) {
    if ($var1 < $var2) {
        $Bname = 'will';
        return array($var1, $Bname);

    } else {
        $Bname = 'Lad';
        return array($var2, $Bname);
    }
}

$var = bestPriceFunc(3, 4);
$bnameoutput = $var[1];
?>

Answer 2

You are getting undefined errors when you attempt to use the $Bname variable because you are already returning $var1 and $var2 .

A return statement ends the function. Once you hit return $var1; or return $var2 the function ends and returns the specified variable. In order to return the $Bname variable, you need to get rid of the other returns.

Answer 3

Once you return $var; the function execution ends, meaning that the rest of the code is ignored, so $Bname is not set.

You also have two "return" in each function, I suggest having a lopk at how functions work