Question

I have a problem with the work with variables from one condition in the second condition. I have something like that:

<form name="exampleForm" method="post">
...
<input type="submit" name="firstSubmit" value="Send">
<input type="submit" name="secondSubmit" value="Show">
</form>

<?php
if(isset($_POST['firstSubmit'])) {

 function a() {
  $a = 5;
  $b = 6;
  $c = $a + $b;
  echo "The result is $c";
  return $c;
 }

 $functionOutput = a();
}

if(isset($_POST['secondSubmit'])) {

 echo $functionOutput;
}
?>

When I need to work with variable $functionOutput from first condition I always get an error message (undefined variable). How I can solve this problem?

Answer 1

<?php
$functionOutput = "";

if(isset($_POST['firstSubmit'])) {

 function a() {
  $a = 5;
  $b = 6;
  $c = $a + $b;
  echo "The result is $c";
  return $c;
 }

 $functionOutput = a();
}

if(isset($_POST['secondSubmit'])) {

 echo $functionOutput;
}
?>

Should fix it. It’s happening because you’re declaring $functionOutput inside your first IF statement.