syntax error, unexpected '$type' (T_VARIABLE), expecting '(' in

Question

  if($_POST) {
  $type = $_POST['roll'];
  if $type == 1 {
    $number = rand(1,100)
    if $number <= 50 {
      echo "Winner!"
    }
    else {
      echo "Loser!"
    }
  }
};

Can't figure out what is wrong. I am kinda new to PHP also. Type (or roll) is always 1 (for now)

here, read this: http://php.net/manual/it/control-structures.if.php — Federkun, Dec 07 '16 at 10:51

J.Aurélien · Accepted Answer · 2016-12-07T10:56:32.467

0

I think you just need to put

if ($type == 1) {

instead of your actual statement, the condition of the if need that. And so you'll need it too on the second "if"

Watch what your error is showing, it's exactly what you need.

edited Dec 07 '16 at 10:56

answered Dec 07 '16 at 10:51

J.Aurélien

46
4

That worked but on the second if it says unexpected if (T_IF). – Old Goldie Dec 07 '16 at 10:54
you forgot the ";" – J.Aurélien Dec 07 '16 at 10:56
Ah, you forgot the ";" of your two echo too – J.Aurélien Dec 07 '16 at 10:57

syntax error, unexpected '$type' (T_VARIABLE), expecting '(' in

1 Answers1