PHP Notice: Undefined variable after decryption

Asked Oct 02 '15 at 20:36

Active Oct 02 '15 at 21:02

Viewed 29 times

I'm trying to bring a old script that was encrypted and shut down back to life but i'm not the best at php but practice makes prefect. The error im getting at the moment is

       <?php
     if(isset($data['parameter.troops.barracks1.1.number='])) // decides $trooptype statement <--- error här
     {
        if(intval($data['parameter.troops.barracks1.1.number=0']))
            $trooptype=' kebab';
          else
            $trooptype=' Unknown troops';
     }
      if($data)
      {
        if(isset($data['parameter.deploy.dead.disable.king'])) // barbarian king
        {
          if(intval($data['parameter.deploy.dead.disable.king']))
            echo('King is enabled = true <br>');
          else
            echo('King is disabled = false <br>');
        }
      if(isset($data['parameter.troops.barracks1.1.number'])) // troop value first barrack
        {
          echo('Barracks 1  = ' . intval($data['parameter.troops.barracks1.1.number']) . $trooptype . ' <br>');
        }
      }
     }
?>

edited Oct 02 '15 at 21:02

asked Oct 02 '15 at 20:36

John Borton

1

You can't be getting the error on that line. `isset()` will never report an error about undefined variables -- that would defeat the purpose of using it to test whether the variable is defined before trying to access the variable. – Barmar Oct 02 '15 at 20:40
What is the complete error message? – Barmar Oct 02 '15 at 20:41
@barmar updated the thread. – John Borton Oct 02 '15 at 20:56
You still haven't put the error message in the question. – Barmar Oct 02 '15 at 20:57
@barmar my bad added it now, thanks – John Borton Oct 02 '15 at 21:02
So the problem is that you're using `$trooptype` later in the program, even though you only assign it when the first `if` succeeds. You should add an `else` clause that gives it a default value. – Barmar Oct 02 '15 at 21:05
@barmar Where should I add an else? – John Borton Oct 02 '15 at 21:25
After the closing brace for the `if`. – Barmar Oct 02 '15 at 21:32
`if (isset($data[...])) { ... } else { $trooptype = 'Unknown troops'; }` – Barmar Oct 02 '15 at 21:32
@barmar the one on line 20 or? I'm not very experienced so a proper explain would be helpful. – John Borton Oct 02 '15 at 21:52
There's only one IF block where you set `$trooptype`. Just think about the problem logically. – Barmar Oct 02 '15 at 22:04
It's the first `if` in the code. – Barmar Oct 02 '15 at 22:06
@barmar I did ass you said here: http://prntscr.com/8n84a4 Now I get this error http://prntscr.com/8n84lu – John Borton Oct 03 '15 at 11:55
You're missing a semicolon after `$trooptype = 'Archers'` – Barmar Oct 04 '15 at 03:00
`isset($data['parameter.troops.barracks1.1.number='])` should probably not have the `=` at the end. And `if(intval($data['parameter.troops.barracks1.1.number=0']))` should be `if(intval($data['parameter.troops.barracks1.1.number']) = 0) – Barmar Oct 04 '15 at 03:02
@barmar when I use if(intval($data['parameter.troops.barracks1.1.number']) = 0) then it returns this error: http://prntscr.com/8o1dwr – John Borton Oct 05 '15 at 16:35
Sorry, it should have been `==`, not `=`. – Barmar Oct 05 '15 at 16:41
@barmar now it this is my code: http://prntscr.com/8o1h0t and this is my error: http://prntscr.com/8o1hb0 What have I missed ? – John Borton Oct 05 '15 at 16:43
Please stop posting screenshots, edit the question. – Barmar Oct 05 '15 at 16:44
You're STILL missing the semicolon after `Archers'`. This is basic PHP syntax, why are you wasting my time with this? – Barmar Oct 05 '15 at 16:45

PHP Notice: Undefined variable after decryption

0 Answers0