php public function is not working outside of the class [duplicate]

Question

I have created a simple class, variable, and public function. When I try to initiate an object and call the public function from the class outside of the class, I get no output that I can see.

<?php
error_reporting(E_ALL);
ini_set("error_reporting", 1);

  class Game {
    var name;

    public function show_game() {
      echo $this->name;
    }
  }

  $game = new Game;
  $game->name = "testing game";

  $game->show_game();
?>

From what I understand, the function should echo out testing game. But I am not seeing any output when I load up the page.

"But I am not seeing any output when I load up the page." - Well, you should with your error reporting set. If that doesn't throw you a parse error, then I have my own views as to what's "not" going on. I'll let you take it up with the answers below. — Funk Forty Niner
– Funk Forty Niner, Commented Jun 14, 2017 at 1:57
You guys down there paying attention up here? Read the question again; completely. Specifically "I get no output". Edit: I guess not. — Funk Forty Niner
– Funk Forty Niner, Commented Jun 14, 2017 at 2:01
"I get no output that I can see." - I don't believe you error_reporting(E_ALL); ini_set("error_reporting", 1);. — Funk Forty Niner
– Funk Forty Niner, Commented Jun 14, 2017 at 2:11
@Fred-ii- I wonder how long it will take to find the next syntax error with no output. I tried running it in the console, and whaddaya know? It didn't complain! Then I looked up error reporting, LOL. Take careful aim at your foot... (the OP's foot, that is) — Tim Morton
– Tim Morton, Commented Jun 14, 2017 at 2:28
@Fred-ii- what are you going on about?? I didn't get an output or an error, and just now I found out the company is having wifi difficulties so I suspect that some of my issue may have been caused by my file not actually uploading to the server. — Dan
– Dan, Commented Jun 14, 2017 at 2:31

Enstage · Accepted Answer · 2017-06-14 01:54:20Z

1

var name;

Is not valid syntax, change to:

var $name;

answered Jun 14, 2017 at 1:54

Enstage

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Comments

Obsidian Age · Accepted Answer · 2017-06-14 01:56:56Z

0

You seem to have simply forgotten the $ in your PHP variable. var name should be var $name:

<?php
  class Game {
    var $name;
    public function show_game() {
      echo $this->name; // Returns "testing game"
    }
  }
  $game = new Game;
  $game->name = "testing game";
  $game->show_game();
?>

Hope this helps! :)

answered Jun 14, 2017 at 1:56

Obsidian Age

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