Notice: undefined variable: vars in C:\xampp\htdocs\joblisting\lib\template.php on line 19

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Please I am new to PHP environment.
I want to create a Template.php class but i am getting an error

What I have tried:

//here is my Template.php file inside a lib folder

$template = $template;
}
public function __get($key)
{
return $this->$vars[$key]; // am getting undefined variable
}
public function __set($key, $value)
{
return $this->$vars[$key] = $value; // I am getting undefined variable
}
public function __toString()
{
extract($this->$vars);
chdir(dirname($this->template));
ob_start();

include basename($this->template);
return ob_get_clean();
}
}

?>

// here is my init.php

Posted 10-Jul-20 3:01am

Codeignite

Updated 10-Jul-20 23:10pm

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Comments

Richard MacCutchan 10-Jul-20 9:39am

You have not defined $vars anywhere.

Codeignite 10-Jul-20 9:49am

I notice that but i have modified the code.

I am still getting error messages
public function __toString()
{
extract($this->vars, EXTR_PREFIX_SAME, "dup");
chdir(dirname($this->template));
ob_start();
include basename($this->template);
return ob_get_clean();
}
// error messages in output
Warning: chdir(): Invalid argument (errno 22) in C:\xampp\htdocs\joblisting\lib\Template.php on line 25

Warning: include(): Filename cannot be empty in C:\xampp\htdocs\joblisting\lib\Template.php on line 27

Warning: include(): Failed opening '' for inclusion (include_path='C:\xampp\php\PEAR') in C:\xampp\htdocs\joblisting\lib\Template.php on line 27

Richard MacCutchan 10-Jul-20 9:55am

That is different code, and it would appear that $this->template is not returning a valid directory name. You need to debug your code and check exactly what values are being returned in the variables.

Codeignite 10-Jul-20 17:58pm

@Richard MacCutchan I am new to PHP I have printed the value it returns using var_dump(), it returns bool(false)

Richard MacCutchan 11-Jul-20 5:17am

Please use the Improve question link above, add the actual code that you are using and the output you see. And please use <pre> tags around your code so it is clear and readable.

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Andre Oosthuizen · Answer 1 · 2020-07-10T23:10:00

You have not declared $var in your class which is why it cannot be found as referenced.

Try the following -

PHP

<?php

  class MyClass {

    private $vars;

    public function __construct(Array $properties=array()){
      $this->vars = $properties;
//No dollar sign...
    }

    // magic methods!
    public function __set($property, $value){
      return $this->vars[$property] = $value;
    }

    public function __get($property){
      return array_key_exists($property, $this->vars)
        ? $this->vars[$property]
        : null
      ;
    }
  }

?>

Usage will be the same -

PHP

// init call to the class name...
$foo = new MyClass(array("hello" => "world"));
$foo->hello;          // => "world"

// set: this calls __set()
$foo->invader = "Codeignite";

// get: this calls __get()
$foo->invader;       // => "Codeignite"

// attempt to get a data[key] that isn't set
$foo->invalid;       // => null