Test if a function is run from a static context

Question

I am writing a PHP class and have included a couple static functions for quick access as they are common for use and are simple in function. However they do use an object in them for database access. I will likely be using these static methods from both a static and non-static context throughout my code so I want to be able to test whether the function was called from a static or non-static context so that I can avoid creating a duplicate object if the function was called from a non-static context (this instance object and the one within the function to be used statically). Is there any way that I can test this in the function so that I can use the instance object if the function is called from a non-static context and create it's own object if the function is called from a static context?

Code Example:

class MyClass {
  private $db;

  function __constuct(){
    $this->db = new DBConnect();
  }

  public static function myFunction(){
    if(/* Is static */){
      $db = new DBConnect();
    } else {
      $db =& $this->db;
    }
    // Do processing with $db, etc.
  }
}

Answer 1

When a method is declared as static, not only is the magic variable $this unavailable (returns NULL), but it is impossible to tell if the function was actually called from a static context. A backtrace implies that for a static method, calling $object->method() is internally translated to className::method() at run time.

http://php.net/manual/en/language.oop5.static.php

Answer 2

You can check for non-static access only if you do not force the method to be static-only. Leave out the static keyword from the function declaration and test with:

public function myFunction(){
    if(!isset($this)) {

(It can still be called as static function, even though you did not declare it as such.)