PHP calling a class method with class name as variable string

Question

I have a class from which I want to call a function, but the instantiated class variable is taken from a string.

class Myclass
{
 public function myFunction($param1,$param2) {
  return 'blah';
 }
}

I am trying the following but it doesn't work. Errors from log file below.

$myclass='myclass';
${$myclass}=new ucfirst($myclass);
$args=array($stringvar,1);
$retval=call_user_func_array(array(${$myclass}, 'myFunction'), $args);

What am I doing wrong ?

Errors :

PHP Warning:  call_user_func_array() expects parameter 1 to be a valid callback, first array member is not a valid class name or object in file.php on line 100
PHP Notice:  Undefined variable: myclass in file.php on line 134
PHP Fatal error:  Call to a member function myFunction() on a non-object in file.php on line 134

I appreciate that you are after an answer to the problem you have. May i ask, what was the original problem that you were trying to solve? I keep remembering this 'maxim': it is hard to remember, when you are up to your as in crocodiles, that the original intention was to drain the swamp. Your code, by its nature, will be hard to maintain and change in the future. — Ryan Vincent
– Ryan Vincent, Commented Apr 11, 2014 at 1:29

Community · Accepted Answer · 2023-11-17 20:38:18Z

2

You're overcomplicating it a bit with the extra ${} bracketing on the class reference.

$myclass='myCLaSs';
$myObject = new $myclass();
$retval = $myObject->myFunction($stringvar, 1);

Or if you need to use call_user_func_array:

$args=array($stringvar,1);
$retval=call_user_func_array(array($myObject, 'myFunction'), $args);

When you use the ${} bracketing, you are referencing variables by a variable name. For example:

$myVariable = "A";
$aVariableThatIsHoldingAVariableName = "myVariable";
echo ${$aVariableThatIsHoldingAVariableName}; // outputs "A".

Applying this to your code shows the following logic happening:

Set the variable $myclass equal to the string 'myclass'

$myclass='myclass';

Set the variable ${$myclass}:

This gets the value of the variable $myclass ('myclass') and uses that as the variable name. In other words, the following variable name resolution happens: ${$myclass} => ${'myclass'} => $myclass.

So this line sets $myclass to a new Myclass(); object:

${$myclass}=new ucfirst($myclass);

The first parameter to call_user_func_array is a callable (See https://www.php.net/manual/en/language.types.callable.php). The callback it looks like you are trying to reference here is Type 3: array($object, $method). But the same variable resolution happens. Now, ${$myclass} is going to resolve differently, because the value of $myclass is a Myclass Object. Variable names have to be strings, not objects (obviously), so ${Myclass Object} is totally invalid.

$args=array($stringvar,1);
$retval=call_user_func_array(array(${$myclass}, 'myFunction'), $args);

But since $myclass at that point is an object already, you can just do the following (as mentioned initially above):

$retval=call_user_func_array(array($myclass, 'myFunction'), $args);

edited Nov 17, 2023 at 20:38

CommunityBot

11 silver badge

answered Apr 10, 2014 at 21:41

Kevin Vaughan

15.4k4 gold badges32 silver badges23 bronze badges

Sign up to request clarification or add additional context in comments.

8 Comments

Palva Over a year ago

I forgot to say that the ${$myclass} is necessary because I'm taking the class name as a string from a DB table.

Kevin Vaughan Over a year ago

In the example I've included, $myclass is a string. You can reference it directly in the new () call.

Kevin Vaughan Over a year ago

I've edited my answer to include an example of how the ${} bracketing syntax works.

Palva Over a year ago

Let's say that I have an array from which I get the various class names which I want to instantiate. Whenever I want to reference these, I have to get the string (myclass, myclass1, myclassn). Am I too tired to understand how to apply this or should I look for a way to store the objects somewhere ?

Kevin Vaughan Over a year ago

Too tired is my bet :) If you dont want to create an object and just want to call the functions on the class (maybe they are static for example), you can pass the string directly in as the class name to the callback parameter for call_user_func_array, e.g., with $myclass being the variable with the class name, call_user_func_array($myclass, 'myFunction'), $args). Copy and paste this code and your class definition in to a test script and play around with it, might help shake the cobwebs off :)

|

Collectives™ on Stack Overflow

PHP calling a class method with class name as variable string

1 Answer 1

8 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

8 Comments

Your Answer

Sign up or log in

Post as a guest

Related