PHP - get object parent

Question

I have a function that is called like this:

foo($object->ID);

and in the function I need to somehow select $object if $object->ID is passed as a variable.

function foo($id = NULL){
  if($id != NULL) ... // here I want to get $object
  else ...
}

How can I do this?

Matěj Zábský · Accepted Answer · 2010-12-14 22:42:39Z

5

That is not possible. You are passing a number without any information about its origin. Do this

foo($object)

function foo($object){
   if($object->ID !== null) ... // work with $object
   else ... // work with ID

}

answered Dec 14, 2010 at 22:42

Matěj Zábský

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Comments

FatherStorm · Accepted Answer · 2010-12-14 22:44:02Z

4

why not:

foo($object);

and

function  foo($localObject){
  if(isset($localObject->id)){
  }
}

answered Dec 14, 2010 at 22:44

FatherStorm

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Comments

simshaun · Accepted Answer · 2010-12-14 22:43:56Z

2

You need to pass the object in as an argument instead of the ID.

answered Dec 14, 2010 at 22:43

simshaun

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1 Comment

Josh Over a year ago

You should just pass the object in, and get ID from that object, not add another parameter.

Demian Brecht · Accepted Answer · 2010-12-14 23:38:38Z

1

If I understand what you're asking correctly, why not just pass the object itself by ref?

function foo(&$obj)
{
    if($obj != NULL && $obj->ID != NULL)
    {
       // ...process your stuff 
    }
}

My PHP's pretty rusty, but I'm fairly sure that's how you pass by ref...

edited Dec 14, 2010 at 23:38

answered Dec 14, 2010 at 22:44

Demian Brecht

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7 Comments

Matěj Zábský Over a year ago

Pass-by-ref is not relevant in this case.

Demian Brecht Over a year ago

I should clarify: You want to make sure that you pass the object by reference so that you don't make an entirely new copy of it.

Alex Over a year ago

well the reason is pretty stupid :) I have a set of functions that all but this one take the $object->id as argument. And it looks ugly if just this one didn't :)

Lee Over a year ago

since php 5 (maybe earlier than that), objects are always passed by reference - no need to explicitly pass objects by ref.

Jonah Over a year ago

There's a typo: it should be $obj, not &obj on line 3.

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