PHP dynamic var from string

Question

I know how PHP dynamic vars work, I know I can access object property like $object->{'somethingWith$var'}; or like $object->$var;

But what I try to accomplish is to access $object->property->subproperty from $object and the string $string = 'property->subproperty';.

I tryed $object->$string, $object->{$string}, $object->$$string hahaha, none worked.

Does anybody know how to do this ? :)

Jiří Brabec · Accepted Answer · 2015-02-11 12:58:40Z

4

You can write simple function, something like this:

function accessSubproperty($object, $accessString) {
    $parts = explode('->', $accessString);
    $tmp = $object;
    while(count($parts)) {
        $tmp = $tmp->{array_shift($parts)};
    }
    return $tmp;
}

answered Feb 11, 2015 at 12:58

Jiří Brabec

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1 Comment

antoni Over a year ago

Thank you, this is what I think is the best then. I will rather use the foreach approach suggested by nl-x bellow for I will use it only once, but the idea is here and you were first. Thanks.

Elon Than · Accepted Answer · 2015-02-11 12:55:09Z

2

There is no way to do this like that.

You have to first assign $property = $object->$propertyName and then access var you wanted to $property->$subpropertyName.

In your examples string property->subproperty will be treated like variable name which obviously doesn't exist.

answered Feb 11, 2015 at 12:55

Elon Than

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1 Comment

antoni Over a year ago

Ok, thank you for the explanation and fast answer, I was looking for a solution though. also, I usually can access directly $object->property->subproperty without assignment, or I didn't get what you meant :)

nl-x · Accepted Answer · 2015-02-11 13:17:34Z

1

It is not working because all you are accomplishing with your tries is getting $object{'property->subproperty'} which off course is not the same as $object->{'property'}->{'subproperty'}.

What you can do is:

$ret = $object;
foreach (explode("->",$string) as $bit)
    $ret = $ret->$bit;

Or you will have to go to the ugly and evil eval() (let the downvoting start):

eval("return \$object->$string;")

answered Feb 11, 2015 at 13:17

nl-x

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3 Comments

Ruan Mendes Over a year ago

The only things this adds to Brabec's answer is the fact that you could use eval which is really not a great suggestion because of the security implications.

antoni Over a year ago

I appreciate the foreach way, and will use it. I don't rely on eval 0:) Thanks

nl-x Over a year ago

@disadb So you used the answer, but accepted the other answer :) ?

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