How does ArrayObject in PHP work?

Question

I'm trying to understand this object but i can't figure out a simple fact. If count method shows public properties and the result is the number of keys in an array that was passed. In the case of an associative array when i try to access a key like a public property is not found. Maybe i misunderstood the interface.

//example
$currentDate = getdate();

//applying print_r() we can see the content
$objectDate = new ArrayObject();

//verifying the public properties- result is 11
$objectDate->count();

//but can't access keys like public properties
$objectDate->hours;

Minus, because this is "I'm too lazy to read docs" question. — Rafał Walczak
– Rafał Walczak, Commented Apr 17, 2014 at 8:54
Beside the fact that you've to use ['hours'], you need a $ in front of objectDate->hours;. — Jurik
– Jurik, Commented Apr 17, 2014 at 8:57
@RafałWalczak, did you read the docs? It's very unclear if you ask me. Upvoting because it's an interesting question and because Rafal is incorrect ;) — Luceos
– Luceos, Commented Apr 17, 2014 at 9:03

ax. · Accepted Answer · 2014-04-17 09:01:23Z

3

You can access array entries as properties (->) by passing the ArrayObject::ARRAY_AS_PROPS flag to the ArrayObject constructor:

//example
$currentDate = getdate();
print_r($currentDate);

// create ArrayObject from array, make entries accessible as properties (read and write).
$objectDate = new ArrayObject($currentDate, ArrayObject::ARRAY_AS_PROPS);

// verifying the public methods - result is 11
print_r($objectDate->count());
print "\n";

// accessing entries like public properties
print_r($objectDate->hours);

edited Apr 17, 2014 at 9:01

answered Apr 17, 2014 at 8:56

ax.

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moonwave99 · Accepted Answer · 2014-04-17 08:53:22Z

2

Such class implements ArrayAccess interface, so you can write:

$objectDate['hours']

With brackets notation, but not with arrow [->] one.

answered Apr 17, 2014 at 8:53

moonwave99

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1 Comment

artgento Over a year ago

Ok! Thanks! I see my problem now

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