ArrayObject::offsetExists
If you're overloading ArrayObject, it's worth noting that while this method (when implemented by the parent) will return a reference, so code like $fakeArray['foobar']['hello'] = 1; will work like you expect.
However, when you overload the offsetGet method, you CANNOT define it as &offsetGet, so the above code falls out (because it returns the 'foobar' variable before you actually work with it).
This is something that the developers broke between 5.0 and 5.1, and was closed as bogus (http://bugs.php.net/bug.php?id=34783). So this is not a big, or question, or request, but just something worth noting.
ArrayObject::offsetGet
(PHP 5 >= 5.0.0)
ArrayObject::offsetGet — Devuelve el valor en el índice especificado
Parámetros
-
index -
El índice con el valor.
Valores devueltos
El valor en el índice especificado o NULL.
Ejemplos
Ejemplo #1 Ejemplo de ArrayObject::offsetGet()
<?php
$arrayobj = new ArrayObject(array('cero', 7, 'ejemplo'=>'e.j.'));
var_dump($arrayobj->offsetGet(1));
var_dump($arrayobj->offsetGet('ejemplo'));
var_dump($arrayobj->offsetExists('noencontrado'));
?>
El resultado del ejemplo sería:
int(7) string(4) "e.j." bool(false)
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User Contributed Notes
ArrayObject::offsetGet - [2 notes]
Sam ¶
5 years ago
Alex Andrienko ¶
4 years ago
Speaking of offsetGet() method overloading, be advised, that if you're iterating through Object via foreach, this method wouldn't be called. Iterator's current() method will be called instead.