Reputation: 303
I need to be able to set my object like this:
$obj->foo = 'bar';
then I need to use it as an array like that:
if($obj['foo'] == 'bar'){
//more code here
}
Upvotes: 30
Views: 29850
Reputation: 10095
Try extending ArrayObject
You'll also need to implement a __get
Magic Method as Valentin Golev mentioned.
Your class will need to looks something like this:
Class myClass extends ArrayObject {
// class property definitions...
public function __construct()
{
//Do Stuff
}
public function __get($n) { return $this[$n]; }
// Other methods
}
Upvotes: 20
Reputation: 18840
You can also use ArrayAccess to access a single array property in your class and leave other properties being accessed in OOP way. Yet still it will work as you requested.
class Foo implements \ArrayAccess
{
/**
* mixed[] now you can access this array using your object
* like a normal array Foo['something'] = 'blablabla'; echo Foo['something']; ... and so on
* other properties will remain accessed as normal: $Foo->getName();
*/
private myArrayOptions = [];
private $name = 'lala';
...
public function offsetExists($offset)
{
return isset($this->myArrayOptions[$offset]);
}
public function offsetGet($offset)
{
if ($this->offsetExists($offset)) {
return $this->myArrayOptions[$offset];
}
return null; // or throw the exception;
}
public function offsetSet($offset, $value)
{
$this->myArrayOptions[$offset] = $value;
}
public function offsetUnset($offset)
{
unset($this->myArrayOptions[$offset]);
}
public function getName()
{
return $this->name;
}
public function __set($offset, $value){
$this->myArrayOptions[$offset] = $value;
}
...
}
The above will work as you expected.
$obj->foo = 'bar';
if($obj['foo'] == 'bar'){
echo "WoWo";
}
Also note that Foo['name'] !== Foo->getName() those a two different variables
Upvotes: 2
Reputation: 28795
You're mixing objects and arrays. You can create and access an object like so:
$obj = new stdClass;
$obj->foo = 'bar';
if($obj->foo == 'bar'){
// true
}
and an array like so:
$obj = new Array();
$obj['foo'] = 'bar';
if($obj['foo'] == 'bar'){
// true
}
You can define a class and add implements ArrayAccess if you want to access your class as both an array and a class.
http://www.php.net/manual/en/language.oop5.php
Upvotes: 9
Reputation: 3093
You can access PHP object as PHP array, but in different ways. Try this:
$obj->{'foo'}
That is similar with accessing array like this:
$arr['foo']
You can also do this:
$propertyName = 'foo';
$obj->$propertyName; // same like first example
Upvotes: 10
Reputation: 180004
You could also cast the object as an array:
if((array)$obj['foo'] == 'bar'){
//more code here
}
Upvotes: 3
Reputation: 96159
ArrayObject implements the ArrayAccess interface (and some more). Using the ARRAY_AS_PROPS flag it provides the functionality you're looking for.
$obj = new ArrayObject(array(), ArrayObject::ARRAY_AS_PROPS);
$obj->foo = 'bar';
echo $obj['foo'];
Alternatively you can implement the ArrayAccess interface in one of your own classes:
class Foo implements ArrayAccess {
public function offsetExists($offset) {
return isset($this->$offset);
}
public function offsetGet($offset) {
return $this->$offset;
}
public function offsetSet($offset , $value) {
$this->$offset = $value;
}
public function offsetUnset($offset) {
unset($this->$offset);
}
}
$obj = new Foo;
$obj->foo = 'bar';
echo $obj['foo'];
Upvotes: 17
Reputation: 316969
Just add implements ArrayAccess
to your class and add the required methods:
See http://php.net/manual/en/class.arrayaccess.php
Upvotes: 23
Reputation: 400952
You'll have to implement the ArrayAccess
interface to be able to do that -- which only means implementing a few (4 to be exact) simple methods :
ArrayAccess::offsetExists
: Whether or not an offset exists. ArrayAccess::offsetGet
: Returns the value at specified offset.ArrayAccess::offsetSet
: Assigns a value to the specified offset. ArrayAccess::offsetUnset
: Unsets an offset. There is a full example on the manual's page I pointed to ;-)
Upvotes: 9
Reputation: 75704
Your object must implement the ArrayAccess
interface, then PHP will allow you to use the square brackets like that.
Upvotes: 3