如何将这样的数组转换为对象?
[128] => Array
(
[status] => "Figure A.
Facebook's horizontal scrollbars showing up on a 1024x768 screen resolution."
)
[129] => Array
(
[status] => "The other day at work, I had some spare time"
)
当前回答
这里有三种方法:
Fake a real object: class convert { public $varible; public function __construct($array) { $this = $array; } public static function toObject($array) { $array = new convert($array); return $array; } } Convert the array into an object by casting it to an object: $array = array( // ... ); $object = (object) $array; Manually convert the array into an object: $object = object; foreach ($arr as $key => $value) { $object->{$key} = $value; }
其他回答
使用我创建的这个函数:
function buildObject($class,$data){
$object = new $class;
foreach($data as $key=>$value){
if(property_exists($class,$key)){
$object->{'set'.ucfirst($key)}($value);
}
}
return $object;
}
用法:
$myObject = buildObject('MyClassName',$myArray);
递归是你的朋友:
function __toObject(Array $arr) {
$obj = new stdClass();
foreach($arr as $key=>$val) {
if (is_array($val)) {
$val = __toObject($val);
}
$obj->$key = $val;
}
return $obj;
}
我也有这个问题，但我注意到json_decode将JSON数组转换为对象。
所以，我通过使用json_encode($PHPArray)来实现我的解决方案，它返回对象的JSON字符串，然后我用Json_decode($string)解码字符串，它将返回一个完美的结构化对象。 速记
$object = json_decode(json_encode($array));
Or
$jsonString = json_encode($array);
$object = json_decode($jsonString);
这里有三种方法:
Fake a real object: class convert { public $varible; public function __construct($array) { $this = $array; } public static function toObject($array) { $array = new convert($array); return $array; } } Convert the array into an object by casting it to an object: $array = array( // ... ); $object = (object) $array; Manually convert the array into an object: $object = object; foreach ($arr as $key => $value) { $object->{$key} = $value; }
简单的方法是
$object = (object)$array;
但这不是你想要的。如果你想要对象，你想要实现一些东西，但这在这个问题中是缺失的。仅仅为了使用对象而使用对象是没有意义的。
