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"
)
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
如果你需要将一个数组强制转换为一个特定的类(在我的例子中,我需要对象类型为Google_Service_AndroidPublisher_Resource_Inappproducts),你可以像这样从stdClass中将类名str_replace为预期的类:
function castArrayToClass(array $array, string $className)
{
//first cast the array to stdClass
$subject = serialize((object)$array);
//then change the class name
$converted = str_replace(
'O:8:"stdClass"',
'O:'.strlen($className).':"'.$className.'"',
$subject
);
unset($subject);
return unserialize($converted);
}
其他回答
一个衬套
$object= json_decode(json_encode($result_array, JSON_FORCE_OBJECT));
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);
它的方法很简单,这将为递归数组创建一个对象:
$object = json_decode(json_encode((object) $yourArray), FALSE);
这个方法对我很管用
function array_to_obj($array, &$obj)
{
foreach ($array as $key => $value)
{
if (is_array($value))
{
$obj->$key = new stdClass();
array_to_obj($value, $obj->$key);
}
else
{
$obj->$key = $value;
}
}
return $obj;
}
function arrayToObject($array)
{
$object= new stdClass();
return array_to_obj($array,$object);
}
用法:
$myobject = arrayToObject($array);
print_r($myobject);
返回:
[127] => stdClass Object
(
[status] => Have you ever created a really great looking website design
)
[128] => stdClass Object
(
[status] => Figure A.
Facebook's horizontal scrollbars showing up on a 1024x768 screen resolution.
)
[129] => stdClass Object
(
[status] => The other day at work, I had some spare time
)
像往常一样,你可以这样循环:
foreach($myobject as $obj)
{
echo $obj->status;
}
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。