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
实际上,如果你想在多维数组中使用这个你就需要使用一些递归。
static public function array_to_object(array $array)
{
foreach($array as $key => $value)
{
if(is_array($value))
{
$array[$key] = self::array_to_object($value);
}
}
return (object)$array;
}
其他回答
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);
我肯定会用这样一种干净的方式:
<?php
class Person {
private $name;
private $age;
private $sexe;
function __construct ($payload)
{
if (is_array($payload))
$this->from_array($payload);
}
public function from_array($array)
{
foreach(get_object_vars($this) as $attrName => $attrValue)
$this->{$attrName} = $array[$attrName];
}
public function say_hi ()
{
print "hi my name is {$this->name}";
}
}
print_r($_POST);
$mike = new Person($_POST);
$mike->say_hi();
?>
如果你提交:
你会得到这个:
我发现这更符合逻辑的比较以上的答案从对象应该用于他们的目的(封装可爱的小对象)。
此外,使用get_object_vars确保在被操纵的对象中没有创建额外的属性(您不希望汽车有姓,也不希望人有4个轮子)。
多维数组转换为对象。此代码用于转换必应搜索API的尝试和捕获方法。
try {
// Perform the Web request and get the JSON response
$context = stream_context_create($options);
$results = file_get_contents($url . "?cc=" . $country . "&category=" . $type, false, $context);
$results = json_decode($results);
return response()->json($results);
} catch (\Exception $e) {
$results = array('value' => array(
(object) array(
"name" => "Unable to Retrive News",
"url" => "http://www.sample.com/",
"image" => (object) array("thumbnail" => (object) array("contentUrl" => "")),
"publishedAt" => "",
"description" => "")
)
);
$results = (object) $results;
return response()->json($results);
}
据我所知,没有内置的方法可以做到这一点,但它就像一个简单的循环一样简单:
$obj= new stdClass();
foreach ($array as $k=> $v) {
$obj->{$k} = $v;
}
如果你需要递归地构建你的对象,你可以详细说明。
一个衬套
$object= json_decode(json_encode($result_array, JSON_FORCE_OBJECT));