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
您可以简单地使用类型强制转换将数组转换为对象。
// *convert array to object* Array([id]=> 321313[username]=>shahbaz)
$object = (object) $array_name;
//now it is converted to object and you can access it.
echo $object->username;
其他回答
我也有这个问题,但我注意到json_decode将JSON数组转换为对象。
所以,我通过使用json_encode($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);
快速攻击:
// assuming $var is a multidimensional array
$obj = json_decode (json_encode ($var), FALSE);
不漂亮,但很好用。
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);
您可以简单地使用类型强制转换将数组转换为对象。
// *convert array to object* Array([id]=> 321313[username]=>shahbaz)
$object = (object) $array_name;
//now it is converted to object and you can access it.
echo $object->username;
你也可以使用ArrayObject,例如:
<?php
$arr = array("test",
array("one"=>1,"two"=>2,"three"=>3),
array("one"=>1,"two"=>2,"three"=>3)
);
$o = new ArrayObject($arr);
echo $o->offsetGet(2)["two"],"\n";
foreach ($o as $key=>$val){
if (is_array($val)) {
foreach($val as $k => $v) {
echo $k . ' => ' . $v,"\n";
}
}
else
{
echo $val,"\n";
}
}
?>
//Output:
2
test
one => 1
two => 2
three => 3
one => 1
two => 2
three => 3