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
容易:
$object = json_decode(json_encode($array));
例子:
$array = array(
'key' => array(
'k' => 'value',
),
'group' => array('a', 'b', 'c')
);
$object = json_decode(json_encode($array));
那么,以下是正确的:
$object->key->k === 'value';
$object->group === array('a', 'b', 'c')
其他回答
通过使用(array)和(object)作为前缀,可以简单地将对象数组转换为标准数组,反之亦然
<?php
//defining an array
$a = array('a'=>'1','b'=>'2','c'=>'3','d'=>'4');
//defining an object array
$obj = new stdClass();
$obj->a = '1';
$obj->b = '2';
$obj->c = '3';
$obj->d = '4';
print_r($a);echo '<br>';
print_r($obj);echo '<br>';
//converting object array to array
$b = (array) $obj;
print_r($b);echo '<br>';
//converting array to object
$c = (object) $a;
print_r($c);echo '<br>';
?>
有点复杂,但很容易扩展的技术:
假设你有一个数组
$a = [
'name' => 'ankit',
'age' => '33',
'dob' => '1984-04-12'
];
假设您有一个Person类,它可能有来自这个数组的或多或少的属性。例如
class Person
{
private $name;
private $dob;
private $age;
private $company;
private $city;
}
如果你还想把数组改成person对象。你可以使用ArrayIterator类。
$arrayIterator = new \ArrayIterator($a); // Pass your array in the argument.
现在你有了迭代器对象。
创建一个扩展FilterIterator class的类;你必须定义抽象方法accept。遵循示例
class PersonIterator extends \FilterIterator
{
public function accept()
{
return property_exists('Person', parent::current());
}
}
上面的实现只在类中存在该属性时才会绑定它。
在类PersonIterator中再添加一个方法
public function getObject(Person $object)
{
foreach ($this as $key => $value)
{
$object->{'set' . underscoreToCamelCase($key)}($value);
}
return $object;
}
确保在类中定义了mutator。 现在,您可以在想要创建对象的地方调用这些函数。
$arrayiterator = new \ArrayIterator($a);
$personIterator = new \PersonIterator($arrayiterator);
$personIterator->getObject(); // this will return your Person Object.
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。
使用json_encode是有问题的,因为它处理非UTF-8数据的方式。值得注意的是,json_encode/json_encode方法也将非关联数组作为数组。这可能是你想要的,也可能不是。我最近需要重新创建这个解决方案的功能,但没有使用json_ functions。这是我想到的:
/**
* Returns true if the array has only integer keys
*/
function isArrayAssociative(array $array) {
return (bool)count(array_filter(array_keys($array), 'is_string'));
}
/**
* Converts an array to an object, but leaves non-associative arrays as arrays.
* This is the same logic that `json_decode(json_encode($arr), false)` uses.
*/
function arrayToObject(array $array, $maxDepth = 10) {
if($maxDepth == 0) {
return $array;
}
if(isArrayAssociative($array)) {
$newObject = new \stdClass;
foreach ($array as $key => $value) {
if(is_array($value)) {
$newObject->{$key} = arrayToObject($value, $maxDepth - 1);
} else {
$newObject->{$key} = $value;
}
}
return $newObject;
} else {
$newArray = array();
foreach ($array as $value) {
if(is_array($value)) {
$newArray[] = arrayToObject($value, $maxDepth - 1);
} else {
$newArray[] = $value;
}
}
return $newArray;
}
}
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