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
function object_to_array($data)
{
if (is_array($data) || is_object($data))
{
$result = array();
foreach ($data as $key => $value)
{
$result[$key] = object_to_array($value);
}
return $result;
}
return $data;
}
function array_to_object($data)
{
if (is_array($data) || is_object($data))
{
$result= new stdClass();
foreach ($data as $key => $value)
{
$result->$key = array_to_object($value);
}
return $result;
}
return $data;
}
其他回答
受到所有这些代码的启发,我尝试创建一个增强版本,支持:特定的类名,避免构造函数方法,'beans'模式和严格模式(仅设置现有属性):
class Util {
static function arrayToObject($array, $class = 'stdClass', $strict = false) {
if (!is_array($array)) {
return $array;
}
//create an instance of an class without calling class's constructor
$object = unserialize(
sprintf(
'O:%d:"%s":0:{}', strlen($class), $class
)
);
if (is_array($array) && count($array) > 0) {
foreach ($array as $name => $value) {
$name = strtolower(trim($name));
if (!empty($name)) {
if(method_exists($object, 'set'.$name)){
$object->{'set'.$name}(Util::arrayToObject($value));
}else{
if(($strict)){
if(property_exists($class, $name)){
$object->$name = Util::arrayToObject($value);
}
}else{
$object->$name = Util::arrayToObject($value);
}
}
}
}
return $object;
} else {
return 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)。
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
根据你需要的位置和访问对象的方式有不同的方法。
例如:只需对它进行类型转换
$object = (object) $yourArray;
然而,最兼容的方法是使用一个实用程序方法(还不是PHP的一部分),它实现了基于指定类型的字符串的标准PHP强制转换(或者忽略它,只是去引用值):
/**
* dereference a value and optionally setting its type
*
* @param mixed $mixed
* @param null $type (optional)
*
* @return mixed $mixed set as $type
*/
function rettype($mixed, $type = NULL) {
$type === NULL || settype($mixed, $type);
return $mixed;
}
您案例中的使用示例(在线演示):
$yourArray = Array('status' => 'Figure A. ...');
echo rettype($yourArray, 'object')->status; // prints "Figure A. ..."
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。