我在PHP中有一个复杂的对象，需要解析它以构建一个Json字符串。我在这里和其他网站上找到了很多例子，但没有一个有效。更进一步的问题是，我的主机工作在PHP 5.2，我不能升级。
这是我的var_dump($myObj)的一个例子:
object(Park)[4]
private 'idObj' => string '60304' (length=5)
private 'name' => string 'AlphaSurf' (length=9)
private 'address' =>
object(Address)[6]
private 'idObj' => string '40304' (length=5)
private 'street' => string 'Champ de la Vigne' (length=17)
private 'number' => string '7' (length=1)
private 'zip' => string '1470' (length=4)
private 'city' => string 'Estavayer-le-Lac' (length=16)
private 'country' =>
object(Country)[8]
private 'idObj' => string '30039' (length=5)
private 'name' => string 'Switzerland' (length=11)
private 'flag' => string 'switzerland.gif' (length=15)
private 'usState' => null
private 'contactInfo' =>
object(ContactInfo)[7]
private 'idObj' => string '70304' (length=5)
private 'phone' => string '' (length=0)
private 'email' => string '' (length=0)
private 'emailcode' => null
private 'confirmed' => string '1' (length=1)
private 'website' => string 'www.alphasurf.ch' (length=16)
private 'mobile' => string '' (length=0)
private 'fax' => string '' (length=0)
private 'newsletter' => string '0' (length=1)
private 'owner' =>
object(User)[9]
private 'idObj' => string '50001' (length=5)
private 'username' => string 'emaborsa' (length=8)
private 'password' => string '5baa61e4c9b93f3f0682250b6cf8331b7ee68fd8' (length=40)
private 'type' => string 'ADMIN' (length=5)
private 'state' => string 'ACTIVE' (length=6)
private 'ip' => string '' (length=0)
private 'time' => string '0' (length=1)
private 'address' => null
private 'contactInfo' =>
object(ContactInfo)[11]
private 'idObj' => string '1' (length=1)
private 'phone' => null
private 'email' => string 'info@emaborsa.com' (length=17)
private 'emailcode' => null
private 'confirmed' => string '1' (length=1)
private 'website' => null
private 'mobile' => null
private 'fax' => null
private 'newsletter' => string '1' (length=1)
private 'logo' => string 'Champ de la Vigne 71470' (length=23)
private 'xcoord' => string '46856912' (length=8)
private 'ycoord' => string '6846918' (length=7)
private 'state' => string 'HIDDEN' (length=6)
private 'detail' =>
object(ParkDetail)[10]
private 'idObj' => string '1' (length=1)
private 'descriptionIT' => string '' (length=0)
private 'descriptionEN' => string '' (length=0)
private 'descriptionDE' => string 'xcxcx' (length=5)
private 'type' => string '' (length=0)
private 'kickers' => string '0' (length=1)
private 'boxes' => string '0' (length=1)
private 'rails' => string '0' (length=1)
private 'specials' => string '0' (length=1)
private 'specialsDescriptionIT' => null
private 'specialsDescriptionEN' => null
private 'specialsDescriptionDE' => null
private 'dimension' => string '0' (length=1)
private 'lastPayment' => null
所有属性都是私有的，但是有公共的getter和setter
试试这个
public function encodeJSON()
{
foreach ($this as $key => $value)
{
if($value instanceOf(stdClass)){
$json->$key = $value->encodeJSON();
}else
$json->$key = $value;
}
return json_encode($json);
}
我正试图将私有成员移动到一个可以由正常json_encode()写入的新对象，并且在第6行中，如果它不是基本类型
我认为你需要暴露私有属性值，这是一个设计错误。
当然，在某些情况下应该这样做。正如PHP Object To JSON格式所指出的那样，这样做的一种方法是通过反射。
下面是一个简单的例子，使用php ReflectionClass来实现你想要的:
function getJson($object)
{
$result = array();
$refl = new ReflectionClass($object);
foreach ($refl->getProperties() as $prop) {
$prop->setAccessible(true);
$result[$prop->name] = $prop->getValue($object);
}
return json_encode($result);
}
当你运行PHP <5.4，我建议在你的对象中创建一个toArray方法，返回一个包含所有属性的数组(如果它们是公共的，私有的或受保护的)。
例如:
public class Park {
private $idObj;
private $address;
public function toArray() {
$toArray = array(
'idObj' => $this->idObj,
'address' => $this->address->toArray() // Assuming address is an Address object
);
}
}
在你所有的类和子类中都这样做，然后你可以使用:
$park = new Park(/* your values to initialize the object */);
echo json_encode($park->toArray());