How do you send a JSON encoded object to a PHP function that takes a PHP class object as a variable?

Question

I have a java aplet that is sending a JSON encoded object to a Zend Framework JSON Server.

the problem that I have is the code is setup like this:

ServerController:

public function jsonAction()
{

$server = new Zend_Json_Server();
$Server->setClass('Application_Model_ClassObject', 'co');

if('GET' == $_SERVER['REQUEST_METHOD'])
{
    $server->setTarget('...')
           ->setEnvelope(Zend_Json_Server_Smd::ENV_JSONRPC_2);
    $smd = $server->getServiceMap();

    header('Content-Type: application/json');

    echo $smd;
    return
}

echo $server->handle();
}

ClassObject function:

/**
* DoSomethign description
* @param ClassObject
*/
public function doSomething(Application_Model_ClassObject $obj)
{

    $someVariable = $obj->getSomeValue();
    ...
}

I get an error response from the server saying that obj needs to be an instance of ClassObject

Answer 1

Ok so this is what we ended up doing (Quick and dirty version)... Although i believe there is a better way.

1. Removed the type-hint so that the function will be able to receive any data.
2. Create a new function in "Application_Model_ClassObject" that will handle an array of data or an object.
3. Added a few extra lines to "doSomething" so that the rest can go on as normal.

Here are some details:

/**
* DoSomethign description
* @param ClassObject
*/
public function doSomething($data)
{
    $obj = new Application_Model_ClassObject();
    $obj->populate($data);

    $someVariable = $obj->getSomeValue();
    ...
}

/**
*
* Application_Model_ClassObject class
*/
class Application_Model_ClassObject
{
    ...

    public function populate($row)
    {
        if (is_array($row)) {
        $row = new ArrayObject($row, ArrayObject::ARRAY_AS_PROPS);
        }

        if (isset ($row->SomeValue)) {
        $this->setSomeValue($row->SomeValue);
        }

        return $this;
    }

}

Answer 2

如果删除类型提示“ ClassObject”，则doSomething方法将接受任何传入的数据。然后由doSomething方法来解码json字符串，并对其进行一些有用的处理。

Answer 3

public function doSomething(ClassObject obj) { ... }

Um, usually you don't write PHP functions like that, as it's not strongly typed. Normally, it looks like this instead:

public function doSomething($obj) {...}

It looks like you've tried to write a PHP function with some Java syntax mixed in.