如何在PHP中从对象属性中获取值并将路径存储在变量中？

Question

For example, I have the following: $ValuePath = "object->data->user_nicename";

And I need to print not the value of the $ValuePath but the value of the $variable->data->user_nicename that is part of a larger call .. as I have the following situation:

echo $objects->$ValuePath->details;

and is not working .. I'm getting the error Class cannot be converted to string or something when I try it like that.

Answer 1

PHP pseudo-code:

function get_by_path($object, $path)
{
    $o = $object;
    $parts = explode('->', $path);
    // if you want to  remove the first dummy object-> reference
    array_shift($parts);
    $l = count($parts);
    while ($l)
    {
        $p = array_shift($parts); $l--;
        if ( isset($o->{$p}) ) $o = $o->{$p};
        else {$o = null; break;}
    }
    return $o;
}

Use like this:

$value = get_by_path($obj, "object->data->user_nicename"); 
// $value = $obj->data->user_nicename;

Check also PHP ArrayAccess Interface which enables to access objects as arrays dynamicaly

Answer 2

NO you cannot do that.

From your declaration $ValuePath is a variable which holds a plain string ('object->data->user_nicename') this string is not an object.
Where as, $objects->object->data->user_nicename->details is a object.

so $objects->object->data->user_nicename->details is not same as $objects->$ValuePath->details