我觉得invariant
旨意为你服务好这里。首先,这可能是让你知道,你可以严格哈克输入一个密钥树:
<?hh // strict
class KeyedTree<+Tk as arraykey, +T> {
public function __construct(
private Map<Tk, KeyedTree<Tk, T>> $descendants = Map{},
private ?T $v = null
) {}
}
(那一定是因为cyclic shape definitions are sadly not allowed类)
我还没有尝试过,但type_structure
s和Fred Emmott's TypeAssert
看起来也有兴趣。如果已知JSON blob的某些部分已修复,那么可以使用invariant
s隔离嵌套的不确定部分并从中构建一棵树。在整个BLOB是未知的极限情况,那么你可以切除TypeAssert
因为没有有趣的固定结构断言:
use FredEmmott\TypeAssert\TypeAssert;
class JSONParser {
const type Blob = shape(
'live' => shape(
'host' => string, // fixed
'somevalue' => string, // fixed
'anobject' => KeyedTree<arraykey, mixed> // nested and uncertain
)
);
public static function parse_json(string $json_str): this::Blob {
$json = json_decode($json_str, true);
invariant(!array_key_exists('anobject', $json), 'JSON is not properly formatted.');
$json['anobject'] = self::DFS($json['anobject']);
// replace the uncertain array with a `KeyedTree`
return TypeAssert::matchesTypeStructure(
type_structure(self::class, 'Blob'),
$json
);
return $json;
}
public static function DFS(array<arraykey, mixed> $tree): KeyedTree<arraykey, mixed> {
$descendants = Map{};
foreach($tree as $k => $v) {
if(is_array($v))
$descendants[$k] = self::DFS($v);
else
$descendants[$k] = new KeyedTree(Map{}, $v); // leaf node
}
return new KeyedTree($descendants);
}
}
沿着这条路走下去,你仍然必须补充的KeyedTree
containsKey
不变,但这是Hack中非结构化数据的现实。
非常感谢您的意见! – Bearzi