Question

我从XML文件中检索了数据

$response = simplexml_load_file($url);

print_r显示以下内容

SimpleXMLElement Object
(
[artist] => SimpleXMLElement Object
    (
        [@attributes] => Array
            (
                [type] => Group
                [id] => b9fb5447-7f95-4a6a-a157-afed2d7b9f4c
            )

        [name] => He Is Legend
        [sort-name] => He Is Legend
        [country] => US
        [area] => SimpleXMLElement Object
            (
                [@attributes] => Array
                    (
                        [id] => 489ce91b-6658-3307-9877-795b68554c98
                    )

                [name] => United States
                [sort-name] => United States
                [iso-3166-1-code-list] => SimpleXMLElement Object
                    (
                        [iso-3166-1-code] => US
                    )

            )

    )

)

所以我可以通过

输出名称和国家/地区等字段

echo $response->artist->name;
echo $response->artist->country;

然而，当涉及到能够访问属性数组中的数据时，我会陷入困境。我怎样才能从第一个属性数组中获取组的类型？

修改

我也试图从像这样的函数中返回细节

func getDetails($id) {

$response = simplexml_load_file('http://musicbrainz.org/ws/2/artist/'.$id);
$data = array();
$data['type'] = $response->artist->attributes()->type;
$data['country'] = $response->artist->country;

return $data;
}

print_r(getDetails());

给我

MusicBrainz Object
(
)

Answer 1

您可以使用attributes()方法访问它们：

echo $response->artist->attributes()->type;

SimpleXML文档中的example #5显示了另一个示例。

如何从Simple XML Object获取某些详细信息？

1 个答案: