如何从last.fm API获取歌手姓名

Question

我想从last.fm获取艺术家名称api HEre是一个代码

$jsonData = file_get_contents('http://ws.audioscrobbler.com/2.0/?method=chart.gettoptracks&limit=56&api_key=MYAPIKEY&format=json');
            $jsonData = json_decode($jsonData, true);
foreach ($jsonData['tracks']['track'] as $track) {
                $title = $track['name'];
                $image = $track['image'][3]['#text'];
                echo '<div id="track_short"><div class="track_short">
                <a href="http://'.$SiteUrl.'/mp3-download-'.cano($title).'/" ><img src="'.$image.'" alt="'.$title.'">
                <div class="description"><p class="description_content">'.$title.'</p>
                </div></a></div></div>';
}

API数据的示例

{"tracks":{"track":[{"name":"Sorry","duration":"0","playcount":"1931615","listeners":"193074","mbid":"","url":"http://www.last.fm/music/Justin+Bieber/_/Sorry","streamable":{"#text":"0","fulltrack":"0"},"artist":{"name":"Justin Bieber","mbid":"e0140a67-e4d1-4f13-8a01-364355bee46e","url":"http://www.last.fm/music/Justin+Bieber"},"image":[{"#text":"http://img2-ak.lst.fm/i/u/34s/d5af34cbc048b190fc7369acdcf8655b.png","size":"small"},{"#text":"http://img2-ak.lst.fm/i/u/64s/d5af34cbc048b190fc7369acdcf8655b.png","size":"medium"},{"#text":"http://img2-ak.lst.fm/i/u/174s/d5af34cbc048b190fc7369acdcf8655b.png","size":"large"},{"#text":"http://img2-ak.lst.fm/i/u/300x300/d5af34cbc048b190fc7369acdcf8655b.png","size":"extralarge"}]}],"@attr":{"page":"1","perPage":"1","totalPages":"253572626","total":"253572626"}}}

Answer 1

使用PHP的json_decode（）方法：

$json = '{"tracks":{"track":[{"name":"Sorry","duration":"0","playcount":"1931615","listeners":"193074","mbid":"","url":"http://www.last.fm/music/Justin+Bieber/_/Sorry","streamable":{"#text":"0","fulltrack":"0"},"artist":{"name":"Justin Bieber","mbid":"e0140a67-e4d1-4f13-8a01-364355bee46e","url":"http://www.last.fm/music/Justin+Bieber"},"image":[{"#text":"http://img2-ak.lst.fm/i/u/34s/d5af34cbc048b190fc7369acdcf8655b.png","size":"small"},{"#text":"http://img2-ak.lst.fm/i/u/64s/d5af34cbc048b190fc7369acdcf8655b.png","size":"medium"},{"#text":"http://img2-ak.lst.fm/i/u/174s/d5af34cbc048b190fc7369acdcf8655b.png","size":"large"},{"#text":"http://img2-ak.lst.fm/i/u/300x300/d5af34cbc048b190fc7369acdcf8655b.png","size":"extralarge"}]}],"@attr":{"page":"1","perPage":"1","totalPages":"253572626","total":"253572626"}}}';
$phpJson = json_decode($json, true, 10); //4 is the depth of the JSON string, but you can make this a big number and it will still work
$phpArtist = $phpJson['tracks']['track'][0]['artist']['name'];

让我知道这是否可行，或者是否需要调整php，我现在没有PHP客户端。

Answer 2

很感谢我有正确的来自： $phpArtist = $phpJson['tracks']['track'][0]['artist']; to： $Artist = $track['artist']['name']; 及其工作，谢谢