Javascript 使用Nodejs解析Tinysong返回

我试图解析从Tinysong的查询中得到的响应。我能够生成get并拥有它console.log，但在这之后我无法对数据做任何处理。我想这是因为我是如何捕捉它的。这是我第一次这样做，所以我只是把很多不同的例子混合在一起

如果导航到URL，则会得到如下结果：

{"Url":"http:\/\/tinysong.com\/1kxG6","SongID":41760691,"SongName":"Begin Again","ArtistID":9472,"ArtistName":"Knife Party","AlbumID":9855703,"AlbumName":"Abandon Ship"}

{"Url":"http:\/\/tinysong.com\/1ksPa","SongID":40778838,"SongName":"Cut Your Teeth (Kygo remix)","ArtistID":2002910,"ArtistName":"Kyla La Grange","AlbumID":9
833902,"AlbumName":"Cut Your Teeth (Kygo remix)"}

这就是我所拥有的

function TinySong (A, S){
    A = A.replace(/ /g,"+");
    S = S.replace(/ /g,"+");
    var url = 'http://tinysong.com/b/' + A + "+" + S + '?format=json&' + TinyKey;
    http.get(url, function(res){
        var bodyChunks = [];
        res.on('data', function(chunk) {
            bodyChunks.push(chunk);
        }).on('end', function() {
            var body = Buffer.concat(bodyChunks);
//             console.log(url)
            console.log('' + body);
        })
    });
}

此控制台记录如下内容：

{"Url":"http:\/\/tinysong.com\/1kxG6","SongID":41760691,"SongName":"Begin Again","ArtistID":9472,"ArtistName":"Knife Party","AlbumID":9855703,"AlbumName":"Abandon Ship"}

{"Url":"http:\/\/tinysong.com\/1ksPa","SongID":40778838,"SongName":"Cut Your Teeth (Kygo remix)","ArtistID":2002910,"ArtistName":"Kyla La Grange","AlbumID":9
833902,"AlbumName":"Cut Your Teeth (Kygo remix)"}

但是如果我不使用

console.log（“”+body）和我使用
console.log（body）
我明白了：

<Buffer 7b 22 55 72 6c 22 3a 22 68 74 74 70 3a 5c 2f 5c 2f 74 69 6e 79 73 6f 6e 67 2e 63 6f 6d 5c 2f 31 6b 73 50 61 22 2c 22 53 6f 6e 67 49 44 22 3a 34 30 37
 37 ...> 



谁能帮我找出我做错了什么。欢迎所有的帮助和教育

谢谢
FPC找到了答案

    http.get(url, function(res){
        var data = '';
        res.on('data', function (chunk){
            data += chunk;
        });
        res.on('end', function(){
            var obj = JSON.parse(data);
            console.log(obj)
        });
    })