我很难使用LastFM API显示搜索艺术家的顶部曲目以获取数据。API返回对象TopTracks。我想从该API数据中获取有关每个顶级曲目的详细信息。
我不确定我是否在正确的轨道上。有人可以看看,让我知道我是否做错了什么?
来自API的示例数据:
{
"toptracks": {
"track": [{
"name": "Best I Ever Had",
"playcount": "3723918",
"listeners": "1086968",
"mbid": "00bde944-7562-446f-ad0f-3d4bdc86b69f",
"url": "https://www.last.fm/music/Drake/_/Best+I+Ever+Had",
"streamable": "0",
"artist": {
"name": "Drake",
"mbid": "b49b81cc-d5b7-4bdd-aadb-385df8de69a6",
},
"@attr": {
"rank": "1"
}
},
{
"name": "Forever",
"playcount": "1713492",
"listeners": "668998",
"url": "https://www.last.fm/music/Drake/_/Forever",
"streamable": "0",
"artist": {
"name": "Drake",
"mbid": "b49b81cc-d5b7-4bdd-aadb-385df8de69a6",
},
"@attr": {
"rank": "2"
}
}
}
function renderTracks(trackArray) {
function createHTML(track){
return `<h1>${track.name}</h1>
<h2>${track.artist[0]}</h2>
<h3>${toptracks[1].rank}</h3>
<h3>${track.playcount}</h3>`;
};
trackHTML = trackArray.map(createHTML);
return trackHTML.join("");
};
var searchString = $(".search-bar").val().toLowerCase();
var urlEncodedSearchString = encodeURIComponent(searchString);
const url = "lastFMwebsite"
axios.get(url + urlEncodedSearchString).then(function(response) {
// createHTML.push(response.data.track);
// $(".tracks-container").innerHTML = renderTracks(response.data.track);
// comented out old code above
createHTML.push(response.toptracks.track);
$(".tracks-container").innerHTML = renderTracks(response.toptracks.track);
})
我已经注意到您尚未解析响应:
axios.get(url + urlEncodedSearchString).then(function(response) {
var parsed = JSON.parse(response);
$(".tracks-container").innerHTML = renderTracks(parsed.toptracks.track)
});
我可以建议的另一种更正是将track.artist[0]
更改为track.artist["name"]
,一旦此属性返回对象而不是数组。关于这个:<h3>${toptracks[1].rank}</h3>
。您将无法访问该属性,因为在您的功能下,您仅提供track
属性。在这种情况下,您有两个选项:提供整个响应数组或添加一个提供此信息的新参数。
function renderTracks(trackArray) {/**...*/};
//...
$(".tracks-container").innerHTML = renderTracks(parsed.toptracks)
或
function renderTracks(trackArray, toptracks) {/**...*/};
//...
$(".tracks-container").innerHTML = renderTracks(parsed.toptracks.track, parsed.toptracks)
我希望这可以帮助您:(
您的输入JSON无效。您需要正确格式化。一旦数据正确:
createHTML.push(response.toptracks.track[0])
or
let i = 0;
for(; i < response.toptracks.track.length; i++){
createHTML.push(response.toptracks.track[i]);
}