Jquery Encoding.getJSON（）API调用

我正在开发一个web应用程序。这与Last.fm API有关。它工作正常，除非艺术家参数包含数字或不寻常的字符（例如，“U2”、“Ke$ha”等），否则如何正确编码参数

   for (var item in billboard) {
        track = billboard[item]['song'];
        artist = billboard[item]['artist'];
    } 
    $.getJSON("http://ws.audioscrobbler.com/2.0/?method=track.search&artist=" + artist + "&track=" + track + "&api_key=(myapikey)&format=json&callback=?", function(data) {
           try {
           var matches = data['results']['trackmatches']['track'][0]
           }
           catch(err) {
            returned = data['results']['opensearch:Query']['searchTerms']
            $('#album-display').find('ul').append(returned + "<br>")
           }
           artist = matches['artist']
           track = matches['name']
       });
    } 

var billboard = {
 "5-23-1987": {"artist": "U2", "song": "With Or Without You"},
 "10-15-1988": {"artist": "UB40", "song": "Red Red Wine"},
 "3-7-2009": {"artist": "Flo Rida Featuring Ke$ha", "song": "Right Round"},
 ...
}

for（广告牌中的var项目）{
轨道=广告牌[项目]['song']；
艺术家=广告牌[项目][“艺术家]；
} 
$.getJSON（“http://ws.audioscrobbler.com/2.0/?method=track.search&artist=“+artist+”&track=“+track+”&api_key=（myapikey）&format=json&callback=？”，函数（数据）{
试一试{
var matches=data['results']['trackmatches']['track'][0]
}
捕捉（错误）{
返回=数据['results']['opensearch:Query']['searchTerms']
$（“#相册显示”）.find（'ul'）.append（返回+“
”）
}
艺术家=匹配项['artist']
track=匹配项['name']
});
} 
var广告牌={
“5-23-1987”：{“艺术家”：“U2”，“歌曲”：“有没有你”}，
“10-15-1988”：{“艺术家”：“UB40”，“歌曲”：“红酒”}，
《3-7-2009》：{《艺术家》：《弗洛·里达》以柯元哈为特色，《歌曲》：《右回合》}，
...
}

您需要在请求中转义URL，其中一些字符不合法或导致问题。改变

$.getJSON("http://ws.audioscrobbler.com/2.0/?method=track.search&artist=" + artist + "&track=" + track + "&api_key=(myapikey)&format=json&callback=?", function(data) {

到

var query = "method=track.search&artist=" + encodeURIComponent(artist) + "&track=" + encodeURIComponent(track) + "&api_key=(myapikey)&format=json&callback=?";
$.getJSON("http://ws.audioscrobbler.com/2.0/?" + query, function(data) {