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我完全陌生,我正在学习这个javascript。我正在构建一个应用程序/机器人,但我坚持这一点。我没有写这个代码,只是在网上找到它。所以当命令开始执行时,我得到这个“catch”错误,但不知道什么是问题。大多数歌曲不会显示,但在浏览器中访问链接可以正常工作:以下是导致错误的代码部分。Javascript捕获错误
if (quizState) {
//Load current song stats
console.log(newMedia.author + " " + newMedia.duration);
var XMLsource = 'http://musicbrainz.org/ws/2/artist/?query=artist:' + newMedia.author.replace(/ /g, "%20") + '&limit=1';
simpleAJAXLib = {
init: function() {
this.fetchJSON(XMLsource);
},
fetchJSON: function(url) {
var root = 'https://query.yahooapis.com/v1/public/yql?q=';
var yql = 'select * from xml where url="' + url + '"';
var proxy_url = root + encodeURIComponent(yql) + '&format=json&diagnostics=false&callback=simpleAJAXLib.display';
document.getElementsByTagName('body')[0].appendChild(this.jsTag(proxy_url));
},
jsTag: function(url) {
var script = document.createElement('script');
script.setAttribute('type', 'text/javascript');
script.setAttribute('src', url);
return script;
},
display: function(results) {
try {
quizCountry = results.query.results.metadata["artist-list"].artist.area.name;
quizYear = results.query.results.metadata["artist-list"].artist["life-span"].begin.match(/\d{4}/);
quizBand = results.query.results.metadata["artist-list"].artist.name;
if (quizCountry != "" && quizYear != "") {
console.log(quizCountry + " " + quizYear);
API.sendChat("U kojoj godini je/su " + quizBand + " osnovan/i?");
}
} catch (e) {
console.log("Error: " + e.description);
API.sendChat("Žao nam je, čini se da musicbrainz ne prepoznaje ovaj bend ili umjetnika. Nastavit ćemo za vrijeme sljedeće pjesme.");
console.log("country or year not known");
}
}
}
simpleAJAXLib.init();
}
什么是在随即出现在浏览器控制台中的错误? – ceejayoz
这是一些疯狂的大缩进 –
@ceejayoz它给这个catch错误,console.log(“国家或一年未知”);它不想执行 –