1
我正在尝试使用我的请求url中的&callback= parameter
返回一个jsonp对象的方式调用BandsInTown api。我很困惑这是如何与Jquery调用结合使用的。我的代码看起来是这样,但并不做任何事情:如何使用api的jsonp回调函数
var shows=[];
var locat = "Denver,CO";
var artists = ["Twenty One Pilots", "Atmosphere", "Vince Staples", "STRFKR", "Rainbow Kitten Surprise", "Mac Demarco", "Hippo Campus", "Drake", "John Legend", "Rihanna", "Flying Lotus", "MGMT", "Jason Derulo", "M83", "Bon Iver", "Childish Gambino"];
for (var i = 0; i < artists.length; i++) {
parseEvent(artists[i], locat, shows);
}
function parseEvent(artist, locat, showsObject) {
var url = 'http://api.bandsintown.com/artists/' + artist + '/events/search.json?api_version=2.0&app_id=music_matcher&location=' + locat + '&radius=150&callback=bandsInInfo';
$.ajax({
url: url,
dataType: 'jsonp',
jsonpCallback: 'bandsInInfo'
});
}
function bandsInInfo(data) {
alert("workingBandy");
alert(data);
var numEvents = Object.keys(data).length;
for (var j = 0; j < numEvents; j++) {
if (!(data[j].venue.latitude == null && data[j].venue.latitude == 'undefined')) {
var element =
{
"location": {"latitude": data[j].venue.latitude, "longitude": data[j].venue.longitude},
"artist": data[j].artists[0].name,
"venue_name": data[j].venue.place,
"date": data[j].datetime,
"ticket_url": data[j].ticket_url,
"ticket_status": data[j].ticket_status,
"title": data[j].title
};
showsObject.push(element);
}
}
}
或者,如果有一种方法可以做到这一点没有jQuery的?这是api文档。除了响应被发送到您指定的回调函数这一事实之外,没有太多关于此的内容。 https://www.bandsintown.com/api/requests#artists-events预先感谢您的任何帮助或建议!