Question

我正在开发一个网络应用程序。与Last.fm API对话。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）。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"},
 ...
}

Answer 1

您需要转义请求中的网址，其中一些字符不合法或导致问题。变化

$.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) {

编码.getJSON（）API调用

1 个答案: