javascript - 在jquery中每个来自json对象的循环如何（使循环从对象中找到一个值并与另一个对象值进行比较以不重复打印）？

Question

大家好，我正在开发一个音乐播放器，我正在使用 Json 来存储我的 ID3 或 AKA“MP3 元数据”。我面临的问题是我正在提取一个对象和 traingf 以查看是否可以比较该值，并且如果两个对象具有相同的值，请不要重复打印。万一我只想让艺术家一次而不是两次无限重复

.

这是 JSON 文件示例；

[
{   
    "Artist": "",
    "Album": "",
    "Year":"",
    "Genre": "", 
    "Song": "",
    "Location": "",
    "Track": "",
    "Img": "",
    "Composer":""
},
    {   
    "Artist": "The Rolling Stones",
    "Album": "Hot Rocks (1964-1971)",
    "Year":"2002",
    "Genre": "Rock", 
    "Song": "Heart of Stone",
    "Location": "scr/music/TheRollingStones/HeartofStone.mp3",
    "Track": "2",
    "Img": "scr/music/TheRollingStones/img/HotRocks.jpg",
    "Record":"Sony Music"
},
    {   
    "Artist": "The Rolling Stones",
    "Album": "Hot Rocks (1964-1971)",
    "Year":"2002",
    "Genre": "Rock", 
    "Song": "Sympathy For The Devil",
    "Location": "scr/music/TheRollingStones/SympathyForTheDevil.mp3",
    "Track": "15",
    "Img": "scr/music/TheRollingStones/img/HotRocks.jpg",
    "Record":"Sony Music"
},

        {   
    "Artist": "Led Zeppelin",
    "Album": "The Complete Led Zeppelin",
    "Year":"2007",
    "Genre": "Rock", 
    "Song": "Good Times Bad Times",
    "Location": "scr/music/LedZeppelin/GoodTimesBadTimes.mp3",
    "Track": "1",
    "Img": "scr/music/LedZeppelin/img/The Complete Led Zeppelin.jpg",
    "Record":"Atlantic Records"
}

]

Javascript 文档

    $.getJSON('scr/json/musicdata.json', function(data){
        var output = '<ol>';
             data.sort(function(a, b){
             return [a.Artist] < [b.Artist] ?  0 : 1;
         });//End of Sort by Artist 

         $.each(data, function(key,val) {

         if(val.Artist != ""){

         if(PreArtis != val.Artist){

        output += '<li><a class="songname" href="#" data-src="'+ val.Location +'">' + val.Artist + '</a></li>';
            var PreSong = val.Artist ;
         }//End of PreSong != Song
        }//End of val.Artist     
        });//End of Each 
         output += '</ol>';
        $("#wrapper").append(output);
        });//End of getJSON 

html

感谢您的帮助

Answer 1

尝试

var songs = {};
$.each(data, function(key,val) {
    if(val.Artist != ""){
        var key = val.Artist + '@' + val.Album;
        if(!songs[key]){
            output += '<li><a class="songname" href="#" data-src="'+ val.Location +'">' + val.Artist + '</a></li>';
            songs[key] = true ;
        }//End of PreSong != Song
    }//End of val.Artist     
});//End of Each 
output += '</ol>';
$("#wrapper").append(output);

演示：小提琴

您可以更改key以找到唯一的组合。例如：如果您想要艺术家、专辑和歌曲的独特组合，请使用var key = val.Artist + '@' + val.Album + val.Song;