Javascript JSON文件只显示最后一个条目,尽管ID不同

Javascript JSON文件只显示最后一个条目,尽管ID不同,javascript,json,Javascript,Json,我的json文件如下所示: var obj = { Dilse : {"name":"Dil Se Re","length":"5:40","songType":"Romantic"}, Dilse : {"name":"Jiya Jale","length":"3:50","songType":"Self Description"}, Dilse : {"name":"Chaiyya Chaiyya","length":"4:30","songType":"Masti"}, Rock

我的json文件如下所示:

var obj = {
 Dilse : {"name":"Dil Se Re","length":"5:40","songType":"Romantic"},
 Dilse : {"name":"Jiya Jale","length":"3:50","songType":"Self Description"},
 Dilse : {"name":"Chaiyya Chaiyya","length":"4:30","songType":"Masti"},

 Rockstar : {"name":"Sadda haq","length":"5:40","songType":"Romantic"},
 Rockstar : {"name":"Tum ho Pas Mere","length":"3:50","songType":"Self Description"},
 Rockstar : {"name":"Sheher me","length":"4:30","songType":"Masti"},
}
我想在单击Dil se时检索Dilse条目,在单击Rockstar时检索Rockstar条目。我的页面上已经有了Dilse和Rockstar的条目

JS函数如下所示:

function dilSe(){

            document.getElementById("div0").innerHTML = ""; var k =0;
            if(xhr.readyState==4){
            if(xhr.status==200){
                var data=xhr.responseText;
                var arr=eval(data);
                for(var i=0;i<3;i++){             //I dont know if I should itearte like this !?

                            var link=document.createElement('a');
                            link.id='name1'+k++;
                            link.innerHTML="Song Name: " +obj.Dilse.name+"<br/>"+"Song Length: "+obj.Dilse.length+"<br/>"+"Song Type: "+obj.Dilse.songType+"<br/><br/>";
                            console.log("link.id is "+link.id);
                            link.setAttribute('href','#');

                            link.addEventListener('click',albums[link.id]);

                            div0.appendChild(link);
                            } 
                }else{
                    div0.innerHTML="Oops, error in communication!";
                    div0.style.color="red";
                }
            }
}


function rockStar(){

            document.getElementById("div0").innerHTML = ""; var k =0; var r=3;
            if(xhr.readyState==4){
            if(xhr.status==200){
                var data=xhr.responseText;
                var arr=eval(data);
                for(var i=0;i<3;i++){       //I dont know if I should itearte like this !?

                            var link=document.createElement('a');
                            link.id='name2'+k++;
                            link.innerHTML="Song Name: " +obj.Rockstar.name+"<br/>"+"Song Length: "+obj.Rockstar.length+"<br/>"+"Song Type: "+obj.Rockstar.songType+"<br/><br/>";
                            console.log("link.id is "+link.id);
                            link.setAttribute('href','#');
                            console.log("obj.rockstar.name is "+obj.Rockstar.name);
                            link.addEventListener('click',albums[link.id]);

                            div0.appendChild(link);
                            } 
                }else{
                    div0.innerHTML="Oops, error in communication!";
                    div0.style.color="red";
                }
            }
}
函数dilSe(){
document.getElementById(“div0”).innerHTML=“”;var k=0;
if(xhr.readyState==4){
如果(xhr.status==200){
var data=xhr.responseText;
var arr=评估(数据);

对于(var i=0;i您的JSON对象是不正确的。显然,您不能有多个具有相同密钥的条目(您所做的是告诉JS:“詹姆斯就是这个家伙。啊,不,詹姆斯就是另一个家伙。啊,不,詹姆斯又是第三个家伙”。当然,他只记得最后一个

您需要使用数组。最简单的解决方案是将JSON修改为如下所示:

var obj = {
 Dilse :
 [
    {"name":"Dil Se Re","length":"5:40","songType":"Romantic"},
    {"name":"Jiya Jale","length":"3:50","songType":"Self Description"},
    {"name":"Chaiyya Chaiyya","length":"4:30","songType":"Masti"},
 ],

 Rockstar :
 [
    {"name":"Sadda haq","length":"5:40","songType":"Romantic"},
    {"name":"Tum ho Pas Mere","length":"3:50","songType":"Self Description"},
    {"name":"Sheher me","length":"4:30","songType":"Masti"},
 ],
}

还有其他可能的选项,例如单个数组(非关联),以前的键是每个对象中的一个值,但这需要您遍历整个数组以找到所需的内容。

谢谢您的回答。如果我必须显示Dilse或Rockstar的所有三个条目,您能告诉我如何在函数中迭代吗?我如何在函数中使用obj.Dilse.name/length/songType来显示g设置所有3个条目?
var list=obj.Dilse;for(var i=0;i