为什么数据仍然为空

时间:2010-09-15 01:26:49

标签: jquery json

我有这个json

{"suggestions":["M.I.A.","M.","Mindless Self Indulgence","The Cure","Telefon Tel Aviv","M","J. Ralph","Jason Mraz","Carbon Based Lifeforms","Cycle of Pain","Chantal Kreviazuk","-M-","ayumi hamasaki","R.E.M.","Donny McCaslin","Penfold","HEALTH","R. Kelly","DJ Khaled","Eminem","Spose","T.I.","The Lonely Island","H.I.M. (His Infernal Majesty)","Dropkick Murphys","Taylor Swift"],"query":"m"}

我从这个ajax电话中获取

$.getJSON('<%= ajax_path("artistName") %>', req, function(data) {
   //create array for response objects
   var suggestions = [];
   console.log(data);
   //process response
   $.each(data, function(i, val){                                
      suggestions.push(val.name);
   });
   console.log(suggestions);
   //pass array to callback
   add(suggestions);
});
},

为什么我的建议仍为空

1 个答案:

答案 0 :(得分:2)

您不必循环,suggestions已经是一个可供使用的数组,所以请替换它:

var suggestions = [];
console.log(data);
//process response
$.each(data, function(i, val){                                
    suggestions.push(val.name);
});
console.log(suggestions);

有了这个:

var suggestions = data.suggestions;
console.log(suggestions);

然后,传递给add()的数组将获得JSON响应的结果。