Question

感谢this post，其中接受的答案包括JS小提琴，我能够验证我可以使用jQuery.grep()过滤我的数据。但是，我无法使用服务器的输出复制效果。我怀疑我在对象引用中忽略了一些明显的东西，但对JavaScript不熟悉我认为我需要有人为我指出它。

这是我的数据输出，一个对象数组：

[{"id":"3","src":"url1","category":"A"},
{"id":"32","src":"url2","category":"D"},
{"id":"38","src":"url3","category":"E"},
{"id":"39","src":"url4","category":"E"},
{"id":"42","src":"url5","category":"F"},
{"id":"49","src":"url6","category":"B"},
{"id":"44","src":"url7","category":"F"}]

我使用以下代码过滤F类中的条目：

＆＃13;

var obj = [{"id":"3","src":"url1","category":"A"},
{"id":"32","src":"url2","category":"D"},
{"id":"38","src":"url3","category":"E"},
{"id":"39","src":"url4","category":"E"},
{"id":"42","src":"url5","category":"F"},
{"id":"49","src":"url6","category":"B"},
{"id":"44","src":"url7","category":"F"}];

obj = $.grep(obj, function(element, index){
  return element.category == "F" // keep elements in category F
});

console.log(obj);

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>

＆＃13;

这是我对服务器的Ajax调用：

var obj = $.post( url, {term: term }, function(data){
}, 'json'
);

执行console.log(obj);时，我在控制台中将我的数据输出视为对象，但似乎无法实现。要使grep函数起作用，我需要以对象数组的形式访问输出。它在那里，等着我，但我该如何参考呢？

Answer 1

当你这样做时：

var obj = $.post( url, {term: term }, function(data){
}, 'json');

obj是承诺而非响应数据

您需要访问成功回调

中的响应数据

$.post( url, {term: term }, function(data){
    var obj = $.grep(data, function(element, index){
        return element.category == "F" // keep elements in category F
    });
    console.log(obj);    
}, 'json' );

引用JSON对象以使用jQuery.grep（）进行过滤

1 个答案: