當對象值是使用jQuery的對象數組時，遍歷JSON

Question

這是我的外部JSON：

{"meta": {"limit": 20, "next": null, "offset": 0, "previous": null, "total_count": 3}, "objects": [{"body": "this is copy text", "id": "1", "pub_date": "2011-05-04T12:23:26", "resource_uri": "/api/v1/entry/1/", "slug": "test-title-number-one", "title": "test title number one", "user": "/api/v1/user/1/"}, {"body": "this is the second test text", "id": "2", "pub_date": "2011-05-04T15:01:16", "resource_uri": "/api/v1/entry/2/", "slug": "second-test", "title": "Second test", "user": "/api/v1/user/1/"}, {"body": "item three", "id": "3", "pub_date": "2011-05-05T12:04:04", "resource_uri": "/api/v1/entry/3/", "slug": "item-3", "title": "item 3", "user": "/api/v1/user/1/"}]}

這是我的JS：

$.ajax({url: "/api/v1/entry/?format=json", 
dataType: "json",
success: function(json) {
    $.each(json.objects[0], function(key, value) { 
      alert(key + ': ' + value); 
    });
}
});

我可以使用$ .each（json.objects [0] ...為數組中的對象建立索引，但是我需要能夠擊中數組中的每個對象，我不知道為什么簡單地$ .each（json。對象...不起作用，謝謝！

Answer 1

只需執行一個普通的JS循環即可：

for(var i = 0; i < json.objects.length; ++i)
{
   $.each(json.objects[i], function(key, value) { 
      alert(key + ': ' + value); 
    });
}

Answer 2

對於完整的jQuery解決方案，您可以執行以下操作：

$.each(json.objects, function(key, value) {
    $.each(json.objects[key], function(key, value){
        alert(key + ': ' + value);
    })
});

jsfiddle演示： http : //jsfiddle.net/LGC9X/當心，很多警報：P