javascript - JSON 获取嵌套数组中的用户 ID

Question

如何获取嵌套数组 JSON 中的所有用户 ID，我使用以下代码来获取 JSON 响应:

var userListObj = JSON.stringify(result.items);
console.log(userListObj);

如有任何帮助，我们将不胜感激，提前谢谢您。

{  
   "current_page":1,
   "per_page":100,
   "total_entries":3,
   "items":[  
      {  
         "user":{  
            "id":12729498,
            "owner_id":51603,
            "full_name":"test1",
            "email":"test1",
            "login":"test1",
            "phone":null,
            "website":null,
            "created_at":"2016-05-16T14:56:24Z",
            "updated_at":"2016-05-17T18:23:35Z",
            "last_request_at":"2016-05-17T19:23:37Z",
            "external_user_id":null,
            "facebook_id":null,
            "twitter_id":null,
            "blob_id":null,
            "custom_data":null,
            "twitter_digits_id":null,
            "user_tags":null
         }
      },
      {  
         "user":{  
            "id":12768158,
            "owner_id":51603,
            "full_name":"test2",
            "email":"",
            "login":"test2",
            "phone":null,
            "website":null,
            "created_at":"2016-05-17T18:25:13Z",
            "updated_at":"2016-05-17T18:25:13Z",
            "last_request_at":null,
            "external_user_id":null,
            "facebook_id":null,
            "twitter_id":null,
            "blob_id":null,
            "custom_data":null,
            "twitter_digits_id":null,
            "user_tags":null
         }
      },
      {  
         "user":{  
            "id":12769692,
            "owner_id":51603,
            "full_name":"test3",
            "email":null,
            "login":"test3",
            "phone":null,
            "website":null,
            "created_at":"2016-05-17T19:22:55Z",
            "updated_at":"2016-05-17T19:22:55Z",
            "last_request_at":null,
            "external_user_id":null,
            "facebook_id":null,
            "twitter_id":null,
            "blob_id":null,
            "custom_data":null,
            "twitter_digits_id":null,
            "user_tags":null
         }
      }
   ]
}

Answer 1

如果您应用@Mike C 的评论，您会得到如下内容:

var formattedJson = JSON.parse(json);

    for (i=0;i<formattedJson.items.length;i++){
        console.log(formattedJson.items[i].user.id);
    }

formattedJson 取决于您获取的 JSON 是字符串还是有效的 JSON，如果是，则不需要使用 JSON.parse，只需访问它即可。