如何从 Firebase 中删除 JSON 数据中的额外层？答案

【问题标题】：How to remove extra layer in JSON data from Firebase?如何从 Firebase 中删除 JSON 数据中的额外层？
【发布时间】：2014-02-16 03:29:31
【问题描述】：

我正在尝试使用 Underscore.js 从 Firebase 返回的 JSON 中获取一些数据，但是返回了很多空值。从 Firebase 返回的 JSON 目前如下所示：

{
 "-JFnc68gIRFohkWKBP05":
{
 "date":"2014-02-15T03:39:16.954Z",
 "description":"Thanks so much!",
 "name":"Rob",
 "role":"Give"
},
 "-JFncNSO4G_hNm0YySTA":
{
 "date":"2014-02-15T03:40:27.858Z",
 "description":"This is fun!",
 "name":"Cobie",
 "role":"Received"
},
 "-JFrhlpgCWxJnDETM1gg":
{
 "date":"2014-02-15T22:42:31.013Z",
 "description":"Brought over some really delicious cookies",
 "name":"John Smith",
 "role":"Gave"
},
 "-JFrjHlV-fsOVHyTXHZJ":
{
 "date":"2014-02-15T22:49:08.448Z",
 "description":"Charles was wonderfully patient.",
 "name":"Charles Darwin",
 "role":"Received"
 },
 "-JFsWZPbL6_j-9nQwP29":
{
 "date":"2014-02-16T02:28:47.950Z",
 "description":"On the Origin of Species... yaddi daddi da....",
 "name":"Charles Darwin",
 "role":"Gave"
},
"-JFsWdH61Y-I01-rqn_n":
{
 "date":"2014-02-16T02:29:07.887Z",
 "description":"Let me off to do my computer work without bugging me.",
 "name":"Cobie",
 "role":"Gave"
}
}

当我使用 Underscore.js 提取数据时，这些是它的输出示例：

javascript：

  $scope.allNames = _.pluck($scope.data, 'name');

输出：

["","","","","","","","","","","","Rob","Cobie","John Smith", “查尔斯·达尔文”、“查尔斯·达尔文”、“科比”]

javascript：

$scope.something = _.groupBy(userRef, 'userRef.name');

输出：

{"未定义":[null,null,null,null,null,null,null,null,null,null,null,{"date":"2014-02-15T03:39:16.954Z"," description":"非常感谢！","name":"Rob","rating":5,"role":"Give"},{"date":"2014-02-15T03:40:27.858Z" ,"description":"这很有趣！","name":"Cobie","rating":4,"role":"Received"},{"date":"2014-02-15T22:42:31.013 Z","description":"带来了一些非常美味的饼干","name":"John Smith","rating":3,"role":"Gave"},{"date":"2014-02- 15T22:49:08.448Z","description":"Charles 非常有耐心。","name":"Charles Darwin","rating":"2","role":"Received"},{"date" :"2014-02-16T02:28:47.950Z","description":"关于物种起源... yaddi daddi da....","name":"Charles Darwin","role":" Gave"},{"date":"2014-02-16T02:29:07.887Z","description":"让我去做我的电脑工作，不要打扰我。","name":"Cobie","角色":"给予"}]}

如何深入了解 JSON 而不返回空值？或者，是否有更好的方法从 JSON 中完全删除标头？

【问题讨论】：

您使用哪个版本的Underscore.js？
1.6.0.今天刚从官网抓了个js文件。
我用 chrome 控制台在underscorejs.org 中测试了你的代码，但我的输出是：["Rob","Cobie","John Smith","Charles Darwin","Charles Darwin","Cobie"]
抱歉，在显示的所有内容的顶部有一个附加层。更像是这样（安卓手机不让我退货）：SOMETHING { "-JFnc68gIRFohkWKBP05": { "date":"2014-02-15T03:39:16.954Z", "description":"非常感谢!", "name":"Rob", "role":"Give" }, "-JFncNSO4G_hNm0YySTA": { "date":"2014-02-15T03:40:27.858Z", "description":"这是有趣！”、“名称”：“Cobie”、“角色”：“收到”}、“-JFrhlpgCWxJnDETM1gg”：{“日期”：“2014-02-15T22：42：31.013Z”、“描述”：“带来在一些非常美味的饼干上”，“name”：“John Smith”，“role”：“Gave”}，...
回到电脑上。看起来是这样的：

标签： javascript json underscore.js firebase

【解决方案1】：

解决了首先使用下划线的 _.without 函数删除空字符串的问题。

$scope.allcontacts = _.without(messy_contacts, "")

【讨论】：

【解决方案2】：

您可以为此使用underscore compact。

$scope.allcontacts = _.compact(messy_contacts);

【讨论】：