Question

我试图自己映射嵌套对象，而不使用像JoinJS这样的东西，这在某种程度上起作用，直到我想让JSON看起来很漂亮。

基于SQL查询，我的输出是：

[
  {
    "id": 1,
    "city_name": "city 1 - some city name",
    "city_category": "city 1 - some category",
    "city_description": "city 1 - some city description",
    "city_id": 1,
    "store_id": 1
  },
  {
    "id": 1,
    "city_name": "city 1 - some city name",
    "city_category": "city 1 - some category",
    "city_description": "city 1 - some city description",
    "city_id": 1,
    "store_id": 2
  },
  {
    "id": 1,
    "city_name": "city 1 - some city name",
    "city_category": "city 1 - some category",
    "city_description": "city 1 - some city description",
    "city_id": 1,
    "store_id": 3
  },
  {
    "id": 2,
    "city_name": "city 2 - some city name",
    "city_category": "city 2 - some category",
    "city_description": "city 2 - some city description",
    "city_id": 2,
    "store_id": 1
  },
  {
    "id": 2,
    "city_name": "city 2 - some city name",
    "city_category": "city 2 - some category",
    "city_description": "city 2 - some city description",
    "city_id": 2,
    "store_id": 2
  }
]

我试图让这个输出看起来像这样：

[
  {
    "city_id": 1,
    "city_name": "city 1 - some city name",
    "city_category": "city 1 - some category",
    "city_description": "city 1 - some city description",
    "store_ids": [1,2,3]
  }
  {
    "city_id": 2,
    "city_name": "city 2 - some city name",
    "city_category": "city 2 - some category",
    "city_description": "city 2 - some city description",
    "store_ids": [1,2]
  }
]

我的第一个猜测是，我需要使用id找到所有唯一的city_id或_.chain(result).uniqBy('city_id').value()。但后来我有点卡在从输出中删除id，最重要的是，如何创建一个具有每个城市所有商店数组的新密钥store_ids。有意义吗？

使用Lodash或只是简单的JS（没有像JoinJS这样的其他模块，除非它们对这样的事情有效），我将如何操纵它？

Answer 1

我会浏览结果，并将它们分组到我的新对象结果中。最后你将得到一个对象的对象，键是item.id，但是如果你想把它转换成数组，那很简单。

let resultsObject = {};

sqlQueryRows.map(function(item) {
   if (item.id in resultsObject) {
       resultsObject[item.id].store_ids.push(item.store_id);
   } else {
       resultsObject[item.id] = {
           city_id: item.city_id,
           city_name: item.city_name,
           city_category: item.city_category,
           city_description: item.city_description,
           store_ids: [item.store_id] 
       }
   }
});

// if you need an array
let resultsArray = Object.values(resultsObject);

Answer 2

这是我的lodash方法：

<!DOCTYPE html>
<html>
<head>
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <link rel="stylesheet" href="api/reponse-api-2.css">
    <link rel="stylesheet" href="css/index.css">
</head>
<body class="nomargin">
    <!--MOBILE VERSION-->
    <div class="size-small">
        <div class="row"><iframe frameBorder="0" src="html/header-small.html" class="col-12" height="100px"/></div>
        <div class="row">
            <div class="col-12 checkbox-mobile">
                <form>
                <input type="checkbox">Embed site in browser</input>
                </form>
            </div>
        </div>
    </div>

    <!--TABLET VERSION-->
    <div class="size-medium">
        tablet
    </div>

    <!--PC VERSION-->
    <div class="size-large">
        pc
    </div>
</body>

这个想法是：

获取独特的城市。
删除存储ID。
通过按唯一城市ID过滤原始数据来创建存储ID数组。

Answer 3

如果你只想删除某些字段，可以使用.map方法，但我不确定我是否完全理解这个问题。

var new_array = array_name.map((val) => {
    //do stuff to array
}

在节点中映射/组织嵌套对象

3 个答案: