如果鍵存在，則使用鍵和值形成JSON

Question

我有一個類似的JSON結構：

    [  
   {  
      "name":"angelinas"
   },
   {  
      "name":"besuto"
   },
   {  
      "name":"catch",
      "cuisine":"Japanese"
   },
   {  
      "name":"center cut"
   },
   {  
      "name":"fedora"
   },
   {  
      "name":"Habanero",
      "cuisine":"Mexican"
   },
   {  
      "name":"Indies"
   },
   {  
      "name":"new"
   },
   {  
      "name":"RazINN"
   },
   {  
      "name":"restaurantTestVenue779"
   },
   {  
      "name":"restaurantTestVenue9703"
   },
   {  
      "name":"Salsa ",
      "cuisine":"Mexican"
   },
   {  
      "name":"Sushi Place",
      "cuisine":"Japanese"
   },
   {  
      "name":"The Ashoka"
   },
   {  
      "name":"The Poboys"
   },
   {  
      "name":"the shogun"
   },
   {  
      "name":"vinyard view"
   }
]

使用上面的JSON，我想確定是否將美食與餐廳相關聯。 如果是，我想構建一個JSON結構，例如：

[  
   {  
      "Mexican":{  
         "venueNames":[  
            "Habanero",
            "Salsa"
         ]
      }
   },
   {  
      "Japanese":{  
         "venueNames":[  
            "Sushi Place",
            "catch"
         ]
      }
   }
]

嘗試使用for循環和.hasProperty構建JSON，但沒有成功。

Answer 1

這是您可以做的！ 首先遍歷數據，然后使用方法“ hasOwnProperty”檢查該美食是否存在，是否存在，然后檢查您的美食對象是否具有該美食，然后將其添加到其中。

const data = [{
        "name": "angelinas"
    },
    {
        "name": "besuto"
    },
    {
        "name": "catch",
        "cuisine": "Japanese"
    },
    {
        "name": "center cut"
    },
    {
        "name": "fedora"
    },
    {
        "name": "Habanero",
        "cuisine": "Mexican"
    },
    {
        "name": "Indies"
    },
    {
        "name": "new"
    },
    {
        "name": "RazINN"
    },
    {
        "name": "restaurantTestVenue779"
    },
    {
        "name": "restaurantTestVenue9703"
    },
    {
        "name": "Salsa ",
        "cuisine": "Mexican"
    },
    {
        "name": "Sushi Place",
        "cuisine": "Japanese"
    },
    {
        "name": "The Ashoka"
    },
    {
        "name": "The Poboys"
    },
    {
        "name": "the shogun"
    },
    {
        "name": "vinyard view"
    }
]

let cuisines = {};


for (const resturant of data) {
    if (resturant.hasOwnProperty('cuisine')) {

        if (cuisines.hasOwnProperty(resturant.cuisine)) {
            cuisines[resturant.cuisine].venueNames.push(resturant.name);
        } else {
            cuisines[resturant.cuisine] = {
                venueNames: [resturant.name]
            };
        }
    }
}

Answer 2

這是對數組的簡單簡化。 如果餐廳有已定義的美食，請檢查結果是否已定義了該美食。 如果沒有，請為其創建一個對象，您可以將餐廳名稱推送到該對象。

 const restaurants = [ { "name":"angelinas" }, { "name":"besuto" }, { "name":"catch", "cuisine":"Japanese" }, { "name":"center cut" }, { "name":"fedora" }, { "name":"Habanero", "cuisine":"Mexican" }, { "name":"Indies" }, { "name":"new" }, { "name":"RazINN" }, { "name":"restaurantTestVenue779" }, { "name":"restaurantTestVenue9703" }, { "name":"Salsa ", "cuisine":"Mexican" }, { "name":"Sushi Place", "cuisine":"Japanese" }, { "name":"The Ashoka" }, { "name":"The Poboys" }, { "name":"the shogun" }, { "name":"vinyard view" } ]; const cuisines = restaurants.reduce((result, restaurant ) => { if ( restaurant.hasOwnProperty( 'cuisine' )) { const { cuisine } = restaurant; if ( !result.hasOwnProperty( cuisine )) { result[ cuisine ] = { venueNames: [] }; } result[ cuisine ].venueNames.push( restaurant.name ); } return result; }, {}); console.log( cuisines ); 

我個人認為，我會使用略有不同的結構。 如果我們用始終相同的對象表示集合，則可以簡化大多數轉換。 這比在一個循環中完成所有操作效率低，但是用於創建轉換的代碼幾乎是可讀的英語：

 const restaurants = [ { "name": "angelinas", "cuisine": null }, { "name": "besuto", "cuisine": null }, { "name": "catch", "cuisine": "japanese" }, { "name": "center cut", "cuisine": null }, { "name": "fedora", "cuisine": null }, { "name": "habanero", "cuisine": "mexican" }, { "name": "Indies", "cuisine": null }, { "name": "new", "cuisine": null }, { "name": "RazINN", "cuisine": null }, { "name": "restaurantTestVenue779", "cuisine": null }, { "name": "restaurantTestVenue9703", "cuisine": null }, { "name": "Salsa ", "cuisine": "mexican" }, { "name": "Sushi Place", "cuisine": "japanese" }, { "name": "The Ashoka", "cuisine": null }, { "name": "The Poboys", "cuisine": null }, { "name": "the shogun", "cuisine": null }, { "name": "vinyard view", "cuisine": null } ]; const create_cuisine = name => ({ name, "venues": [] }); const unique = () => { const seen = {}; return item => { const json = JSON.stringify( item ); return seen.hasOwnProperty( json ) ? false : ( seen[ json ] = true ); }; }; // Filter away all the restaurants without a cuisine value. const restaurants_with_cuisine = restaurants.filter( restaurant => restaurant.cuisine ); const cuisines = restaurants_with_cuisine // Extract the cuisine anmes from the restaurants. .map( restaurant => restaurant.cuisine ) // Filter aways all the duplicates. .filter( unique() ) // Create a new cuisine object. .map( cuisine_name => create_cuisine( cuisine_name )); // Finally add all the restaurant names to the right cuisine. restaurants_with_cuisine.forEach( restaurant => cuisines.find( cuisine => cuisine.name === restaurant.cuisine ).venues.push( restaurant.name )); console.log( cuisines ); 

Answer 3

您可以在下面的一個循環中使用。

data.forEach(function(item) {
    // if item has cuisine and cuisine not exist in new array
    if(item["cuisine"] != null && typeof newArr.find(v => v[item.cuisine] != null) == 'undefined') {
    // create new object with structure
    let obj = {};
    obj[item.cuisine] = {
         "venueNames":[item.name]
      };

    newArr.push(obj);
  }
  else {
    // else find existing cuisine and add new venue
    let obj = newArr.find(v => v.hasOwnProperty(item.cuisine));
    if(typeof obj != 'undefined') {
        obj[item.cuisine].venueNames.push(item.name);
    }
  }
});

的jsfiddle

Answer 4

使用一些es6功能，我們可以使用Set ， map和filter生成此列表。

我們將首先映射美食列表，並刪除無效的美食，例如undefined 。 這樣，我們將使用Set來創建獨特的美食列表。

接下來，我們將通過過濾與烹飪與當前迭代匹配的原始對象，獲取該列表並將其再次映射以返回最終對象。 最后，我們將過濾后的結果映射為僅將名稱返回給venueNames對象。

我們的結果將如下所示：

 function getItems(places) { // Get a unique list of cuisines return [...new Set(places.map(p => p.cuisine).filter(c => c))] // Build the result .map(c => { return { [c]: { // Get a list of cuisines that match the current cuisine venueNames: places.filter(p => p.cuisine == c).map(c => c.name) } } }) } const places = [ {"name": "angelinas"}, {"name": "besuto"}, {"name": "catch","cuisine": "Japanese"}, {"name": "center cut"}, {"name": "fedora"}, {"name": "Habanero","cuisine": "Mexican"}, {"name": "Indies"}, {"name": "new"}, {"name": "RazINN"}, {"name": "restaurantTestVenue779"}, {"name": "restaurantTestVenue9703"}, {"name": "Salsa ","cuisine": "Mexican"}, {"name": "Sushi Place","cuisine": "Japanese"}, {"name": "The Ashoka"}, {"name": "The Poboys"}, {"name": "the shogun"}, {"name": "vinyard view"} ] console.log(getItems(places))