如果存在键，则使用键和值形成JSON

Question

我有一个JSON结构，类似于：

    [  
   {  
      "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，我想确定一种菜肴是否与餐馆联系在一起。如果是，我想构建一个JSON结构，类似于：

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

已经尝试使用for循环和.hasProperty构建JSON，但没有成功。

Answer 1

您可以在下面的一个循环中使用。

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);
    }
  }
});

JSFIDDLE

Answer 2

这就是你能做的！首先遍历数据，并使用"hasOwnProperty“方法来检查菜系是否存在，以及它是否存在，然后检查您的菜系对象是否有该菜系，然后再将其添加到其中。

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 3

这是数组的一个简单的缩减。如果餐厅有明确的菜肴，检查结果是否已经定义了这种菜肴。如果不是，为它创建一个对象，您可以在其中推送餐厅名称。

​

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 );

​

在我个人看来，我会使用稍微不同的结构。如果我们用始终相同的对象表示集合，则可以简化大多数转换。这比在一个循环中完成所有操作效率低，但是用于创建转换的代码几乎是可读的英语：

​

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 );

​