Получить ключ верхнего уровня из значения во вложенном объекте JSON, используя JavaScript?

Question

Если мне известно значение Africola для ключа name во вложенном JSON ниже, как я могу получить соответствующий ему ключ верхнего уровня ' barID1 ' в JavaScript?

{
  "barID1": {
    "address": "4 East Terrace, Sydney NSW 2000",
    "appStoreURL": "http://itunes.apple.com/app/idXXXXXXXXX",
    "description": “description text”,
    "imgURLs": [ "Https:url1”,  "https:url2”, "https:url3” ],
    "lat": -34.810585,
    "lon": 138.616739,
    "name": "Africola",
    "phone": "(08) 8223 3885",
    "status": "active",
    "venueImgURL": "https:url”
  },
  "barID2": {
    "address": "138/140 Gouger St, Sydney NSW 2000",
    "appStoreURL": "http://itunes.apple.com/app/idXXXXXXXXX",
     "description": “description text”,
    "imgURLs": [ "Https:url1”,  "https:url2”, "https:url3” ],
    "lat": -34.848082,
    "lon": 138.599813,
    "name": "Disco Mexico Taqueria",
    "phone": "0416 855 108",
    "status": "active",
    "venueImgURL": "https:url”
  }
}

Answer 1

Одним из решений является использование Object.entries () в сочетании с Array.find () , как в следующем примере:

const input = {
  "barID1": {
    "address": "4 East Terrace, Sydney NSW 2000",
    "appStoreURL": "http://itunes.apple.com/app/idXXXXXXXXX",
    "description": "description text",
    "imgURLs": [ "Https:url1",  "https:url2", "https:url3" ],
    "lat": -34.810585,
    "lon": 138.616739,
    "name": "Africola",
    "phone": "(08) 8223 3885",
    "status": "active",
    "venueImgURL": "https:url"
  },
  "barID2": {
    "address": "138/140 Gouger St, Sydney NSW 2000",
    "appStoreURL": "http://itunes.apple.com/app/idXXXXXXXXX",
     "description": "description text",
    "imgURLs": [ "Https:url1",  "https:url2", "https:url3" ],
    "lat": -34.848082,
    "lon": 138.599813,
    "name": "Disco Mexico Taqueria",
    "phone": "0416 855 108",
    "status": "active",
    "venueImgURL": "https:url"
  }
};

let [key, val] = Object.entries(input).find(
    ([k, v]) => v.name === "Africola"
);

console.log("key is :", key);
console.log("value is :", val);

Answer 2

Я бы использовал для этого Object.entries() и Array.find().

const data = {
  "barID1": {
    "address": "4 East Terrace, Sydney NSW 2000",
    "appStoreURL": "http://itunes.apple.com/app/idXXXXXXXXX",
    "description": "description text",
    "imgURLs": [ "Https:url1",  "https:url2", "https:url3" ],
    "lat": -34.810585,
    "lon": 138.616739,
    "name": "Africola",
    "phone": "(08) 8223 3885",
    "status": "active",
    "venueImgURL": "https:url"
  },
  "barID2": {
    "address": "138/140 Gouger St, Sydney NSW 2000",
    "appStoreURL": "http://itunes.apple.com/app/idXXXXXXXXX",
     "description": "description text",
    "imgURLs": [ "Https:url1",  "https:url2", "https:url3" ],
    "lat": -34.848082,
    "lon": 138.599813,
    "name": "Disco Mexico Taqueria",
    "phone": "0416 855 108",
    "status": "active",
    "venueImgURL": "https:url"
  }
};

const [key, obj] = Object.entries(data).find(([key, obj]) => {
  return obj.name === 'Africola';
});

console.log(key, obj);

Fiddle: https://jsfiddle.net/zty6fgcp/

Answer 3

Здесь вы JSON были искажены недействительными символами.Используйте поиск по ключам

let obj = {
  "barID1": {
    "address": "4 East Terrace, Sydney NSW 2000",
    "appStoreURL": "http://itunes.apple.com/app/idXXXXXXXXX",
    "description": "description text",
    "imgURLs": [ "Https:url1",  "https:url2", "https:url3" ],
    "lat": -34.810585,
    "lon": 138.616739,
    "name": "Africola",
    "phone": "(08) 8223 3885",
    "status": "active",
    "venueImgURL": "https:url"
  },
  "barID2": {
    "address": "138/140 Gouger St, Sydney NSW 2000",
    "appStoreURL": "http://itunes.apple.com/app/idXXXXXXXXX",
     "description": "description text",
    "imgURLs": [ "Https:url1",  "https:url2", "https:url3" ],
    "lat": -34.848082,
    "lon": 138.599813,
    "name": "Disco Mexico Taqueria",
    "phone": "0416 855 108",
    "status": "active",
    "venueImgURL": "https:url"
  }
};

let result = Object.keys(obj).find(key => obj[key].name === "Africola");
console.log(result);

Как функция:

let obj = {
  "barID1": {
    "address": "4 East Terrace, Sydney NSW 2000",
    "appStoreURL": "http://itunes.apple.com/app/idXXXXXXXXX",
    "description": "description text",
    "imgURLs": [ "Https:url1",  "https:url2", "https:url3" ],
    "lat": -34.810585,
    "lon": 138.616739,
    "name": "Africola",
    "phone": "(08) 8223 3885",
    "status": "active",
    "venueImgURL": "https:url"
  },
  "barID2": {
    "address": "138/140 Gouger St, Sydney NSW 2000",
    "appStoreURL": "http://itunes.apple.com/app/idXXXXXXXXX",
     "description": "description text",
    "imgURLs": [ "Https:url1",  "https:url2", "https:url3" ],
    "lat": -34.848082,
    "lon": 138.599813,
    "name": "Disco Mexico Taqueria",
    "phone": "0416 855 108",
    "status": "active",
    "venueImgURL": "https:url"
  }
};

const findKeyByName = (obj, search) => Object.keys(obj).find(key => obj[key].name === search);

console.log(findKeyByName(obj, 'Africola'));
console.log(findKeyByName(obj,'Disco Mexico Taqueria'));