JSON - 純粋なJSで値を取得する方法

Question

JSONオブジェクトから値を取得できません。

私がしようとしていた：

console.log(players[0].player.info.position) 

それはundefinedを返します。 何が間違っていますか？

バリデーターJSONが有効であると言ってください。

{"players": [ 
{ 
    "player": { 
    "info": { 
     "position": "aaa", 
     "shirtNum": 1, 
     "positionInfo": "aaa" 
    }, 
    "nationalTeam": { 
     "isoCode": "aaa", 
     "country": "aaa", 
     "demonym": "aaa" 
    }, 
    "age": "221", 
    "name": { 
     "first": "aaa", 
     "last": "aaa" 
    }, 
    "id": 111, 
    "currentTeam": { 
     "name": "aaa", 
     "teamType": "aaa", 
     "shortName": "aaa", 
     "id": 21 
    } 
    } 
}]} 

Answer 1

私がtheThingと呼ぶ変数のjson。 jsonにアクセスするには、theThingを使用する必要があります。

var theThing = JSON.parse('{"players":[{"player":{"info":{"position":"aaa","shirtNum":1,"positionInfo":"aaa"},"nationalTeam":{"isoCode":"aaa","country":"aaa","demonym":"aaa"},"age":"221","name":{"first":"aaa","last":"aaa"},"id":111,"currentTeam":{"name":"aaa","teamType":"aaa","shortName":"aaa","id":21}}}]}'); 
alert(theThing.players[0].player.info.position); 

Answer 2

JSON.parse()をお試しください： はここでJSONオブジェクトです。 JSON自体は文字列です。あなたのJSON文字列と

obj = JSON.parse('{"players":[{"player":{"info":{"position":"aaa","shirtNum":1,"positionInfo":"aaa"},"nationalTeam":{"isoCode":"aaa","country":"aaa","demonym":"aaa"},"age":"221","name":{"first":"aaa","last":"aaa"},"id":111,"currentTeam":{"name":"aaa","teamType":"aaa","shortName":"aaa","id":21}}}]}'); 
 

 
console.log(obj['players'][0].player.info.position);

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/JSON/parse

Answer 3

使用JSON.parse：あなたはおそらく持っている

obj = JSON.parse('{"players":[{"player":{"info":{"position":"aaa","shirtNum":1,"positionInfo":"aaa"},"nationalTeam":{"isoCode":"aaa","country":"aaa","demonym":"aaa"},"age":"221","name":{"first":"aaa","last":"aaa"},"id":111,"currentTeam":{"name":"aaa","teamType":"aaa","shortName":"aaa","id":21}}}]}'); 
 

 
console.log(obj['players'][0].player.info.position);