子を置換せずにJavascriptオブジェクトに追加

Question

forループを実行して新しいjavascriptオブジェクトを作成し、置換せずに既存のjavascriptオブジェクトに追加しようとしています。

例えば、私は（[i]のデータだけ番号を追加する）グループなしで出力としてこれを作成しようとしている[i]をするたびに上書きされています：

{ 
    "group1": { 
     "number1": "data", 
     "number2": "data", 
     "number3": "data" 
    }, 
    "group2": { 
     "number1": "data", 
     "number2": "data" 
    } 
} 

私のループは次のようになります。

var array = [{ 
    "groupName": "group1", 
    "number": "number1", 
    "data": "data" 
}, { 
    "groupName": "group1", 
    "number": "number2", 
    "data": "data" 
}, { 
    "groupName": "group1", 
    "number": "number3", 
    "data": "data" 
}, { 
    "groupName": "group2", 
    "number": "number1", 
    "data": "data" 
}, { 
    "groupName": "group2", 
    "number": "number2", 
    "data": "data" 
}] 

var groupLinks = {}; 

for(i = 0; i < array.length; i++) { 

    groupLinks[array[i].groupName] = { 
     [array[i].number]: array[i].data 

} 

このコードでわかるように、私のループは毎回 "group1"と "group2"を置き換えるだけです。

Answer 1

これは私がこれに近づいて行くだろうかです：

ここ

var array = [{ 
 
    "groupName": "group1", 
 
    "number": "number1", 
 
    "data": "data" 
 
}, { 
 
    "groupName": "group1", 
 
    "number": "number2", 
 
    "data": "data" 
 
}, { 
 
    "groupName": "group1", 
 
    "number": "number3", 
 
    "data": "data" 
 
}, { 
 
    "groupName": "group2", 
 
    "number": "number1", 
 
    "data": "data" 
 
}, { 
 
    "groupName": "group2", 
 
    "number": "number2", 
 
    "data": "data" 
 
}]; 
 

 
var groupLinks = {}; 
 

 
array.forEach(
 
    function (element) 
 
    { 
 
     var groupName = element.groupName; 
 
     (groupLinks[groupName] = groupLinks[groupName] || {}) 
 
      [element.number] = element.data; 
 

 
    } 
 
); 
 

 
console.log(groupLinks);

私は各繰り返しで、既存のgroupLinks[groupName]を上書きし、何も存在しない場合、新しいグループを作成するには||演算子を使用配列要素をループするにはforEach構造を使用しました。

Answer 2

これを試してみてください：

var array = [{ 
     "groupName": "group1", 
     "number": "number1", 
     "data": "data" 
    }, { 
     "groupName": "group1", 
     "number": "number2", 
     "data": "data" 
    }, { 
     "groupName": "group1", 
     "number": "number3", 
     "data": "data" 
    }, { 
     "groupName": "group2", 
     "number": "number1", 
     "data": "data" 
    }, { 
     "groupName": "group2", 
     "number": "number2", 
     "data": "data" 
    }] 

    var groupLinks = {}; 

    array.forEach(function(el){ 
     if(groupLinks[el.groupName]){ 
     groupLinks[el.groupName][el.number] = el.data 
     } 
     else{ 
     groupLinks[el.groupName] = {}; 
     groupLinks[el.groupName][el.number] = el.data 
     } 

    }); 

    console.log(groupLinks); 

Answer 3

をあなたは少しES2015を恐れていないなら、これはあなたが

let array = [{"groupName":"group1","number":"number1","data":"data"},{"groupName":"group1","number":"number2","data":"data"},{"groupName":"group1","number":"number3","data":"data"},{"groupName":"group2","number":"number1","data":"data"},{"groupName":"group2","number":"number2","data":"data"}]; 
 

 
let groupLinks = array.reduce((gl, group) => { 
 
    gl[group.groupName] = Object.assign({ 
 
     [group.number]: group.data 
 
    }, gl[group.groupName]); 
 
    return gl; 
 
}, Object.create(null)); 
 

 
console.log(groupLinks);

Answer 4

これを試してみてくださいために働く必要があります仕事を減らす：

を

var groupLinks = {}; 

    for(i = 0; i < array.length; i++) { 
     var groupLinks2 = groupLinks[array[i].groupName]; 
     if(!groupLinks2){ 
      groupLinks2 ={}; 
      var a = array[i].number; 
      groupLinks2[a] = array[i].data; 
     }else{ 
     var a = array[i].number; 
     groupLinks2[a] = array[i].data; 
     } 
     groupLinks[array[i].groupName] = groupLinks2; 

    }