回答編集履歴
2
修正
answer
CHANGED
|
@@ -2,5 +2,8 @@
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|
|
2
2
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const obj3=Object.fromEntries(Object.entries(obj1).filter(x=>Object.keys(obj2).includes(x[0])));
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|
3
3
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もしくは
|
|
4
4
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const obj3=Object.keys(obj2).reduce((x,y)=>(x[y]=obj1[y],x),{});
|
|
5
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+
これでもいけます
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6
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+
const obj3=Object.keys(obj2).reduce((x,y)=>Object.defineProperty(x,y,{enumerable:true,value:obj1[y]}),{});
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5
7
|
|
|
8
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+
|
|
6
9
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```
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1
ちょうせい
answer
CHANGED
|
@@ -1,5 +1,6 @@
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|
|
1
1
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```javascript
|
|
2
2
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const obj3=Object.fromEntries(Object.entries(obj1).filter(x=>Object.keys(obj2).includes(x[0])));
|
|
3
3
|
もしくは
|
|
4
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-
const obj3=Object.keys(obj2).reduce((x,y)=>(x
|
|
4
|
+
const obj3=Object.keys(obj2).reduce((x,y)=>(x[y]=obj1[y],x),{});
|
|
5
|
+
|
|
5
6
|
```
|