JavaScript / ES2015-保留對象的所有屬性，除一個白色的Rest / Spread運算符外

Question

我有這個對象：

const value = {
  "template": "<span translate>myRecommendations</span>",
  "badge": "<span>{{ notifications }}</span>",
  "sref": "recommendation.list"
}

我想使用Rest / Spread運算符來創建另一個具有相同名稱（ value ）的對象，並具有除badge屬性之外的所有屬性

Answer 1

因此，如果要在當前變量中保留另一個形狀，只需從對象中刪除該屬性。

 const value = { "template": "<span translate>myRecommendations</span>", "badge": "<span>{{ notifications }}</span>", "sref": "recommendation.list" }; delete value.badge; console.log(value); 

如果您需要另一個對象，則可以使用對象銷毀

 const value = { "template": "<span translate>myRecommendations</span>", "badge": "<span>{{ notifications }}</span>", "sref": "recommendation.list" }; const { badge, ...anotherValue } = value; console.log(anotherValue); 

或遍歷對象鍵並獲取所需的屬性

 const value = { "template": "<span translate>myRecommendations</span>", "badge": "<span>{{ notifications }}</span>", "sref": "recommendation.list" }; const anotherValue = Object.keys(value).reduce((obj, key) => { if(key !== 'badge') { obj[key] = value[key] ; } return obj; }, {}); console.log(anotherValue); 

Answer 2

您不能有兩個名稱相同的變量。 但是，如果您不想變異原始對象而仍然使用spread syntax ，則可以考慮以不同的方式命名新對象，例如value2 。

 const value = { "template": "<span translate>myRecommendations</span>", "badge": "<span>{{ notifications }}</span>", "sref": "recommendation.list" }; const { badge, ...value2 } = value; // ^^^^^^ prop you want NOT to include in your new object console.log(value2); 

Answer 3

干得好：

 const value = { "template": "<span translate>myRecommendations</span>", "badge": "<span>{{ notifications }}</span>", "sref": "recommendation.list" } const { badge, ...updatedValue } = value; console.log(updatedValue)