Question

var color3 = {
    name: 'Black',
    type: 'special',
    rating: 1
};

var rateColor3 = (colorName, Nrating) => ({ ...colorName, Nrating });
console.log(color3.rating); ///// 1
console.log(rateColor3(color3, 6).rating);//////1
console.log(color3.rating);//////1

我想在保持color3不可变的同时使用对象散布运算符，但是我第二个console.log的输出是1而不是6，我在做什么错了？

Answer 1

您正在将6的值NRating分配给对象，而不是现有的rating。

所以您的对象看起来像这样：

{
    name: 'Black',
    type: 'special',
    rating: 1,
    Nrating: 6
}

要覆盖现有的rating属性，您必须执行以下操作：

var rateColor3 = (colorName, Nrating) => ({ ...colorName, rating: Nrating });

或将参数Nrating更改为rating。

var color3 = {
    name: 'Black',
    type: 'special',
    rating: 1
};

var rateColor3 = (colorName, rating) => ({ ...colorName, rating });

console.log(rateColor3(color3, 6));

Answer 2

首先，您不是在改变原始的color3对象，而是返回一个新的对象。其次：

({ ...colorName, Nrating });

将返回带有第四个道具Nrating: 6的新对象，因为它是对象分配的简写。您只需要简单地将值分配给rating键即可。

var color3 = {
    name: 'Black',
    type: 'special',
    rating: 1
};

var rateColor3 = (colorName, Nrating) => ({ ...colorName, rating: Nrating });

console.log(rateColor3(color3, 6));

Answer 3

Nrating需要一个属性名称

var color3 = {
   name: 'Black',
   type: 'special',
   rating: 1
};

var rateColor3 = (colorName, Nrating) => ({ ...colorName, rating: Nrating });
console.log(color3.rating); ///// 1
console.log(rateColor3(color3, 6).rating);//////6
console.log(color3.rating);//////1

Answer 4

您已经创建了一个新属性Nrating。将您的rating更改为Nrating。

var color3 = {
    name: 'Black',
    type: 'special',
    rating: 1
};

var rateColor3 = (colorName, Nrating) => ({ ...colorName, Nrating });
console.log(color3.rating); ///// 1
console.log(rateColor3(color3, 6).Nrating);//////6
console.log(color3.rating);//////1

我的对象散布运算符用法有什么问题？

4 个答案: