我的最终目标是能够在分配默认值(如果它们不存在)的情况下,使用spread operator对象destructured分配新对象。
看起来我可能不喜欢这样。这是我的期望和尝试:
//Expected beginning object
const a1 = {
key1: "test1",
key2: "test2",
};
//Expected end result
const b1 = {
key1: "test1",
key2: "test2",
key3: {
nestedKey1: "nestedVal1",
},
};
//Expected beginning object
const a2 = {
key1: "test1",
key2: "test2",
key3: {
nestedKey1: "actualValue",
}
}
//Expected end result
const b2 = {
key1: "test1",
key2: "test2",
key3: {
nestedKey1: "actualValue",
},
};

代码段1:不分配default
值。
const a = {
key1: "test1",
key2: "test2",
}
const {...b} = {
key1,
key2,
key3: {
nestedKey1: nestedKey1 = "nestedVal1",
} = {},
key4 = 'someDefault'
} = a
console.log(b); // does not have key3, or key4
console.log(key4); //this exists as a const, key3 does not

Snippet 2:功能齐全,但如果需要多级嵌套,则可能会出现问题。
const a = {
key1: "test1",
key2: "test2",
}
let {
key1,
key2,
key3: {
nestedKey1: nestedKey1 = "nestedVal1",
} = {},
} = a
const key3 = a.key3 || {}; // is it possible to include this in the destructuring? Specifically in the destructuring which defines the default.
const b = {
key1,
key2,
key3: {
...key3,
nestedKey1,
}
}
console.log(b);

小部件2(显示嵌套对象未被覆盖)
const a = {
key1: "test1",
key2: "test2",
key3: {
someOtherKey: "itWorks",
}
}
let {
key1,
key2,
key3: {
nestedKey1: nestedKey1 = "nestedVal1",
} = {},
} = a
const key3 = a.key3 || {};
const b2 = {
key1,
key2,
key3: {
...key3,
nestedKey1,
}
}
console.log(b2);

答案 0 :(得分:1)
您的困惑来自解构语法与对象声明语法的相似性。
这是对象声明:
const a = {
key1: "test1",
key2: "test2",
}
这是破坏性的
const { key1, key2 } = a
console.log(key1) // => test1
console.log(key2) // => test2
接下来,您需要记住赋值运算符=
在JavaScript中是正确的关联。像
let a = b = 1
是指将值1
分配给b
,然后分配给a
。
接下来,如果将所有值都分散到一个var中,则实际上是在使用花哨的ES6语法进行简单分配。所以:
const {...b} = a
// is same as
const b = a
结合以上3种效果,您得到的实际上是多次分配:
const {...b} = a
const {key1, key2} = a
多次分配可能会造成混淆,并且存在禁止使用此类规则的规则:https://eslint.org/docs/rules/no-multi-assign
所以您的问题的答案很简单,不! :)
这样做至少需要两行-一个用于在销毁时收集具有默认值的道具,另一个用于使用这些道具创建下一个对象。