Javascript中用于从多个索引中销毁项的快捷方式

Question

给定一个少于160项的数组，我想提取三个字段并复制它们；

const item1 = {name:'Item-1'};
const item2 = {name:'Item-1'};
const item1 = {name:'Item-1'};
...........
const item_N_minus_one = {name:'Item-N_minus_one'};
const item_N = {name:'Item-N'};

const itemsList = [{item1}, {item2} ..... upto {itemN}]

// Where n <= 160

下面是我的方法，它正在工作

const index_X = 23, index_Y = 45, index_Z= 56; // Can not exceed beyond 160 in my case

const item_XCopy = {...itemsList[index_X]};
const item_YCopy = {...itemsList[index_Y]};
const item_ZCopy = {...itemsList[index_Z]};

我想要的：寻找一个单行快捷方式解决方案，我可以在一个javascript语句中传递索引，并在另一个数组中返回字段(我知道我可以生成一个函数，但只是想知道是否有javascript快捷方式解决方案(

Answer 1

您可以获得项目，而不是副本，如下所示：

const itemsList = [{name:'Item-1'}, {name:'Item-2'}, {name:'Item-3'}];
const index_X = 0, index_Y = 1, index_Z= 2;
const { [index_X]: index_X_Item, [index_Y]: index_Y_Item, [index_Z]: index_Z_Item } 
= itemsList;
console.log(index_X_Item, index_Y_Item, index_Z_Item);

要获得副本，您可以使用Array#map:

const itemsList = [{name:'Item-1'}, {name:'Item-2'}, {name:'Item-3'}];
const index_X = 0, index_Y = 1, index_Z= 2;
const [index_X_Copy, index_Y_Copy, index_Z_Copy] = 
[index_X, index_Y, index_Z].map(index => ({ ...itemsList[index] }));
console.log(index_X_Copy, index_Y_Copy, index_Z_Copy);

Answer 2

您可以直接寻址并分配给新的变量名。

const
index_X = 23,
index_Y = 45,
index_Z = 56,
{
[index_X]: { ...item_XCopy },
[index_Y]: { ...item_YCopy },
[index_Z]: { ...item_ZCopy }
} = itemsList;