javascript-传播语法以删除

我不会删除一把钥匙.看看这个

console.log(state);

我得到{1：{这里是下一个对象}},
下一个

const { 1: deletedValue, ...newState } = state;
console.log(newState);
console.log(state);

我得到

{1: {here is next object}}
{1: {here is next object}}

删除无效.我不理解为什么

邀请您在评论中描述数据的外观如何更准确：

state: {1: {id: 1, content: {name: "xyz", surname: "dsd"}},
2: {id: 2, content: {name: "abc", surname: "dsq"}}
}

解决方法:

看起来像是babeljs问题.

数字作为destructuring assignment的属性的问题.

var object = { 1: 40, foo: 41, bar: 42, baz: 43 },
    { 1: y, foo: z, ...x } = object;
    //^
    
console.log(x);
console.log(y);
console.log(z);

以字符串数字作为目标属性,而不仅仅是数字.

var object = { 1: 40, foo: 41, bar: 42, baz: 43 },
    { '1': y, foo: z, ...x } = object;
    //^^^
    
console.log(x);
console.log(y);
console.log(z);

标签：spread-syntax,ecmascript-6,babeljs,javascript
来源： https://codeday.me/bug/20191025/1925067.html