浅拷贝对象在 ES6/ES7 中遗漏了一个或多个属性？

Question

我就是这样做的：

var props = { id: 1, name: 'test', children: [] }

//copy props but leave children out
var newProps = { ...props }
delete newProps.children

console.log(newProps) // { id: 1, name: 'test' }

有没有更干净、更简单的方法？

Answer 1

var props = { id: 1, name: 'test', children: [] }

function clone(orig, blacklistedProps) {
    var newProps = {};
    Object.keys(props).forEach(function(key) {
        if (!blacklistedProps || blacklistedProps.indexOf(key) == -1) {
            newProps[key] = props[key];
        }
    });
    return newProps;
}
var newProps = clone(props, ['children']); 
console.log(newProps) // { id: 1, name: 'test' }
var newProps1 = clone(props); 
console.log(newProps1) // { id: 1, name: 'test', children:[] }

Answer 2

您可以使用 destructuring assignment:

var props = { id: 1, name: 'test', children: [] }

var {children:_, ...newProps} = props;
console.log(newProps) // { id: 1, name: 'test' }
console.log(_) // [] - as an "empty" placeholder

（与您已经使用的 rest/spread properties proposal for ES7 相同）