如何在隐式返回中使用扩展运算符,在隐式返回中,我将从对象的数组中移除一个键值。我可以使用显式返回,但我需要短代码和其他可能的解决方案。
let array = [
{"sales":2341,"targetMet":false,"advertisment":true},
{"sales":981,"advertisment":true},
{"sales":3423,"targetMet":true,"advertisment":false},
{..},
{..}
];
let expectedArray = array.map(({targetMet,...rest}) => {...rest});
console.log(expectedArray) // should remove all targetMet keys
发布于 2019-04-15 02:31:10
只需返回rest
即可,无需再次传播。
注意:在()
中从箭头函数中返回一个对象。但在这里你不需要它。
let array = [
{"sales":2341,"targetMet":false,"advertisment":true},
{"sales":981,"advertisment":true},
{"sales":3423,"targetMet":true,"advertisment":false},
];
let expectedArray = array.map(({targetMet,...rest}) => rest);
console.log(expectedArray)
https://stackoverflow.com/questions/55678603
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