ES6对象破坏默认参数 [英] ES6 Object Destructuring Default Parameters

问题描述

我想弄清楚是否有办法使用默认参数的对象解构，而不用担心对象被部分定义。请考虑以下内容：

 （function test（{a，b} = { a：foo，b：bar}）{console.log（a ++ b）;}）（）;  

/ div>

例如，当我以 {a：qux} code> qux undefined 在控制台中，我真正想要的是 qux bar 。有没有手段来检查所有对象的属性？

解决方案

是的。您可以在解构中使用defaults：

 （function test {a =foo，b =bar} = {}）{console.log（a ++ b）;}）（）;  

/ div>

这不限于函数参数，而是适用于每个解构表达式。

I'm trying to figure out if there's a way to use object destructuring of default parameters without worrying about the object being partially defined. Consider the following:

(function test({a, b} = {a: "foo", b: "bar"}) {
  console.log(a + " " + b);
})();

When I call this with {a: "qux"}, for instance, I see qux undefined in the console when what I really want is qux bar. Is there a way to achieve this without manually checking all the object's properties?

解决方案

Yes. You can use "defaults" in destructuring as well:

(function test({a = "foo", b = "bar"} = {}) {
  console.log(a + " " + b);
})();

This is not restricted to function parameters, but works in every destructuring expression.

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