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javascript - How to destructure option argument with all default values in ES6?

问题描述:

I use ES6 features with babel compiler.

I have a function which takes option object as an argument:

function myFunction({ option1 = true, option2 = 'whatever' }) {

console.log(option1, option2);

// do something...

}

When I call it, destructuring happens and everything works well.

I want to call it with default options most of the time, so I do:

myFunction({}); // true 'whatever'

but it looks little bit strange. It would much more cleaner just call:

myFunction(); // TypeError: Cannot read property 'option1' of undefined

Is it possible?

网友答案:

Yes, you just have to provide a default value for the complete argument:

function myFunction({option1 = true, option2 = 'whatever'} = {}) {
//                                                         ^^^^
    console.log(option1, option2);
    // do something...
}
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