How do I destructure all properties into the current scope/closure in ES2015?

风流意气都作罢 提交于 2019-11-26 15:32:02
Bergi

I think you're looking for the with statement, it does exactly what you are asking for:

const vegetableColors = {corn: 'yellow', peas: 'green'};
with (vegetableColors) {
    console.log(corn);// yellow
    console.log(peas);// green
}

However, it is deprecated (in strict mode, which includes ES6 modules), for good reason.

destructure all properties into the current scope

You cannot in ES61. And that's a good thing. Be explicit about the variables you're introducing:

const {corn, peas} = vegetableColors;

Alternatively, you can extend the global object with Object.assign(global, vegetableColors) to put them in the global scope, but really, that's worse than a with statement.

1: … and while I don't know whether there is a draft to allow such things in ES7, I can tell you that any proposal will be nuked by the TC :-)

T.J. Crowder

I think you're looking for:

const {corn, peas} = vegetableColors;

Live on Babel's REPL


If Pointy's right that you're asking how to do this without knowing the names corn and peas, you can't with destructuring assignment.

You can at global scope only, using a loop, but I'm sure you don't want to do this at global scope. Still, just in case:

// I'm sure you don't really want this, just being thorough
Object.keys(vegetableColors).forEach((key) => {
    Object.defineProperty(this, key, {
        value: vegetableColors[key]
    });
});

(Throw enumerable: true on there if you want these pseudo-constants to be enumerable.)

That works at global scope because this refers to the global object.

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