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

I think you're looking for:

const {corn, peas} = vegetableColors;

Live on Babel's REPL

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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.

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 :-)}

I wouldn't recommend it, but you can use eval() to accomplish something similar:

vegetableColors = {corn: 'yellow', peas: 'green'};

function test() {
    for ( let i=0; i < Object.keys(vegetableColors).length; i++ ) {
        let k = Object.keys(vegetableColors)[i];
        eval(`var ${k} = vegetableColors['${k}']`);
    }

    console.log(corn); // yellow
}

test();

console.log(corn); // undefined (out of scope)