ES6 destructuring object, default value on assignment when not used as a function parameter?

Question

I've got:

const fs = require('fs');

const packageConfig = JSON.parse(fs.readFileSync('./package.json'));
const { jspm: { configFile: jspmConfigFile }} = packageConfig;

but packageConfig's value for configFile could be undefined. In that case, I'd like to default jspmConfigFile to 'config.js

Is it possible to do this without creating an extended packageConfig object?

I realize I could do something like:

const { jspm: { configFile: jspmConfigFile }} = _.extend({
    jspm: { config: 'config.js'}
}, packageConfig);

but that's pretty messy just to get a sensible default with destructuring.

What am I missing?

How about const jspmConfigFile = packageConfig.jspm.configFile || 'config.js';? — Alexey Lebedev
– Alexey Lebedev, Commented Oct 4, 2015 at 19:59
Well, I thought about this, but I was trying to learn more about destructuring + leverage the fact I can safely access nested properties without checking for the existence of parent properties. — Sean Anderson
– Sean Anderson, Commented Oct 4, 2015 at 20:05

Rob W · Accepted Answer · 2015-10-04 20:10:01Z

5

The syntax for default values in object destructuring is:

const { jspm: { configFile: jspmConfigFile = 'config.js' }} = packageConfig;

But this is not yet implemented in Firefox: https://bugzilla.mozilla.org/show_bug.cgi?id=932080

answered Oct 4, 2015 at 20:10

Rob W

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1 Comment

Rob W Over a year ago

No problem. And hi @SeanAnderson!

Benjamin Gruenbaum · Accepted Answer · 2015-10-04 20:12:52Z

2

I'm not really sure ES2015 destructuring actually buys you anything here. At the end of the day since you're only destructuring a single element. You're pretty much writing a fancy way to do:

const jspmConfigFile = packageConfig.jspm.configFile;

You can assign a default value with = 'defaultValue' in the innermost section but I'm not convinced that's actually better than the ES5 || 'default' version of it.

As a side note, you can require JSON files in Node:

const jspmConfigFile = require("./package.json").jspm.configFile || "default";

Is a one liner which I think is shorter. You can import it in a one liner directly too as:

import  { jspm: { configFile: jspmConfigFile = 'config.js' }} from "./package.json"

But I honestly don't think that's clearer.

edited Oct 4, 2015 at 20:12

answered Oct 4, 2015 at 20:11

Benjamin Gruenbaum

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9 Comments

Sean Anderson Over a year ago

Well, it does get me the benefit of not having to check if "jspm" is undefined without erroring.

Benjamin Gruenbaum Over a year ago

@SeanAnderson yes, but IMO destructuring a single nested property often creates confusing code. I've considered whether or not to add this answer for a bit and decided to add it when Rob added his answer to provide an alternative to the "ES2015" way :)

Sean Anderson Over a year ago

Yeah, it's certainly a valid argument. I think after seeing the resulting code I'll stick with the old way, but I'm glad I at least checked out other ways.

Rob W Over a year ago

@SeanAnderson Why do you think that you can use it without checking whether jspm is defined? ({ jspm: { configFile: jspmConfigFile }} = {}); throws "TypeError: can't convert undefined to object" in Firefox 41.0.

Sean Anderson Over a year ago

Oh. I thought that was a benefit? Clearly still super new to ES6. Alrighty, I dont' see the point in using it at all for this scenario then!

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