"export default myFn" vs "export { myFn as default }" - javascript

const myFn = () => {
console.log('hello world');
}
Is there any benefits on using below export methods one over the another?
export default myFn
vs
export { myFn as default }
I tried compiling it in babel, I got this results:
// export { myFn as default }
// this throws error if function not found
// /repl: Export 'myFn' is not defined
exports.default = myFn;
// export default myFn // this doesn't error like mentioned above
var _default = myFn;
exports.default = _default;
There seems no benefit other than save a line of code var _default = myFn; :) , is that correct?

They are same.
I write these two statements into two typescript file. Then compile them into javascript. No matter to what version of javascript, there is no difference between the two result js file.

Related

How to export multiple functions as aliases in ES6

I want to export multiple functions as aliases in ES6.
I exported it like this because I want to use export default but "as" doesn't seem to work.
const fn1 = () => {};
const fn2 = () => {};
export default = {
fn1 as function1,
fn2 as function2,
} // -> not working
How can I do what I want?
What follows export default should be an expression (not a =). So all you need to do after that is use the right syntax for a normal Javascript object:
export default {
function1: fn1,
function2: fn2,
};
I think that is a wrong syntax. Correct use would be without '=' before "{".
export default {
fn1 as function1,
fn2 as function2
}

How to import anonymous functions in ES6

In ES6, we can import exported modules like this:
import { Abc } from './file-1'; // here Abc is a named export
import Def from './file-2'; // here Def is the default export
But how can we import anonymous functions? Consider this code:
file-3.js:
export function() {
return 'Hello';
}
export function() {
return 'How are you doing?';
}
How can I import above two functions in another file, given that neither are they default exports nor are they named exports (anonymous functions don't have names!)?
Single anonymous function can be exported as default (default export value can be anything). Multiple anonymous functions cannot be exported from a module - so they cannot be imported, too.
export statement follows strict syntax that supports either named or default exports. This will result in syntax error:
export function() {
return 'Hello';
}
These functions should be named exports:
export const foo = function () {
return 'Hello';
}
export const bar = function () {
return 'How are you doing?';
}
So they can be imported under same names.

How can I export all functions from a file in JS?

I'm creating a unit converter, and I want to put all of the conversion functions into their own file. Using ES6 export, is there any way to export all of the functions in the file with their default names using only one line? For example:
export default all;
The functions are all just in the file, not within an object.
No, there's no wildcard export (except when you're re-exporting everything from another module, but that's not what you're asking about).
Simply put export in front of each function declaration you want exported, e.g.
export function foo() {
// ...
}
export function bar() {
// ...
}
...or of course, if you're using function expressions:
export var foo = function() {
// ...
};
export let bar = () => {
// ...
};
export const baz = value => {
// ...
};
I think there are a lot of solutions to this. And as has been answered, there's no wildcard export. But, you can 'wildcard' the import. So, I much prefer the one putting export before each of the functions you want to expose from the file:
//myfile.js
export function fn1() {...}
export function fn2() {...}
and then import it like so:
import * as MyFn from './myfile.js'
Afterwards you could use it like so:
MyFn.fn1();
MyFn.fn2();
You can also use module.exports as follows:
function myFunction(arg) {
console.debug(arg);
}
function otherFunction(arg) {
console.error(arg);
}
module.exports = {
myFunction: myFunction,
otherFunction: otherFunction,
};
Then you can import it:
import {myFunction, otherFunction} from "./Functions.js";
In my use case, I do have three reusable functions in one file.
utils/reusables.js
export const a = () => {}
export const b = () => {}
export const c = () => {}
In order to point the root folder instead of individual file names, I created a file called index.js which will comprise of all the functions that are listed in individual files.
utils/index.js
export * from './reusables'
Now, when I want to use my a function, I will have to simply import it like this
import { a } from '../utils'
Rather than calling it from its individual files
import { a } from '../utils/reusables'
You could also export them at the bottom of your script.
function cube(x) {
return x * x * x;
}
const foo = Math.PI + Math.SQRT2;
var graph = {
options: {
color:'white',
thickness:'2px'
},
draw: function() {
console.log('From graph draw function');
}
}
export { cube, foo, graph };
You can also aggregate submodules together in a parent module so that they are available to import from that module.
// In parentModule.js
export { myFunction, myVariable } from 'childModule1.js';
export { myClass } from 'childModule2.js';
// In top-level module
import { myFunction, myVariable, myClass } from 'parentModule.js'
For Node.js environment, what I did to export functions was this.
UserController.js
module.exports = {
signUp: () => {
return "user"
},
login: () => {
return "login"
}
}
UserRouter.js
const UserController = require('./UserController')
then login and signUp functions could be used inside UserRouter as UserController.signUp() and UserController.login()
I think there's a missing common solution, which is exporting in index.js file:
myModule/myFunctions.js
export const foo = () => { ... }
export const bar = () => { ... }
then in myModule/index.js
export * from "./myFunctions.js";
This way you can simply import and use it with:
import { foo, bar } from "myModule";
foo();
bar();
functions.js
function alpha(msj) {
console.log('In alpha: ' + msj);
}
function beta(msj) {
console.log('In beta: ' + msj);
}
module.exports = {
alpha,
beta
};
main.js
const functions = require('./functions');
functions.alpha('Hi');
functions.beta('Hello');
Run
node main.js
Output
In alpha: Hi
In beta: Hello
In case anyone still needs an answer to this in modern JavaScript:
const hello = () => "hello there"
const bye = () => "bye bye"
export default { hello, bye }
What I like to do is to export all functions within an object:
//File.js
export default {
testFunction1: function testFunction1(){
console.log("Hello World")
},
//a little bit cleaner
testFunction2: () => {
console.log("Nothing here")
}
}
Now you can access the functions with calling the key value of the object:
//differentFile.js
import file from 'File.js'
file.testFunction1()
//Hello World
file.testFunction2()
//Nothing here

Correct way to export object in es6 module

I'm trying to export an my module as an object but exporting it seems an 'anti pattern' (https://medium.com/#rauschma/note-that-default-exporting-objects-is-usually-an-anti-pattern-if-you-want-to-export-the-cf674423ac38)
So I was wondering what's the correct way to export an object and then use it as
import utils from "./utils"
`
utils.foo()
Currently I'm doing like so
/** a.js **/
function foo(){
//...
}
export {
foo
}
/** b.js **/
import * as utils from "a";
utils.foo()
is it correct like so? Do I maintain the tree-shaking feature?
thanks
If the object you want to import/export only contains some functions (as I assume due to the Utils name), you can export the functions separately as follows:
export function doStuff1() {}
export function doStuff2() {}
And import like this:
import {doStuff1, doStuff2} from "./myModule";
However, if the object you want to export holds state in addition to methods, you should stick to a simple export default myObject. Otherwise, calling the imported methods won't work as intended, since the context of the object is lost.
As an example, the following object should be exported as a whole, since the properties of the object should stay encapsulated. Only importing and calling the increment function would not mutate myObject since the context of the object cannot be provided (since it's not imported as a whole).
const myObject = {
counter: 0,
increment: function() {
this.counter++;
}
}
export default myObject;
es6 native way to do this:
// file1.es6
export const myFunc = (param) => {
doStuff(param)
}
export const otherFunc = ({ param = {} }) => {
doSomething({ ...param })
}
// file2.es6
import { otherFunc } from './file1.es6'
import * as MyLib from './file1.es6'
MyLib.myfunc(0)
MyLib.otherFunc({ who: 'Repley' })
otherFunc({ var1: { a1: 1 } })
And so on.

How to use es6 syntax to import a function module in Typescript

I have a simple module. Use for checking variable type.
index.js
'use strict';
var typeOf = function (variable) {
return ({}).toString.call(variable).match(/\s([a-zA-Z]+)/)[1].toLowerCase();
};
module.exports = typeOf;
index.d.ts
export default typeOf;
declare function typeOf(value:any):string;
Here how I use it.
import typeOf from 'lc-type-of';
typeOf(value);
But the code dosen't work as expect. The typeOf function came out undefined error. Do I miss something?
when you export node like with Javascript:
module.exports = something;
in Typescript import it like :
import * as something from "./something"
and in the definition
// Tell typescript about the signature of the function you want to export
declare const something: ()=> void ;
// tell typescript how to import it
declare module something {
// Module does Nothing , it simply tells about it's existence
}
// Export
export = something;

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