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";