I can:
import * as foo from './foo'
But can't seem to export the same:
export * as foo from './foo'
This doesn't seem to work either...:
import * as fooImport from './foo';
export const foo = fooImport;
Any ideas?
--- UPDATE ---
What are you trying to achieve?
Basically, I am working on implementing an ngrx/store
backend for my app. I want to organize my code like this:
app/core/
index.ts
viewer/
index.ts
viewer-actions.ts
viewer-reducer.ts
view-data/
index.ts
view-data-actions.ts
view-data-reducer.ts
And I want to use my index.ts
files to chain up all the exports from each subset (common paradigm).
However, I want to keep things namespaced. Each of my xxx-reducer.ts
and xxx-actions.ts
files have exports of the same name (reducer
, ActionTypes
, Actions
, ...) so normal chaining would result in a name collision. What I am trying to do is allow for all of the exports from xxx-actions
and xxx-reducer
to be re-exported as xxx
. This would allow me to:
import { viewer, viewerData } from './core';
...
private viewer: Observable<viewer.Viewer>;
private viewData: Observable<viewData.ViewData>;
ngOnInit() {
this.viewer = this.store.let(viewer.getViewer());
this.viewData = this.store.let(viewData.getViewData());
}
Instead of the more verbose:
import * as viewer from './core/viewer';
import * as viewerData from './core/viewer-data';
...
Thats the gist anyway...
See https://stackoverflow.com/a/59712387/1108891 for details.
Is it possible to export * as foo in typescript
Nope. You can, however, use a two step process:
src/core/index.ts
import * as Foo from "./foo";
import * as Bar from "./bar";
export {
Foo,
Bar,
}
src/index.ts
import { Foo, Bar } from "./core";
function FooBarBazBop() {
Foo.Baz;
Foo.Bop;
Bar.Baz;
Bar.Bop;
}
src/core/foo/index.ts and src/core/bar/index.ts
export * from "./baz";
export * from "./bop";
src/core/foo/baz.ts and src/core/bar/baz.ts
export class Baz {
}
src/core/foo/bop.ts and src/core/bar/bop.ts
export class Bop {
}
See also: https://www.typescriptlang.org/docs/handbook/modules.html