typescript 是否可以在打字稿中将 * 导出为 foo
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Is it possible to export * as foo in typescript
提问by Lucas
I can:
我能：
import * as foo from './foo'
But can't seem to export the same:
但似乎无法导出相同的内容：
export * as foo from './foo'
This doesn't seem to work either...:
这似乎也不起作用...：
import * as fooImport from './foo';
export const foo = fooImport;
Any ideas?
有任何想法吗？
--- UPDATE ---
- - 更新 - -
What are you trying to achieve?
你想达到什么目的？
Basically, I am working on implementing an ngrx/storebackend for my app. I want to organize my code like this:
基本上，我正在ngrx/store为我的应用程序实现后端。我想像这样组织我的代码：
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.tsfiles to chain up all the exports from each subset (common paradigm).
我想使用我的index.ts文件来链接每个子集（通用范例）的所有导出。
However, I want to keep things namespaced. Each of my xxx-reducer.tsand xxx-actions.tsfiles 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-actionsand xxx-reducerto be re-exportedas xxx. This would allow me to:
但是，我想保留命名空间。每个 myxxx-reducer.ts和xxx-actions.tsfiles 都有相同名称的导出 ( reducer, ActionTypes, Actions, ...) 所以正常的链接会导致名称冲突。我所试图做的是让所有从出口xxx-actions和xxx-reducer被复出口的xxx。这将使我能够：
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:
而不是更冗长：
import * as viewer from './core/viewer';
import * as viewerData from './core/viewer-data';
...
Thats the gist anyway...
无论如何，这就是要点......
回答by Shaun Luttin
Is it possible to export * as foo in typescript
是否可以在打字稿中将 * 导出为 foo
Nope. You can, however, use a two step process:
不。但是，您可以使用两步过程：
src/core/index.ts
src/核心/index.ts
import * as Foo from "./foo";
import * as Bar from "./bar";
export {
Foo,
Bar,
}
src/index.ts
源代码/索引.ts
import { Foo, Bar } from "./core";
function FooBarBazBop() {
Foo.Baz;
Foo.Bop;
Bar.Baz;
Bar.Bop;
}
src/core/foo/index.ts andsrc/core/bar/index.ts
src/core/foo/index.ts和src/core/bar/index.ts
export * from "./baz";
export * from "./bop";
src/core/foo/baz.ts andsrc/core/bar/baz.ts
src/core/foo/baz.ts和src/core/bar/baz.ts
export class Baz {
}
src/core/foo/bop.ts andsrc/core/bar/bop.ts
src/core/foo/bop.ts和src/core/bar/bop.ts
export class Bop {
}
See also: https://www.typescriptlang.org/docs/handbook/modules.html
另见：https: //www.typescriptlang.org/docs/handbook/modules.html
回答by Przemek Struciński
Since TypeScript 3.8has been released you are able to add alias for your exports.
由于TypeScript 3.8已发布，您可以为导出添加别名。
Example:
例子：
export * as utilities from "./utilities.js";
Ref: https://www.typescriptlang.org/docs/handbook/release-notes/typescript-3-8.html
参考：https: //www.typescriptlang.org/docs/handbook/release-notes/typescript-3-8.html
