我可以:

import * as foo from './foo'

但似乎不能出口同样的产品:

export * as foo from './foo'

这似乎也不管用…:

import * as fooImport from './foo';
export const foo = fooImport;

有什么 idea 吗？

---更新---

你想要实现什么？

基本上，我正在为我的应用程序实现一个ngrx/store后端.我想这样组织我的代码:

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

我想用我的index.ts个文件来链接每个子集的所有导出(通用范例).

然而，我想保留一些东西.我的xxx-reducer.ts个和xxx-actions.ts个文件中的每个文件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.我想做的是让xxx-actions和xxx-reducer的所有出口都变成re-exported和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());
    }

而不是更冗长的:

import * as viewer from './core/viewer';
import * as viewerData from './core/viewer-data';

...

这就是要点...