如何从打字稿中的模块重新导出所有导入?

Pat*_*gan 4 typescript

我正在将 JS 代码库迁移到 TS。在 JS 中,我们有一些 index.js 文件,它们重新导出模块中的所有导入:

export * as users from './users';

这个等效的 TS 是多少?

Aja*_*les 18

从 TypeScript 3.8 开始就有可能

\n

这是再出口的摘要*

\n
// Re-export *\nexport * as default from \'./example1\' // as default-import\nexport * from \'./example2\' // separately\nexport * as ex3 from \'./example3\' // as a named-object\n\n// usage:\n// import ex1, { example2Const, ex3 } from \'./reexport\'\n// import * as all from \'./reexport\'\n
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并提醒您选择再出口

\n
// Re-export chosen\nexport { a, b as newB, c as default } from "./example4"; // pick what to export\n\n// usage:\n// import c, { a, newB } from \'./reexport\'\n// import * as all from \'./reexport\'\n
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文档:
\n\xc2\xb7重新导出所有
\n\xc2\xb7重新导出为命名对象

\n


Pat*_*gan 6

我还没有找到一种方法来做到这一点,但一位同事指出这种方法可以工作:

import * as users from './users';

export {
  users,
};
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