This question already has answers here:
Is it possible to export * as foo in typescript
                            
                                (2个答案)
                            
                    
                去年关闭。
        

    

以以下文件夹结构为例:

example.ts
outer/
|--- inner/
   |--- index.ts
   |--- file1.ts
   |--- file2.ts
|--- index.ts


file1.ts的内容

export class FileOneClass{/**/}
export interface IFileOneInterface{/**/}


file2.ts的内容:

export class FileTwoClass{/**/}
export interface IFileTwoInterface{/**/}


outer/inner/index.ts的内容:

export { FileOneClass, IFileOneInterface } from "./file1.ts"
export { FileTwoClass, IFileTwoInterface } from "./file2.ts"


outer/index.ts的内容:

import * as InnerImport from "./inner";
export const Inner = InnerImport;


example.ts的内容:

import { Inner } from "./outer"

    function exampleUsage(){
      console.log(Inner.FileOneClass)
      console.log(Inner.FileOneInterface)

      console.log(Inner.FileTwoClass)
      console.log(Inner.FileTwoInterface)
    }


我的问题:

有没有一种方法可以将outer/index.ts中的export语句写入一行?

// Ugly:
import * as InnerImport from "./inner";
export const Inner = InnerImport;

// Desired is something like this:
export * as Inner from "./inner"; // This line leads to a error!

最佳答案

好吧,当然(开个玩笑):

import * as InnerImport from "./inner"; export const Inner = InnerImport;


但是不幸的是,没有一种方法可以重命名出口并将出口包装起来。

关于javascript - 是否可以单行通配通配符? ,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/55105690/

10-13 07:55
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