2017-08-12 92 views
0

我想知道是否有導出模塊中所有定義的簡寫。沒有辦法做到這一點在文檔中建議。我正在尋找類似export*的東西來導出模塊中的所有定義。速記在Typescript中導出模塊中的所有定義

/* this is one way according to docs*/ 
//module1 
const foo =() => {console.log('function foo')} 
const bar =() => {console.log('function bar')} 
export {foo, bar} 

// module2 
import * as m1 from './module1' 
m1.foo() 
m1.bar() 

/* this is another way according to docs*/ 
// module1 
export const foo =() => {console.log('function foo')} 
export const bar =() => {console.log('function bar')} 

// module2 
import * as m1 from './module1' 
m1.foo() 
m1.bar() 

/* looking for something like this */ 
// module1 
const foo =() => {console.log('function foo')} 
const bar =() => {console.log('function bar')} 
export * // looking for something like this 

// module2 
import * as m1 from './module1' 
m1.foo() //doesn't work 
m1.bar() // doesn't work 
+0

這不是由ECMAScript的 –

+0

打字稿支持, 沒有辦法出口單表達'所有屬性*' –

回答

0

看到這個答案https://stackoverflow.com/a/30714301/977206

// export the default export of a legacy (`export =`) module 
export import MessageBase = require('./message-base'); 

// export the default export of a modern (`export default`) module 
export { default as MessageBase } from './message-base'; 

而且這應該工作,也許只有某些類型的模塊?不知道

export * from 'foo';