Ben*_*uer 5 module export class typescript systemjs
关于如何将 TypeScript 模块与单独文件中的类分开存在几个问题,但到目前为止还没有适用于我的问题的解决方案。
我有一个模块 ( store),其中包含两个类 ( Person& SerialisedRecord)。当两个类都在一个文件中时,编译和导出工作正常。
现在,我希望每个类 ( Person.ts& SerialisedRecord.ts) 都有一个文件,遵循我已有的相同导出模式。但我不知道如何实现这一目标。
这是我最初的情况:
商店.ts
export module store {
export class Person {
public fullName: string = '';
constructor(firstName: string, lastName: string) {
this.fullName = `${firstName} ${lastName}`;
}
}
export class SerialisedRecord {
constructor(public serialised: string, public id: string) {}
}
}
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当我编译store.ts到store.js(ES5)时,我得到了我想要的(一个 SystemJS 模块在一个模块中导出两个类):
System.register([], function(exports_1, context_1) {
"use strict";
var __moduleName = context_1 && context_1.id;
var store;
return {
setters:[],
execute: function() {
(function (store) {
var Person = (function () {
function Person(firstName, lastName) {
this.fullName = '';
this.fullName = firstName + " " + lastName;
}
return Person;
}());
store.Person = Person;
var SerialisedRecord = (function () {
function SerialisedRecord(serialised, id) {
this.id = id;
this.serialised = serialised;
}
return SerialisedRecord;
}());
store.SerialisedRecord = SerialisedRecord;
})(store = store || (store = {}));
exports_1("store", store);
}
}
});
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现在我尝试这样做:
export module store {
export {Person} from "./Person";
export {SerialisedRecord} from "./SerialisedRecord";
}
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但它没有告诉我:
错误 TS1194:命名空间中不允许导出声明。
你能告诉我我做错了什么吗?
这是我的tsconfig.json:
{
"compilerOptions": {
"module": "system",
"moduleResolution": "node",
"noEmitOnError": true,
"noImplicitAny": false,
"noImplicitReturns": true,
"removeComments": true,
"target": "es5"
},
"exclude": [
"node_modules",
"typings/browser",
"typings/browser.d.ts"
]
}
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如果删除商店模块,效果会很好:
商店.ts:
export { Person } from "./Person";
export { SerialisedRecord } from "./SerialisedRecord";
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索引.ts:
import { Person, SerialisedRecord } from "./store";
let p = new Person("first", "last");
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如果您必须保留名称空间结构,您可以尝试以下操作:
import Person from "./Person";
import SerialisedRecord from "./SerialisedRecord";
export default {
store: {
Person,
SerialisedRecord
}
}
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