尝试按文件夹组织:

baseTypes.ts

export class Animal {
    move() { /* ... */ }
}

export class Plant {
    photosynthesize() { /* ... */ }
}

dog.ts

import b = require('./baseTypes');

export class Dog extends b.Animal {
    woof() { }
}   

tree.ts

import b = require('./baseTypes');

class Tree extends b.Plant {
}

LivingThings.ts

import dog = require('./dog')
import tree = require('./tree')

export = {
    dog: dog,
    tree: tree
}

main.ts

import LivingThings = require('./LivingThings');
console.log(LivingThings.Tree)
console.log(LivingThings.Dog)

这个想法是你的模块本身不应该关心/知道它们正在参与一个命名空间，但是这以一种紧凑、合理的方式将你的API暴露给消费者，而不知道你正在为项目使用哪种类型的模块系统。

白化病小改善答案:

base.ts

export class Animal {
move() { /* ... */ }
}

export class Plant {
  photosynthesize() { /* ... */ }
}

dog.ts

import * as b from './base';

export class Dog extends b.Animal {
   woof() { }
} 

things.ts

import { Dog } from './dog'

namespace things {
  export const dog = Dog;
}

export = things;

main.ts

import * as things from './things';

console.log(things.dog);

尝试按文件夹组织:

baseTypes.ts

export class Animal {
    move() { /* ... */ }
}

export class Plant {
    photosynthesize() { /* ... */ }
}

dog.ts

import b = require('./baseTypes');

export class Dog extends b.Animal {
    woof() { }
}   

tree.ts

import b = require('./baseTypes');

class Tree extends b.Plant {
}

LivingThings.ts

import dog = require('./dog')
import tree = require('./tree')

export = {
    dog: dog,
    tree: tree
}

main.ts

import LivingThings = require('./LivingThings');
console.log(LivingThings.Tree)
console.log(LivingThings.Dog)

这个想法是你的模块本身不应该关心/知道它们正在参与一个命名空间，但是这以一种紧凑、合理的方式将你的API暴露给消费者，而不知道你正在为项目使用哪种类型的模块系统。

试试这个命名空间模块

namespaceModuleFile.ts

export namespace Bookname{
export class Snows{
    name:any;
    constructor(bookname){
        console.log(bookname);
    }
}
export class Adventure{
    name:any;
    constructor(bookname){
        console.log(bookname);
    }
}
}





export namespace TreeList{
export class MangoTree{
    name:any;
    constructor(treeName){
        console.log(treeName);
    }
}
export class GuvavaTree{
    name:any;
    constructor(treeName){
        console.log(treeName);
    }
}
}

bookTreeCombine.ts

——编译部分

import {Bookname , TreeList} from './namespaceModule';
import b = require('./namespaceModule');
let BooknameLists = new Bookname.Adventure('Pirate treasure');
BooknameLists = new Bookname.Snows('ways to write a book'); 
const TreeLis = new TreeList.MangoTree('trees present in nature');
const TreeLists = new TreeList.GuvavaTree('trees are the celebraties');

伙计，我支持你。 同样，在300多票的情况下，这个答案没有错，但我的观点是:

what is wrong with putting classes into their cozy warm own files individually? I mean this will make things looks much better right? (or someone just like a 1000 line file for all the models) so then, if the first one will be achieved, we have to import import import... import just in each of the model files like man, srsly, a model file, a .d.ts file, why there are so many *s in there? it should just be simple, tidy, and that's it. Why I need imports there? why? C# got namespaces for a reason. And by then, you are literally using "filenames.ts" as identifiers. As identifiers... Come on its 2017 now and we still do that? Ima go back to Mars and sleep for another 1000 years.

因此，遗憾的是，我的回答是:不，如果不使用所有这些导入或使用这些文件名作为标识符(我认为这真的很愚蠢)，您就不能使“名称空间”东西起作用。另一种选择是:将所有这些依赖项放入一个名为filenameasidentifier的框中。t和用法

export namespace(or module) boxInBox {} .

包装它们，这样它们就不会试图访问具有相同名称的其他类，当它们只是简单地试图从它们上面的类获得引用时。

你可以使用*作为wrapper_var语法，使所有导入的方法都可以在wrapper_var下访问:

import * as validator from "./ZipCodeValidator";
let myValidator = new validator.ZipCodeValidator();