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I am using TypeScript 1.6 with ES6 modules syntax.
My files are:
test.ts:
module App {
export class SomeClass {
getName(): string {
return 'name';
}
}
}
main.ts:
import App from './test';
var a = new App.SomeClass();
When I am trying to compile the main.ts
file I get this error:
Error TS2306: File 'test.ts' is not a module.
How can I accomplish that?
I am using TypeScript 1.6 with ES6 modules syntax.
My files are:
test.ts:
module App {
export class SomeClass {
getName(): string {
return 'name';
}
}
}
main.ts:
import App from './test';
var a = new App.SomeClass();
When I am trying to compile the main.ts
file I get this error:
Error TS2306: File 'test.ts' is not a module.
How can I accomplish that?
Share Improve this question edited Feb 13, 2022 at 1:38 Paolo 21k21 gold badges76 silver badges121 bronze badges asked Sep 27, 2015 at 7:24 BazingaBazinga 11.2k7 gold badges41 silver badges66 bronze badges 2- I had this issue , I did not have a constructor in the class , added one and problem went away – dorriz Commented May 22, 2019 at 12:00
- I was simply importing the wrong file... – Daniel Danielecki Commented Mar 27, 2021 at 11:27
10 Answers
Reset to default 201Extended - to provide more details based on some comments
The error
Error TS2306: File 'test.ts' is not a module.
Comes from the fact described here http://exploringjs.com/es6/ch_modules.html
17. Modules
This chapter explains how the built-in modules work in ECMAScript 6.
17.1 Overview
In ECMAScript 6, modules are stored in files. There is exactly one module per file and one file per module. You have two ways of exporting things from a module. These two ways can be mixed, but it is usually better to use them separately.
17.1.1 Multiple named exports
There can be multiple named exports:
//------ lib.js ------ export const sqrt = Math.sqrt; export function square(x) { return x * x; } export function diag(x, y) { return sqrt(square(x) + square(y)); } ...
17.1.2 Single default export
There can be a single default export. For example, a function:
//------ myFunc.js ------ export default function () { ··· } // no semicolon!
Based on the above we need the export
, as a part of the test.js file. Let's adjust the content of it like this:
// test.js - exporting es6
export module App {
export class SomeClass {
getName(): string {
return 'name';
}
}
export class OtherClass {
getName(): string {
return 'name';
}
}
}
And now we can import it in these three ways:
import * as app1 from "./test";
import app2 = require("./test");
import {App} from "./test";
And we can consume imported stuff like this:
var a1: app1.App.SomeClass = new app1.App.SomeClass();
var a2: app1.App.OtherClass = new app1.App.OtherClass();
var b1: app2.App.SomeClass = new app2.App.SomeClass();
var b2: app2.App.OtherClass = new app2.App.OtherClass();
var c1: App.SomeClass = new App.SomeClass();
var c2: App.OtherClass = new App.OtherClass();
and call the method to see it in action:
console.log(a1.getName())
console.log(a2.getName())
console.log(b1.getName())
console.log(b2.getName())
console.log(c1.getName())
console.log(c2.getName())
Original part is trying to help to reduce the amount of complexity in usage of the namespace
Original part:
I would really strongly suggest to check this Q & A:
How do I use namespaces with TypeScript external modules?
Let me cite the first sentence:
Do not use "namespaces" in external modules.
Don't do this.
Seriously. Stop.
...
In this case, we just do not need module
inside of test.ts
. This could be the content of it adjusted test.ts
:
export class SomeClass
{
getName(): string
{
return 'name';
}
}
Read more here
Export =
In the previous example, when we consumed each validator, each module only exported one value. In cases like this, it's cumbersome to work with these symbols through their qualified name when a single identifier would do just as well.
The
export =
syntax specifies a single object that is exported from the module. This can be a class, interface, module, function, or enum. When imported, the exported symbol is consumed directly and is not qualified by any name.
we can later consume it like this:
import App = require('./test');
var sc: App.SomeClass = new App.SomeClass();
sc.getName();
Read more here:
Optional Module Loading and Other Advanced Loading Scenarios
In some cases, you may want to only load a module under some conditions. In TypeScript, we can use the pattern shown below to implement this and other advanced loading scenarios to directly invoke the module loaders without losing type safety.
The compiler detects whether each module is used in the emitted JavaScript. For modules that are only used as part of the type system, no require calls are emitted. This culling of unused references is a good performance optimization, and also allows for optional loading of those modules.
The core idea of the pattern is that the import id = require('...') statement gives us access to the types exposed by the external module. The module loader is invoked (through require) dynamically, as shown in the if blocks below. This leverages the reference-culling optimization so that the module is only loaded when needed. For this pattern to work, it's important that the symbol defined via import is only used in type positions (i.e. never in a position that would be emitted into the JavaScript).
Above answers are correct. But just in case... Got same error in VS Code. Had to re-save/recompile file that was throwing error.
How can I accomplish that?
Your example declares a TypeScript < 1.5 internal module, which is now called a namespace. The old module App {}
syntax is now equivalent to namespace App {}
. As a result, the following works:
// test.ts
export namespace App {
export class SomeClass {
getName(): string {
return 'name';
}
}
}
// main.ts
import { App } from './test';
var a = new App.SomeClass();
That being said...
Try to avoid exporting namespaces and instead export modules (which were previously called external modules). If needs be you can use a namespace on import with the namespace import pattern like this:
// test.ts
export class SomeClass {
getName(): string {
return 'name';
}
}
// main.ts
import * as App from './test'; // namespace import pattern
var a = new App.SomeClass();
In addition to A. Tim's answer there are times when even that doesn't work, so you need to:
- Rewrite the import string, using the intellisense. Sometimes this fixes the issue
- Restart VS Code
I had this issue and I had forgotten to export the Class.
Just in case this may works for you as it did form me, i had this files
//server.ts
class Server{
...
}
exports.Server = Server
//app.ts
import {Server} from './server.ts'
And this actually raised an error but i changed server.ts
to
//server.ts
export class Server{
...
}
and it worked
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