I am looking at a user service, my understanding is it's similar to a user service in Nest, but not really.
In it I see the following:
export class UsersService {
private usersDao: UsersDao
constructor() {
this.usersDao = UsersDao.getInstance();
}
}
static getInstance(): UsersService {
if (!UsersService.instance) {
UsersService.instance = new UsersService();
}
return UsersService.instance;
}
What is that getInstance() doing exactly? And why not just:
export class UsersService {
constructor(private usersDao: UsersDao) {}
}
What is the goal of getInstance()?
Usually this is part of the singleton pattern. Basically one class that, once instantiated, any subsequent classes will refer to that instance, rather than creating a fresh instance each time.
https://en.wikipedia.org/wiki/Singleton_pattern
Its useful for a class where something complex needs to happen when it is first constructed, but all following calls just need access to the properties.
I'd also like to mention that you can (in JavaScript specifically) export an instance, and all modules that import the module will have access to the same instance.
Hello i am using some module behavior and it ask me a question.
in one.js :
import { Two } from './two.mjs';
export class One extends Two<<<ReferenceError {
}
in two.js :
import { One } from './one.mjs';
export class Two {
constructor() {
let one = new One();
}
}
Ok it's not a casual case and it's not really my code but i have simplified.
I know why this error append (the interpreter go in file two then import the file one and in the file one the class Two is not initisialized)
But the question is : How can i find (like in java) a way to avoid this (use only one file, avoid class, idk). Give me your opinion.
Thx a lot.
I've got a conundrum for you. I'm fairly new to Typescript, but I've enjoyed the experience thus far. Wanting to be a good developer and logically group my modules into namespaces I've come across a rare? issue. Here's the simplest example to recreate the problem.
Lets say you got a class... we'll call it ClassA, which references a class aptly named ClassB which resides in a child namespace.
namespace Foo {
export class ClassA {
public classB: Bar.ClassB;
public constructor() {
}
}
}
and
namespace Foo.Bar {
export class ClassB {
public constructor() {
}
}
}
Typescript couldn't be happier with this. Well when I want to import things for ClassA, I'll put it inside the namespace like so:
namespace Foo {
import SomeClass = SomewhereElse.SomeClass;
export class ClassA {
public classB: Bar.ClassB;
public constructor() {
}
}
}
Again... we cool. But what if I wanted to import an ambient module?
namespace Foo {
import * as SockJS from "sockjs-client"; // TS1147 GRUMBLE! CAN'T DO DAT HERE!
export class ClassA {
public classB: Bar.ClassB;
public constructor() {
}
}
}
Ok, that's fine. I'll just move it outside the namespace like it's telling me to.
import * as SockJS from "sockjs-client";
namespace Foo {
export class ClassA {
public classB: Bar.ClassB; // TS2503:Now you're just a Bar that I used to know
public constructor() {
}
}
}
Suddenly Typescript gets amnesia and doesn't know what Bar is. It doesn't recognize child namespaces. I've tried several ways to make it remember like nesting the namespaces or exporting them, but nothing will make it recognize this. What's the dilly yo? Any of you cool cats know what's going on or how I can resolve it and keep my child namespaces?
Suddenly Typescript gets amnesia and doesn't know what Bar is. It doesn't recognize child namespaces. I've tried several ways to make it remember like nesting the namespaces or exporting them, but nothing will make it recognize this. What's the dilly yo
Its because namespaces are global. And modules are modular. So as soon as you import a module into your file, your file becomes a module. https://basarat.gitbooks.io/typescript/content/docs/project/modules.html
Suggestion
Please use modules. There are plenty of reasons : https://github.com/TypeStrong/atom-typescript/blob/8d43dd1b930a6df0ce62454a1560acfb7eee24c9/docs/out.md
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?
Extended - 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.
In addition to Tim's answer, this issue occurred for me when I was splitting up a refactoring a file, splitting it up into their own files.
VSCode, for some reason, indented parts of my [class] code, which caused this issue. This was hard to notice at first, but after I realised the code was indented, I formatted the code and the issue disappeared.
for example, everything after the first line of the Class definition was auto-indented during the paste.
export class MyClass extends Something<string> {
public blah: string = null;
constructor() { ... }
}
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 😎👌
Note: i am using this config
"target": "esnext",
"module": "commonjs",
I faced the same issue in a module that has no exports. I used it for side-effects only. This is what the TypeScript docs say about importing side-effects modules:
Though not recommended practice, some modules set up some global state that can be used by other modules. These modules may not have any exports, or the consumer is not interested in any of their exports. To import these modules, use:
import "./my-module.js";
In that situation, you can fix the "File is not a module" error by simply exporting an empty object:
// side-effects stuff
export default {};
I faced the same issue ("File is not a module error") for import js in vue component
import handleClientLoad from "../../../public/js/calendar.js"
I do this and solve it
// #ts-ignore
import handleClientLoad from "../../../public/js/calendar.js"
The file needs to add Component from core hence add the following import to the top
import { Component } from '#angular/core';
I use Mocha/Chai for testing JavaScript front-end code and now we switched to TypeScript. I have several functions I want to test. But they shouldn't be exportable. Can I get an access to this functions and test it without adding export to them?
There is no way to access a non-exported module function.
module MyModule {
function privateFunction() {
alert("privateFunction");
}
}
MyModule.privateFunction(); // Generates a compiler error
However, leaving aside the question of validity of private method testing, here is what you can do.
Group your functions into an utility class, and then leverage the fact that private class members can be accessed via square bracket notation.
module MyModule {
export class UtilityClass {
private privateFunction() {
alert("privateFunction");
}
}
}
var utility = new MyModule.UtilityClass();
//utility.privateFunction(); Generates a compiler error
utility["privateFunction"](); // Alerts "privateFunction"
While it is not possible to access non-exported function directly there is still a way to export them in a "semi-hidden" way. One possible approach would be:
// In your library module declare internal functions as non-exported like normal.
function someInternalFunctionA(x: number): number {
return x;
}
function someInternalFunctionB(x: number): number {
return x;
}
// At the bottom, offer a public escape hatch for accessing certain functions
// you would like to be available for testing.
export const _private = {
someInternalFunctionA,
someInternalFunctionB,
};
On test side you can do:
import { _private } from "./myModule";
test("someInternalFunctionA", () => {
expect(_private.someInternalFunctionA(42)).toEqual(42);
});
What I like about the approach:
No need to mark someInternalFunctionA with export directly.
It should still be very obvious that the stuff under _private isn't officially part of the public interface.
As you can see in related questions the issue of testing private functions inside classes or modules is heavily debated on StackOverflow - The following might be an architectural solution to not even have that discussion:
If these functions feel important enough to be tested separately, but should not be accessible as part of a module, should they maybe be placed in their own module?
This would solve your issue of accessibility - they now are public functions in one module, and you can easily use them from within another module and not expose them as part of that module.
Best solution I found is to export the private function under different name so this name will remind you not to use this function anywhere else.
export const getPrice__test = getPrice;
function getPrice(): number {
return 10 + Math.Random() * 50;
}
But they shouldn't be exportable. Can I get an access to this functions and test it without adding export to them?
In general, no. It's possible to access private class members, but not unexported members of modules.
I would echo #Katana314's comments -- unit tests should not be concerning themselves with non-public methods. Trying to do so is an indication that you're testing the implementation details of the module rather than the contract the module claims to implement.
This is a total hack, but hey...
window.testing = {};
Then in your module:
module.exports = {
myPublicFunction: myPublicFunction
};
window.testing.myModule = {
myPublicFunction: myPublicFunction,
myPrivateFunction: myPrivateFunction
};