Typescript, require module as class - javascript

I'm using an existing js library that uses AMD modules in my typescript code. I want to use a Javascript class as the base for my Typescript class. This is what I'm trying to do:
famous.js
define('famous/core/View',['require','exports','module'],function(require, exports, module) {
function View() {
...
}
...
module.exports = View;
});
View.d.ts
declare module "famous/core/View" {
}
AppView.ts
import View = require('famous/core/View');
class AppView extends View {
}
export = AppView;
But it says "Cannot find name 'View'". I suppose it's logical it doesn't work since a module is not a class, but I don't know another way.

You need to define a class and use export = in View.d.ts. For example:
declare module "famous/core/View" {
class View {
// TODO define members of View
}
export = View;
}

On further analysis it seems like purely a require configuration problem. It shouldn't have anything to do with class vs module. Are you sure your paths are correct and you have a config.ts require configuration file?

Related

Javascript import of Typescript - exporting default object

This may have been answered somehow already, but wasn't sure how to ask it.
I have a Typescript file that exports a single class:
export class MyClass {
...
}
I have a Javascript file that imports the transpiled version of that typescript and uses it:
define('random-source', ['src/my-class'], function(MyClassMod) {
...
var myClass = new MyClassMod.MyClass();
...
});
My question is this: Is there a way to write the Typescript file so that the only object exported is that 'MyClass' definition? As you see from the define in the Javascript, it's imported as a module name, but I have to say '.MyClass' to access it. When I have a lot of Typescript files with a single class defined like this, it makes the Javascript code a bit messy. Ideally I'd like something like:
define('random-source', ['src/my-class'], function(MyClass) {
...
var myClass = new MyClass();
...
});
Is that possible? Thanks in advance!
Actually you are looking for an export assignment, which will make MyClass the value of the entire module:
class MyClass {
// ...
}
export = MyClass;
Seems like you are looking for a default export:
export default class MyClass {
...
}

Add new JS class and template to aurelia project

I am new to Aurelia, but am working on an existing project and learning as I go.
I want to add a javascript class file and have NPM include it in the build, but I cannot find clear documentation on how to do that.
It is not a complex class and does not require an html template.
Just add the .js file containing your class to your src folder and add the following to import it in the file you intend to use the class.
import * as MyClass from './my-classs';
following the convention, you can create a class with the below...
export class MyClass {
myProperty = 'foo';
myOtherProperty = { key:'value' };
constructor() {
// constructor stuff (optional of course)
}
myMethod() {
// do something
}
}
Then in your ViewModel, for example...
import { inject } from 'aurelia-framework';
import { MyClass } from 'path/to/my-class';
#inject(MyClass)
export class MyViewModel {
constructor(myClass){
this.myClass = myClass;
}
attached() {
console.log(this.myClass.myProperty);
this.myClass.myMethod();
}
}
The way you import your class does depend on whether it's written as a module or not. If it's not a module, you'll have to write it out as #john-little mentioned.
The MyClass will automatically be a singleton until you make it transient (see https://aurelia.io/docs/fundamentals/dependency-injection#object-lifetime-child-containers-and-default-behavior)

How to Distinguish Between Private and Public Classes in Webpack

I'm using web pack to package a library.
We have multiple ES6 classes, in this fashion:
/src/Lib.js
import HelperClass from './HelperClass.js';
class Lib {
method1() {...}
}
/src/HelperClass.js
class HelperClass {
doSth() {...}
}
Packaging with webpack works, we end up with one file lib.js that contains Lib and HelperClass as var Lib = ....
How can I hide the HelperClass from the global namespace (e.g. make it a private class) with webpack?
UPDATE:
Now I'm running into an issue with importing the HelperClass! I uploaded a sample project https://github.com/benmarten/webpack_es6_test
This line:
__WEBPACK_IMPORTED_MODULE_0__Helper_js___default.a.doSth();
results in:
[Error] TypeError: __WEBPACK_IMPORTED_MODULE_0__Helper_js___default.a.doSth is not a function. (In '__WEBPACK_IMPORTED_MODULE_0__Helper_js___default.a.doSth()', '__WEBPACK_IMPORTED_MODULE_0__Helper_js___default.a.doSth' is undefined)
method1 (lib.js:92)
Global Code (index.htm:6)
When creating a library with webpack you expose everything that is exported in your entry point, everything else is not accessible from outside but you can use it within your code. If you want to use anything from another file, you still need to export it, because the files are still modules. Just because there is an export, does not mean it becomes a global. Only the exports in the entry specified in the webpack.config.js will be exposed.
Export the Helper in HelperClass.js:
class Helper {
static doSth() {
console.log('helper:doSth');
}
}
export default Helper;
Then import it in Lib.js:
import Helper from './HelperClass.js';
class Lib {
static method1() {
Helper.doSth();
}
}
export default Lib;
Now the default export of your bundle will be the Lib class and you can use Helper inside it without exposing it.
You should also read the Authoring Libraries Guides.

TypeScript ES6 import module "File is not a module error"

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';

Typescript and requirejs

I have a factory module which use require to create the desired objects
getProviderManager(providername: string): VideoProviderManager {
var providerManager = require(providername);
}
the VideoProviderManager is decalred as `
export class VideoProviderManager
However I have lots of errors that VideoProviderManager isn't known and
I have tried `
import VideoProviderManager = require("VideoProvider/VideoProviderManager");
But without sucess. I am trying to use combination of require and export classes with namespace of a module
is it possible?`
If you want to export VideoProviderManager as the value of the module, you need to use the export = syntax:
class VideoProviderManager {
// ...
}
export = VideoProviderManager;
Using export class VideoProviderManager means that the class will be exposed on the VideoProviderManager property of the module instead.
See The Definitive Guide to TypeScript for more detail on this.
Additionally, for getProviderManager, the synchronous require syntax you are using necessitates the module having already been loaded by RequireJS.

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