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Import regular CSS file in SCSS file?
(15 answers)
Closed 7 years ago.
I recently found out about this feature in node-sass named importer, which allows you to handle any encounter with #import in your sass files.
I want to take advantage of this feature to import .css files from a npm package easily, similarly to Browserify.
Here's my sass file. (index.scss)
#import "normalize.css";
When I run this file through Sass and my importer, it should output this (supposed index.css):
/*! normalize.css v3.0.3 | MIT License | github.com/necolas/normalize.css */
/**
* 1. Set default font family to sans-serif.
* 2. Prevent iOS and IE text size adjust after device orientation change,
* without disabling user zoom.
*/
html {
...
Instead, it outputs this (actual index.css):
#import url(/Users/me/workspace/project/node_modules/normalize.css/normalize.css);
What gives? Why is it that my importer only changes the file path instead of taking the file and combining it with the original file?
Here's my importer and gulp task that runs sass (gulpfile.babel.js):
import gulp from 'gulp';
import sass from 'gulp-sass';
import rename from 'gulp-rename';
import path from 'path';
import fs from 'fs';
let aliases = {};
function importNpmModule(url, file, done) {
// check if the path was already found and cached
if (aliases[url]) {
return done({ file: aliases[url] });
}
// look for modules installed through npm
try {
const basePath = path.resolve('node_modules', url);
const pkgInfo = path.join(basePath, 'package.json');
const info = JSON.parse(fs.readFileSync(pkgInfo));
const newPath = path.join(basePath, info.style);
aliases[url] = newPath; // cache this request
return done({ file: newPath });
} catch(e) {
// If module could not be found, return the original url
aliases[url] = url;
return done({ file:url });
}
}
gulp.task('sassify', () => {
return gulp.src('./index.scss')
.pipe(sass({ importer: inportNpmModule }))
.pipe(rename('index.css'))
.pipe(gulp.dest('public'));
});
EDIT: I also tried using an npm package named sass-module-importer, but it has the exact same problem that I have.
It's not your importer problem. You have this error because there is .css in Sass import. Just omit it.
If you have problems after that may be you should update to last version of node-sass or use some hacks like rename .css to .scss before compiling.
Currently in our Sass files we have something like the following:
#import "../../node_modules/some-module/sass/app";
This is bad, because we're not actually sure of the path: it could be ../node_modules, it could be ../../../../../node_modules, because of how npm installs stuff.
Is there a way in Sass that we can search up until we find node_modules? Or even a proper way of including Sass through npm?
If you are looking for a handy answer in 2017 and are using Webpack, this was the easiest I found.
Suppose your module path is like:
node_modules/some-module/sass/app
Then in your main scss file you can use:
#import "~some-module/sass/app";
Tilde operator shall resolve any import as a module.
As Oncle Tom mentioned, the new version of Sass has this new importer option, where every "import" you do on your Sass file will go first through this method. That means that you can then modify the actual url of this method.
I've used require.resolve to locate the actual module entry file.
Have a look at my gulp task and see if it helps you:
'use strict';
var path = require('path'),
gulp = require('gulp'),
sass = require('gulp-sass');
var aliases = {};
/**
* Will look for .scss|sass files inside the node_modules folder
*/
function npmModule(url, file, done) {
// check if the path was already found and cached
if(aliases[url]) {
return done({ file:aliases[url] });
}
// look for modules installed through npm
try {
var newPath = path.relative('./css', require.resolve(url));
aliases[url] = newPath; // cache this request
return done({ file:newPath });
} catch(e) {
// if your module could not be found, just return the original url
aliases[url] = url;
return done({ file:url });
}
}
gulp.task("style", function() {
return gulp.src('./css/app.scss')
.pipe(sass({ importer:npmModule }))
.pipe(gulp.dest('./css'));
});
Now let's say you installed inuit-normalize using node. You can simply "require" it on your Sass file:
#import "inuit-normalize";
I hope that helps you and others. Because adding relative paths is always a pain in the ass :)
You can add another includePaths to your render options.
Plain example
Snippet based on example from Oncle Tom.
var options = {
file: './sample.scss',
includePaths: [
path.join(__dirname, 'bower_components'), // bower
path.join(__dirname, 'node_modules') // npm
]
};
sass.render(options, function(err, result){
console.log(result.css.toString());
});
That should do. You can include the files from package using #import "my-cool-package/super-grid
Webpack and scss-loader example
{
test: /\.scss$/,
loader: 'style!css!autoprefixer?browsers=last 2 version!sass?outputStyle=expanded&sourceMap=true&sourceMapContents=true&includePaths[]=./node_modules'
},
Notice the last argument, includePaths has to be array. Keep in mind to use right format
You can use a Sass importer function to do so. Cf. https://github.com/sass/node-sass#importer--v200.
The following example illustrates node-sass#3.0.0 with node#0.12.2:
Install the bower dependency:
$ bower install sass-mq
$ npm install sass/node-sass#3.0.0-pre
The Sass file:
#import 'sass-mq/mq';
body {
#include mq($from: mobile) {
color: red;
}
#include mq($until: tablet) {
color: blue;
}
}
The node renderer file:
'use strict';
var sass = require('node-sass');
var path = require('path');
var fs = require('fs');
var options = {
file: './sample.scss',
importer: function bowerModule(url, file, done){
var bowerComponent = url.split(path.sep)[0];
if (bowerComponent !== url) {
fs.access(path.join(__dirname, 'bower_components', bowerComponent), fs.R_OK, function(err){
if (err) {
return done({ file: url });
}
var newUrl = path.join(__dirname, 'bower_components', url);
done({ file: newUrl });
})
}
else {
done({ file: url });
}
}
};
sass.render(options, function(err, result){
if (err) {
console.error(err);
return;
}
console.log(result.css.toString());
});
This one is simple and not recursive. The require.resolve function could help to deal with the tree – or wait until npm#3.0.0 to benefit from the flat dependency tree.
I made the sass-npm module specifically for this.
npm install sass-npm
In your SASS:
// Since node_modules/npm-module-name/style.scss exists, this will be imported.
#import "npm-module-name";
// Since just-a-sass-file isn't an installed npm module, it will be imported as a regular SCSS file.
#import "just-a-sass-file";
I normally use gulp-sass (which has the same 'importer' option as regular SASS)
var gulp = require('gulp'),
sass = require('gulp-sass'),
sassNpm = require('sass-npm')();
Then, in your .pipe(sass()), add the importer as an option:
.pipe(sass({
paths: ['public/scss'],
importer: sassNpm.importer,
}))
For dart-sass and commandline user at 2022, just use the --load-path option:
$ npx sass --load-path=node_modules
Important: the whole node_modules folder contains so much, just set it launch extremely slow in watch mode. Your should only set your package paths, eg:
$npx sass -w --load-path=node_modules/foo --load-path=node_modules/bar/scss
From offical docuumentation of Sass, adding ~ to imports should do the job.
However, for some reason it did'nt work for me, and sass compiler still complains that the module cannot be found.
Hence, I tried another method which worked for me without any issues. Here's the solution:
If you are compiling sass files directly from CLI try this:
sass src/main.scss dist/main.css --load-path=node_modules
If you are using npm and/or webpack for compiling sass files, add something like this to the scripts of package.json:
"scripts": {
...
"build": "sass src/main.scss dist/main.css --load-path=node_modules",
...
}
Then Run:
npm run build
Finally, import your modules like this:
#import "some-module/sass/app";
To wrap it up, adding --load-path=node_modules flag solved the issue permanently. For more information you can check:
sass --help
Project Structure
root
wwwroot <-- files under this location are static files public to the site
css
lib
bootstrap/js/bootstrap.js
jquery/js/jquery.js
knockout/knockout.js
requires/require.js
scripts
modules ┌───────────────┐
global.js <--│ Built modules │
dropdown.js └───────────────┘
modules
global.js ┌────────────────┐
dropdown <--│ Source modules │
dropdown.js └────────────────┘
gruntfile.js
global.cs Contents (pre-built version at ~/modules/global.js)
require.config({
baseUrl: "scripts/modules",
paths: {
jquery: "../../lib/jquery/js/jquery",
bootstrap: "../../lib/bootstrap/js/bootstrap",
knockout: "../../lib/knockout/knockout"
},
shims: {
bootstrap: {
deps: ['jquery']
}
},
});
define(function (require) {
var $ = require('jquery');
var ko = require('knockout');
var bootstrap = require('bootstrap');
});
dropdown.js Contents (pre-built version at ~/modules/dropdown.js)
define(function () {
console.log('dropdown initialized');
return 'foo';
});
HTML Page
Contains this script tag in the <head> of the page for loading requires config:
<script src="~/lib/requirejs/require.js" data-main="scripts/modules/global"></script>
In the body of the HTML page, I have the following:
<script>
require(['global'], function () {
require(['dropdown'], function (dropdown) {
console.log(dropdown);
});
});
</script>
Issue
The dropdown callback is undefined instead of the expected "foo" string that I'm returning from the defined module.
In fact, the console does not contain a log item for "dropdown initialized" either. This makes me believe the module is not being invoked somehow? However, it's strange the dropdown.js is present in F12 debugger as a script loaded into the page. Therefore, requires did make a call to load it, but did not run the contents of the define?
Noteworthy mentions
I'm using r.js to optimize and build. Both global.js and dropdown.js are processed over.
The name assigned to the dropdown module by r.js processing is "modules/dropdown/dropdown.js". I'm unsure if I should be using this somehow, or if I'm referring to the module correctly as just dropdown and relying on my baseUrl config having the correct path.
Edit #1
I have added the r.js build configuration used with grunt per commenter request. In conjunction, I updated the file structure to include the overall project structure, instead of just the runtime public wwwroot structure.
The r.js process will compile built forms of global.js + other modules in ~/wwwroot/scripts/modules from the source location ~/modules in summary.
function getRequireJsConfiguration() {
var baseUrl = './';
var paths = {
jquery: "wwwroot/lib/jquery/js/jquery",
bootstrap: "wwwroot/lib/bootstrap/js/bootstrap",
knockout: "wwwroot/lib/knockout/knockout"
};
var shims = {
bootstrap: {
deps: ['jquery']
}
};
var optimize = 'none';
var configuration = {};
var jsFilePaths = grunt.file.expand('modules/**/*.js');
jsFilePaths.forEach(function (jsFilePath) {
var fileName = jsFilePath.split('/').pop();
if (configuration[fileName]) {
throw 'Duplicate module name conflict: ' + fileName;
}
configuration[fileName] = {
options: {
baseUrl: './',
name: jsFilePath,
out: 'wwwroot/scripts/modules/' + fileName,
paths: paths,
shims: shims,
optimize: optimize,
exclude: ['jquery', 'knockout', 'bootstrap']
}
};
});
configuration['global'] = {
options: {
baseUrl: './',
name: 'modules/global.js',
out: 'wwwroot/scripts/modules/global.js',
paths: paths,
shims: shims,
optimize: optimize,
}
};
return configuration;
}
Edit #2
Thought it'd be a good idea to include the versions of requirejs packages I'm using:
requirejs: 2.1.15
grunt-contrib-requirejs: 0.4.4
Thanks.
The name assigned to the dropdown module by r.js processing is "modules/dropdown/dropdown.js". I'm unsure if I should be using this somehow, or if I'm referring to the module correctly as just dropdown and relying on my baseUrl config having the correct path.
In a sense, yes, you should be using that full path. That's what Require refers to as the module id - "modules/dropdown/dropdown" (if the .js in the above output was real, I suggest stripping that extension in the "name" config. .js is assumed by RequireJS, you don't want that string in your module ids). The basePath is used, when given IDs, to transform some unknown ID to a file path (e.g. 'bootstrap' id -> (applying path config) -> '../../lib/bootstrap/js/bootstrap' -> (applying base URL) -> 'scripts/modules/../../lib/bootstrap/js/bootstrap').
Really, though, just allowing r.js to concatenate everything into one file
is the preferred way to go. You could use the include option to include modules un-referenced by global.js in with the optimized bundle, too ( https://github.com/jrburke/r.js/blob/master/build/example.build.js#L438 )
As to your specific problem: your lazy require(['dropdown']) call is misleading you. By combining the requested module id with the basePath, RequireJS comes up with the URL you want - scripts/modules/dropdown - which defines a module with the module id scripts/module/dropdown - but since you requested the module id dropdown, you get nothing. (I would've guessed you'd get a RuntimeError instead of undefined, but I suppose that's how things go). One way or another you need to address the id/path mismatches.
Although I have resolved my issue with the hints wyantb's answer provided, I've since changed my approach to a single file concat due to the simplicity it brings. I still wanted to post the specifics of how I solved this question's issue for anyone else to happens along it.
In the grunt build configuration options, I added the onBuildWrite field to transform the content, so my assigned module IDs lined up with how I was lazily loading them.
onBuildWrite: function (moduleName, path, contents) {
return contents.replace(/modules\/global.js/, 'global');
}
This code is specifically for the global.js file. I implemented a similar onBuildWrite for the other module files (in the foreach loop). The transformation will essentially strip the path and extension from the module name that r.js assigns.
Here are some examples of before and after:
Before After
/modules/global.js global
/modules/dropdown/dropdown.js dropdown
/modules/loginButton/loginButton.js loginButton
Therefore, when I load the modules using the HTML script from my original question, requirejs resolves and finds a match.
Either require by path or define global and dropdown in global.cs
require(['./global'], function () {
require(['./dropdown'], function (dropdown) {
console.log(dropdown);
});
});
I am trying to use gulp-requirejs to build a demo project. I expect result to be a single file with all js dependencies and template included. Here is my gulpfile.js
var gulp = require('gulp');
var rjs = require('gulp-requirejs');
var paths = {
scripts: ['app/**/*.js'],
images: 'app/img/**/*'
};
gulp.task('requirejsBuild', function() {
rjs({
name: 'main',
baseUrl: './app',
out: 'result.js'
})
.pipe(gulp.dest('app/dist'));
});
// The default task (called when you run `gulp` from cli)
gulp.task('default', ['requirejsBuild']);
The above build file works with no error, but the result.js only contains the content of main.js and config.js. All the view files, jquery, underscore, backbone is not included.
How can I configure gulp-requirejs to put every js template into one js file?
If it is not the right way to go, can you please suggest other method?
Edit
config.js
require.config({
paths: {
"almond": "/bower_components/almond/almond",
"underscore": "/bower_components/lodash/dist/lodash.underscore",
"jquery": "/bower_components/jquery/dist/jquery",
"backbone": "/bower_components/backbone/backbone",
"text":"/bower_components/requirejs-text/text",
"book": "./model-book"
}
});
main.js
// Break out the application running from the configuration definition to
// assist with testing.
require(["config"], function() {
// Kick off the application.
require(["app", "router"], function(app, Router) {
// Define your master router on the application namespace and trigger all
// navigation from this instance.
app.router = new Router();
// Trigger the initial route and enable HTML5 History API support, set the
// root folder to '/' by default. Change in app.js.
Backbone.history.start({ pushState: false, root: '/' });
});
});
The output is just a combination this two files, which is not what I expected.
gulp-requirejs has been blacklisted by the gulp folks. They see the RequireJS optimizer as its own build system, incompatible with gulp. I don't know much about that, but I did find an alternative in amd-optimize that worked for me.
npm install amd-optimize --save-dev
Then in your gulpfile:
var amdOptimize = require('amd-optimize');
var concat = require('gulp-concat');
gulp.task('bundle', function ()
{
return gulp.src('**/*.js')
.pipe(amdOptimize('main'))
.pipe(concat('main-bundle.js'))
.pipe(gulp.dest('dist'));
});
The output of amdOptimize is a stream which contains the dependencies of the primary module (main in the above example) in an order that resolves correctly when loaded. These files are then concatenated together via concat into a single file main-bundle.js before being written into the dist folder.
You could also minify this file and perform other transformations as needed.
As an aside, in my case I was compiling TypeScript into AMD modules for bundling. Thinking this through further I realized that when bundling everything I don't need the asynchronous loading provided by AMD/RequireJS. I am going to experiment with having TypeScript compile CommonJS modules instead, then bundling them using webpack or browserify, both of which seem to have good support within gulp.
UPDATE
My previous answer always reported taskReady even if requirejs reported an error. I reconsidered this approach and added error logging. Also I try to fail the build completely as described here gulp-jshint: How to fail the build? because a silent fail really eats your time.
See updated code below.
Drew's comment about blacklist was very helpfull and gulp folks suggest using requirejs directly. So I post my direct requirejs solution:
var DIST = './dist';
var requirejs = require('requirejs');
var requirejsConfig = require('./requireConfig.js').RJSConfig;
gulp.task('requirejs', function (taskReady) {
requirejsConfig.name = 'index';
requirejsConfig.out = DIST + 'app.js';
requirejsConfig.optimize = 'uglify';
requirejs.optimize(requirejsConfig, function () {
taskReady();
}, function (error) {
console.error('requirejs task failed', JSON.stringify(error))
process.exit(1);
});
});
The file at ./dist/app.js is built and uglified. And this way gulp will know when require has finished building. So the task can be used as a dependency.
My solution works like this:
./client/js/main.js:
require.config({
paths: {
jquery: "../vendor/jquery/dist/jquery",
...
},
shim: {
...
}
});
define(["jquery"], function($) {
console.log($);
});
./gulpfile.js:
var gulp = require('gulp'),
....
amdOptimize = require("amd-optimize"),
concat = require('gulp-concat'),
...
gulp.task('scripts', function(cb) {
var js = gulp.src(path.scripts + '.js')
.pipe(cached('scripts'))
.pipe(jshint())
.pipe(jshint.reporter('default'))
.pipe(remember('scripts'))
.pipe(amdOptimize("main",
{
name: "main",
configFile: "./client/js/main.js",
baseUrl: './client/js'
}
))
.pipe(concat('main.js'));
.pipe(gulp.dest(path.destScripts));
}
...
This part was important:
configFile: "./client/js/main.js",
baseUrl: './client/js'
This allowed me to keep my configuration in one place. Otherwise I was having to duplicate my paths and shims into gulpfile.js.
This works for me. I seems that one ought to add in uglification etc via gulp if desired. .pipe(uglify()) ...
Currently I have to duplicate the config in main.js to run asynchronously.
....
var amdOptimize = require("amd-optimize");
...
var js = gulp.src(path.scripts + '.js')
.pipe(cached('scripts'))
.pipe(jshint())
.pipe(jshint.reporter('default'))
.pipe(remember('scripts'))
.pipe(amdOptimize("main",
{
name: "main",
paths: {
jquery: "client/vendor/jquery/dist/jquery",
jqueryColor: "client/vendor/jquery-color/jquery.color",
bootstrap: "client/vendor/bootstrap/dist/js/bootstrap",
underscore: "client/vendor/underscore-amd/underscore"
},
shim: {
jqueryColor : {
deps: ["jquery"]
},
bootstrap: {
deps: ["jquery"]
},
app: {
deps: ["bootstrap", "jqueryColor", "jquery"]
}
}
}
))
.pipe(concat('main.js'));
Try this code in your gulpfile:
// Node modules
var
fs = require('fs'),
vm = require('vm'),
merge = require('deeply');
// Gulp and plugins
var
gulp = require('gulp'),
gulprjs= require('gulp-requirejs-bundler');
// Config
var
requireJsRuntimeConfig = vm.runInNewContext(fs.readFileSync('app/config.js') + '; require;'),
requireJsOptimizerConfig = merge(requireJsRuntimeConfig, {
name: 'main',
baseUrl: './app',
out: 'result.js',
paths: {
requireLib: 'bower_modules/requirejs/require'
},
insertRequire: ['main'],
// aliases from config.js - libs will be included to result.js
include: [
'requireLib',
"almond",
"underscore",
"jquery",
"backbone",
"text",
"book"
]
});
gulp.task('requirejsBuild', ['component-scripts', 'external-scripts'], function (cb) {
return gulprjs(requireJsOptimizerConfig)
.pipe(gulp.dest('app/dist'));
});
Sorry for my english. This solution works for me. (I used gulp-requirejs at my job)
I think you've forgotten to set mainConfigFile in your gulpfile.js. So, this code will be work
gulp.task('requirejsBuild', function() {
rjs({
name: 'main',
mainConfigFile: 'path_to_config/config.js',
baseUrl: './app',
out: 'result.js'
})
.pipe(gulp.dest('app/dist'));
});
In addition, I think when you run that task in gulp, require can not find its config file and
This is not gulp-requirejs fault.
The reason why only main.js and config.js is in the output is because you're not requiring/defining any other files. Without doing so, the require optimizer wont understand which files to add, the paths in your config-file isn't a way to require them!
For example you could load a main.js file from your config file and in main define all your files (not optimal but just a an example).
In the bottom of your config-file:
// Load the main app module to start the app
requirejs(["main"]);
The main.js-file: (just adding jquery to show the technique.
define(["jquery"], function($) {});
I might also recommend gulp-requirejs-optimize instead, mainly because it adds the minification/obfuscation functions gulp-requirejs lacks: https://github.com/jlouns/gulp-requirejs-optimize
How to implement it:
var requirejsOptimize = require('gulp-requirejs-optimize');
gulp.task('requirejsoptimize', function () {
return gulp.src('src/js/require.config.js')
.pipe(requirejsOptimize(function(file) {
return {
baseUrl: "src/js",
mainConfigFile: 'src/js/require.config.js',
paths: {
requireLib: "vendor/require/require"
},
include: "requireLib",
name: "require.config",
out: "dist/js/bundle2.js"
};
})).pipe(gulp.dest(''));
});
Ok, I've been reading a lot of questions and answers about this, and a lot of it is rubbish.
I have a very simple question. How do I do the equivalent of this:
require.config({
paths: {
"blah": '/libs/blah/blah',
}
});
require(['blah'], function(b) {
console.log(b);
});
In typescript?
This doesn't work:
declare var require;
require.config({
paths: {
"blah": '/libs/blah/blah',
}
});
import b = require('blah');
console.log(b);
s.ts(8,1): error TS2071: Unable to resolve external module ''blah''.
s.ts(8,1): error TS2072: Module cannot be aliased to a non-module type.
error TS5037: Cannot compile external modules unless the '--module' flag is provided.
Compiling with the --module flag, with a dummy blah.ts shim compiles, but the output is:
define(["require", "exports", 'blah'], function(require, exports, b) {
require.config({
paths: {
"blah": '/libs/blah/blah'
}
});
console.log(b);
});
Looks like it might work, but actually no, the require.config is inside the require block, it is set after it is already needed.
SO! I've ended up so far with this as a solution:
class RequireJS {
private _r:any = window['require'];
public config(config:any):void {
this._r['config'](config);
}
public require(reqs:string[], callback:any):void {
this._r(reqs, callback);
}
}
var rjs = new RequireJS();
rjs.config({
paths: {
"jquery": '/libs/jquery/jquery',
"slider": '/js/blah/slider'
}
});
rjs.require(['slider'], function(slider) {
console.log(slider);
});
Which seems terrible.
So be clear, inside modules that depend on each other, this sort of thing works perfectly fine:
import $ = require('jquery');
export module blah {
...
}
I just need a proper way to setting the requirejs config at a top level, so that the imported paths for the various named modules are correct.
(...and this is because, largely, 3rd party dependencies are resolved using bower, and installed in the /lib/blah, where as the shim files I have for their definitions are in src/deps/blah.d.ts, so the default import paths are incorrect after moving the generated modules files into /js/ on the site)
NB. I've mentioned jquery here, but the problem is not that jquery doesn't work property as an AMD module; I have a shim jquery.ts.d file for this. The issue here is the requirejs paths.
Yesterday I wrote up a solution to this exact issue on my blog - http://edcourtenay.co.uk/musings-of-an-idiot/2014/11/26/typescript-requirejs-and-jquery:
TL;DR - create a config file config.ts that looks something like:
requirejs.config({
paths: {
"jquery": "Scripts/jquery-2.1.1"
}
});
require(["app"]);
and ensure your RequireJS entry point points to the new config file:
<script src="Scripts/require.js" data-main="config"></script>
You can now use the $ namespace from within your TypeScript files by simply using
import $ = require("jquery")
Hope this helps
This post is 3 years old, and there's a lot of changes that have been made when using Typescript. Anyway, after some search on the web,some research on TS documentation-these guys made some good job, I found something that could help.
so this can apply to the latest current of TS (2.2.1)
you probably know that you can use
npm install --save #types/jquery
do the same for your 3rd party JS librairies such as require
now you need to define what your TypeScript Compiler has to do, and how, so create a tsconfig.json file that contains:
// tsconfig.json file
{
"compilerOptions": {
"allowJs": true,
"baseUrl": "./Scripts",//Same as require.config
"module": "amd",
"moduleResolution": "Node",//to consider node_modules/#types folders
"noImplicitAny": false,
"target": "es5", // or whatever you want
"watch": true
}
now, let's focus on require's configuration
// require-config.ts
declare var require: any;
require.config({
baseUrl: "./Scripts/",
paths: {
"jquery": "vendor/jquery/jquery.x.y.z"
// add here all the paths your want
// personnally, I just add all the 3rd party JS librairies I used
// do not forget the shims with dependencies if needed...
}
});
so far so good
now focus on your module written in TS that would use jquery and that is located in Scripts/Module folder:
// module/blah.ts
/// <amd-module name="module/blah" />
import $ = require("jquery");
export function doSomething(){
console.log("jQuery version :", $.version);
}
So this answer looks the same as Ed Courtenay's, doesn't it?
and user210757 mentioned that it does NOT work!!!
and it does not! if you type in your console tsc -w --traceResolution, you'll see that tsc cannot find any definition for jquery.
Here's how to alleviate assuming you previously launch npm install --save #types/jquery by doing this, in a folder named node_modules\#types, you should get the TS definition for jquery
select the package.json file in jquery subfolder
look for the "main" property
set it to "jquery", the same as the alias you are using in your require.config
and done! your module would be transpiled as
define("module/blah", ["require", "exports", "jquery"], function (require, exports, $) {
"use strict";
Object.defineProperty(exports, "__esModule", { value: true });
function doSomething() {
console.log("jQuery version:", $.fn.jQuery);
}
exports.doSomething = doSomething;
});
and that JS code looks good to me!
I just don't like the fact that our module dependencies list "require" & "exports", that sounds like a TS issue, but anyway IT WORKS!
if you want to use import for javascript modules you need to tell typescript about it so,
declare var require;
require.config({
paths: {
"blah": '/libs/blah/blah',
}
});
// Important, place in an external.d.ts:
declare module 'blah'{
// your opportunity to give typescript more information about this js file
// for now just stick with "any"
var mainobj:any;
export = mainobj;
}
import b = require('blah');
console.log(b);
alternatively you could simply do:
var b = require('blah'); and it should work as well