gulp-typescript

A typescript compiler for gulp with incremental compilation support.

gulp-typescript

A gulp plugin for handling TypeScript compilation workflow. The plugin exposes TypeScript's compiler options to gulp using TypeScript API.

  • Incremental compilation (so faster builds)
  • Error reporting
  • Different output streams for .js, .d.ts files.
  • Support for sourcemaps using gulp-sourcemaps
  • Compile once, and filter different targets
npm install --global gulp
npm install gulp
npm install gulp-typescript
  • out (string) - Generate one javascript and one definition file. Only works when no module system is used.
  • outDir (string) - Move output to a different (virtual) directory. Note that you still need gulp.dest to write output to disk.
  • removeComments (boolean) - Do not emit comments to output.
  • noImplicitAny (boolean) - Warn on expressions and declarations with an implied 'any' type.
  • noLib (boolean) - Don't include the default lib (with definitions for - Array, Date etc)
  • noEmitOnError (boolean) - Do not emit outputs if any type checking errors were reported.
  • target (string) - Specify ECMAScript target version: 'ES3' (default), 'ES5' or 'ES6'.
  • module (string) - Specify module code generation: 'commonjs' or 'amd'.
  • declarationFiles (boolean) - Generates corresponding .d.ts files.
  • isolatedCompilation (boolean) - Compiles files seperately and doesn't check types, which causes a big speed increase. You have to use gulp-plumer and TypeScript 1.5 (or higher, 1.5-beta is not fully working).
  • noExternalResolve (boolean) - Do not resolve files that are not in the input. Explanation below.
  • sortOutput (boolean) - Sort output files. Usefull if you want to concatenate files (see below).
  • typescript (object) - Use a different version / fork of TypeScript (see below). Use it like: typescript: require('typescript') or typescript: require('my-fork-of-typescript')

You can use almost all other options that TypeScript supports too. Only these two are not supported:

  • sourceRoot - Use sourceRoot option of gulp-sourcemaps instead.
  • rootDir - Use base option of gulp.src() instead.

Below is a minimal gulpfile.js which will compile all TypeScript file in folder src and emit a single output file called output.js in built/local. To invoke, simple run gulp.

var gulp = require('gulp');
var ts = require('gulp-typescript');
 
gulp.task('default', function () {
  var tsResult = gulp.src('src/**/*.ts')
    .pipe(ts({
        noImplicitAny: true,
        out: 'output.js'
      }));
  return tsResult.js.pipe(gulp.dest('built/local'));
});

Another example of gulpfile.js. Instead of creating default task, the file specifies custom named task. To invoke, run gulp scripts instead of gulp. As a result, the task will generate both javascript files and typescript definition files.

var gulp = require('gulp');
var ts = require('gulp-typescript');
var merge = require('merge2');  // Require separate installation 
gulp.task('scripts', function() {
  var tsResult = gulp.src('lib/**/*.ts')
    .pipe(ts({
        declarationFiles: true,
        noExternalResolve: true
      }));
 
  return merge([
    tsResult.dts.pipe(gulp.dest('release/definitions')),
    tsResult.js.pipe(gulp.dest('release/js'))
    ]);
});

Instead of calling ts(options), you can create a project first, and then call ts(project). An example:

var gulp = require('gulp');
var ts = require('gulp-typescript');
var merge = require('merge2');
 
var tsProject = ts.createProject({
    declarationFiles: true,
    noExternalResolve: true
});
 
gulp.task('scripts', function() {
    var tsResult = gulp.src('lib/*.ts')
                    .pipe(ts(tsProject));
 
    return merge([ // Merge the two output streams, so this task is finished when the IO of both operations are done. 
        tsResult.dts.pipe(gulp.dest('release/definitions')),
        tsResult.js.pipe(gulp.dest('release/js'))
    ]);
});
gulp.task('watch', ['scripts'], function() {
    gulp.watch('lib/*.ts', ['scripts']);
});

When you run gulp watch, the source will be compiled as usual. Then, when you make a change and save the file, your TypeScript files will be compiled in about half the time.

Make sure you create the project outside of a task! Otherwise it won't work.

To use tsconfig.json, you have to use ts.createProject:

var tsProject = ts.createProject('tsconfig.json');

If you want to add or overwrite certain settings in the tsconfig.json file, you can use:

var tsProject = ts.createProject('tsconfig.json', { sortOutput: true });

The task will look like:

gulp.task('scripts', function() {
    var tsResult = tsProject.src() // instead of gulp.src(...) 
        .pipe(ts(tsProject));
    
    return tsResult.js.pipe(gulp.dest('release'));
});

Note: you can only use tsProject.src() if your tsconfig.json file has a files property. If it doesn't, you should use gulp.src('**/**.ts').

You can use a custom version of TypeScript. Add the version you want (1.4+) to your package.json file as a devDependency. You can also use the master from GitHub to get the latest features. You can use this in your package.json to get the master from GitHub:

{
    "devDependencies": {
        "gulp-typescript": "*",
        "typescript": "Microsoft/TypeScript"
    }
}

And add this to your gulpfile:

[...].pipe(ts({
    typescript: require('typescript')
}));

You can use 1.5.0-beta of TypeScript if you write this in your package.json file: "typescript": "1.5.0-beta"

It's also possible to use a fork of TypeScript. Add an extra option to the options object like this:

[...].pipe(ts({
    typescript: require('my-fork-of-typescript')
}));

There are two ways to filter files:

gulp.task('scripts', function() {
    var tsResult = gulp.src('lib/*.ts')
                       .pipe(ts(tsProject, filterSettings));
    
    ...
});

And

gulp.task('scripts', function() {
    var tsResult = gulp.src('lib/*.ts')
                       .pipe(ts(tsProject));
    
    tsResult.pipe(ts.filter(filterSettings)) ... ;
});

The first example doesn't add files (that don't pass the filter) to the compiler, the second one does add them to the compiler, but removes them later from the stream. You can put as much pipes between compilation and filtering as you want, as long as the filename doesn't change.

At the moment there is only one filter available:

  • referencedFrom (string[]) Only files that are referenced (using /// <reference path="..." />) by the files in this array pass this filter.

By default, gulp-typescript will try to resolve the files you require and reference. These files are parsed, but not emitted (so you will not see them in the output stream).

If you set the option noExternalResolve to true, gulp-typescript will not resolve all the requires and references. It assumes that all the necessary files are in the input stream. For example, if you have your .ts files in the lib folder, and the .d.ts files in the definitions folder, you must use gulp.src(['lib/**.ts', 'definitions/**.ts']) instead of gulp.src(['lib/**.ts']) in your gulpfile if you use the option noExternalResolve.

Advantage of noExternalResolve: faster compilation. Disadvantage of noExternalResolve: won't work when you forgot some input files. Advice: turn it on, and make sure you list all the input files.

Files that are resolved when noExternalResolve is off, won't be pushed to the output stream, unless you are using the out option.

The tsc command has the ability to concatenate using the --out parameter. There are two approaches to do that in gulp-typescript.

You can use the out option. This is fine for small projects, but for big projects it's not always sufficient.

The other option is to use gulp-concat. The tsc command sorts the files using the <reference> tags. gulp-typescript does this when you enable the sortOutput option. You can use the referencedFrom filter to only include files that are referenced from certain files.

Example of gulpfile.js which will compile typescript to javascript as well as generate associated sourcemap.

var gulp = require('gulp')
var ts = require('gulp-typescript');
var concat = require('gulp-concat');
var sourcemaps = require('gulp-sourcemaps');
 
gulp.task('scripts', function() {
    var tsResult = gulp.src('lib/*.ts')
                       .pipe(sourcemaps.init()) // This means sourcemaps will be generated 
                       .pipe(ts({
                           sortOutput: true,
                           // ... 
                       }));
    
    return tsResult.js
                .pipe(concat('output.js')) // You can use other plugins that also support gulp-sourcemaps 
                .pipe(sourcemaps.write()) // Now the sourcemaps are added to the .js file 
                .pipe(gulp.dest('release/js'));
});

For more information, see gulp-sourcemaps.

You can specify a custom reporter as the 3rd argument of the main function:

ts(optionsOrProject, filters, reporter);

You can set options, project or filter to undefined if you don't want to set them. Available reporters are:

  • nullReporter (ts.reporter.nullReporter()) - Don't report errors
  • defaultReporter (ts.reporter.defaultReporter()) - Report basic errors to the console
  • longReporter (ts.reporter.longReporter()) - Extended version of default reporter, intelliJ link functionality + file watcher error highlighting should work using this one
  • fullReporter (ts.reporter.fullReporter(showFullFilename?: boolean)) - Show full error messages, with source.

If you want to build a custom reporter, you take a look at lib/reporter.ts, in that file is an interface which a reporter should implement.

The plugin uses itself to compile. There are 2 build directories, release and release-2. release must always contain a working build. release-2 contains the last build. When you run gulp compile, the build will be saved in the release-2 directory. gulp test will compile the source to release-2, and then it will run some tests. If these tests give no errors, you can run gulp release. The contents from release-2 will be copied to release.

gulp-typescript is licensed under the MIT license.