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JavaScript Gulp Basics Compile Sass with Gulp Add Source Maps to your SaSS

Simon Merrick
Simon Merrick
18,305 Points

Will the source map still work if I minify the css output?

For my project, I also want to minify the css after it has been compiled from sass. Will the source map still work after minification or would I have have to do the minification in the same task that I compile the sass, in between maps.init() and maps.write('./').

What about javascript?

3 Answers

Sam Donald
Sam Donald
36,305 Points

You can do it from a single task

If you're going to serve minified files there isn't really any reason to generate individual css files for each sass file, generate maps for each, and then compress to a single minified file that further extends it's map to these individual css files.


If you need to make changes you're still going to make them in your sass files. Because any changes to these generated css files would just be overridden/lost the next time you run these Gulp tasks (i.e. rebuild).


Just concatenate all your sass into one minified css file that gets exported to production. But still maps to each individual sass file.

const gulp = require("gulp"),
      sass = require("gulp-sass"),
      rename = require("gulp-rename"),
      minCss = require("gulp-clean-css"),
      maps = require("gulp-sourcemaps");

gulp.task("compileSass", function(){
    return  gulp.src("_sass/main.scss")
Ryan Field
Ryan Field
Courses Plus Student 21,241 Points

Hi, Simon. I believe it will, but because minification removes space/line breaks, I think you'll only see filename.scss:1 because technically it will all be on one line.

Simon Merrick
Simon Merrick
18,305 Points

I have done a bit more work on my gulp build pipeline and managed to resolve my problem, at least when dealing with my scripts. I found that passing {loadMaps: true} into sourcemaps.init() during the minify task seems to make sourcemaps use the first sourcemap as a reference to relate the minified file back to the very orignal source file. Bellow is an extract of my scripts section of my gulp build process

  gulp.task("concatScripts", ['lintScripts'], function() {
    return gulp.src([
      config.paths.dep + '/jquery-1.11.3/jquery-1.11.3.js',
      config.paths.dep + '/bootstrap/bootstrap.js',
      config.paths.src + '/js/plugins.js',
      config.paths.src + '/js/script.js'
    .pipe(gulp.dest(config.paths.dist + '/js'));

  gulp.task("minifyScripts", ["concatScripts"], function() {
    return gulp.src(config.paths.dist + "/js/script.js")
    .pipe($.sourcemaps.init({loadMaps: true}))
    .pipe(gulp.dest(config.paths.dist + '/js'));

I can't say this is perfect, but...

When compiling scss partials to css, if you init your maps in that process and then load the maps (as in the previous example) during the minification process, the final maps will show where your source scss rules are (partial name and line).

Here's my working version:

    return gulp.src('./scss/style.scss')
        .pipe(maps.init())          // create maps from scss partials
        .pipe(maps.write('./'))     // this path is relative to the output directory
        .pipe(gulp.dest('./css'));  // this is the output directory

// a task in here called concat-css  concats all css files and changes name to output.css

gulp.task('minify-css',['concat-css'], function(){
    return gulp.src('css/output.css')
        .pipe(maps.init({loadMaps:true}))   // create maps from scss *sourcemaps* not the css