How to run Gulp tasks sequentially one after the other

0 votes

Snippet like this:

gulp.task "coffee", ->
    gulp.src("src/server/**/*.coffee")
        .pipe(coffee {bare: true}).on("error",gutil.log)
        .pipe(gulp.dest "bin")

gulp.task "clean",->
    gulp.src("bin", {read:false})
        .pipe clean
            force:true

gulp.task 'develop',['clean','coffee'], ->
    console.log "run something else"

In develop task I want to run clean and after it's done, run coffee and when that's done, run something else. But I can't figure that out. This piece doesn't work. Please advise.

Nov 26, 2020 in Node-js by kartik
• 37,510 points
337 views

1 answer to this question.

0 votes

Hello @kartik,

By default, gulp runs tasks simultaneously, unless they have explicit dependencies. This isn't very useful for tasks like clean, where you don't want to depend, but you need them to run before everything else.

After you install it, use it like this:

var runSequence = require('run-sequence');

gulp.task('develop', function(done) {
    runSequence('clean', 'coffee', function() {
        console.log('Run something else');
        done();
    });
});

You can read the full instructions on the package README — it also supports running some sets of tasks simultaneously.

answered Nov 26, 2020 by Niroj
• 82,680 points

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