How to run Gulp tasks sequentially one after the other

0 votes

In the 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. Any Help?.

Jul 12, 2020 in Node-js by kartik
• 37,530 points
424 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.

Run-sequence plugin specifically to fix this issue with gulp. 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();
    });
});

Hope it helps!!

Thank You!

answered Jul 12, 2020 by Niroj
• 82,880 points

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