如何按顺序一个接一个地运行 Gulp 任务

Question

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. This piece doesn't work. Please advise.

转载于:https://stackoverflow.com/questions/22824546/how-to-run-gulp-tasks-sequentially-one-after-the-other

Answer 1

Try this hack :-) Gulp v3.x Hack for Async bug

I tried all of the "official" ways in the Readme, they didn't work for me but this did. You can also upgrade to gulp 4.x but I highly recommend you don't, it breaks so much stuff. You could use a real js promise, but hey, this is quick, dirty, simple :-) Essentially you use:

var wait = 0; // flag to signal thread that task is done
if(wait == 0) setTimeout(... // sleep and let nodejs schedule other threads

Check out the post!