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

The only good solution to this problem can be found in the gulp documentation which can be found here

var gulp = require('gulp');

// takes in a callback so the engine knows when it'll be done
gulp.task('one', function(cb) {
  // do stuff -- async or otherwise
  cb(err); // if err is not null and not undefined, the orchestration will stop, and 'two' will not run
});

// identifies a dependent task must be complete before this one begins
gulp.task('two', ['one'], function() {
  // task 'one' is done now
});

gulp.task('default', ['one', 'two']);
// alternatively: gulp.task('default', ['two']);