如何一个接一个地顺序运行Gulp任务

问题描述

如下所示：

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"

在 / code>任务我想运行 clean ，完成后，运行 coffee 运行别的东西。但我不能想出来。这件不工作。请指教。

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.

推荐答案

这个问题的唯一好的解决方案可以在gulp文档找到

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']);

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