如何从另一个 gulp file.js 导入所有任务

2022-01-12 00:00:00 node.js javascript gulp

是否可以有一个主要的 gulpfile.js 来调用其他 gulp files.js 的任务?将子 gulpfile.js 简单地要求"到主文件中是行不通的.我有一个平台项目，其中包括几个带有单独 gulpfile 的子项目，所以我需要一个解决方案来管理主要项目中的所有子 gulpfile

Is it possible to have one main gulpfile.js from which to call tasks from other gulp files.js? Simple "require" of child gulpfile.js into main one doesn't work. I have a platform project which includes several sub projects with separate gulpfiles, so I need a solution to manage all child gulpfiles from within main one

推荐答案

如果我想让它面向未来并且不想安装怎么办另一个包.

And what if I want to make it future-proof and don't want to install another package for it.

以下内容适用于 gulp 4，没有任何额外的插件.

The following works for me with gulp 4, without any extra plugins.

在taskfile.js中:

const { src, dest } = require('gulp');

const mytask = function () {
  return src('assets/**/*')
    .pipe(dosomething())
    .pipe(dest('dest');
}

module.exports = {
  mytask
}

在 gulpfile.js 中:

const { mytask } = require('taskfile.js');

// use in other tasks
gulp.task('manythings', gulp.series(..., mytask, ...));

// or use directly as 'gulp mytask'
module.exports = {
  mytask
}

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