从一个 gulpfile.js 从另一个 gulpfile.js 运行 gulp 任务

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

也许我的方法有问题，但我有以下情况:

Perhaps it's something wrong with my approach but I have a following situation:

  1. 我有一个包含 gulpfile 的 component-a.它的一项任务(例如构建)构建组件并在 dist 文件夹中创建一个组合的 js 文件
  2. 我有一个包含 gulpfile 的 component-b.它的一项任务(例如构建)构建组件并在 dist 文件夹中创建一个组合的 js 文件
  3. 我有一个使用这两个组件的项目.这个项目也有一个 gulpfile，我想在其中编写一个任务:
    • 从/components/component-a/gulpfile.js 执行构建任务
    • 从/components/component-b/gulpfile.js 执行构建任务
    • concats/components/component-a/dist/build.js 和/components/component-b/dist/build.js(我知道怎么做)
  1. I have a component-a that has a gulpfile. One of its tasks (eg. build) builds the component and creates a combined js file in dist folder
  2. I have a component-b that has a gulpfile. One of its tasks (eg. build) builds the component and creates a combined js file in dist folder
  3. I have a project that uses both components. This project has a gulpfile as well and in it I would like to write a task that:
    • executes build task from /components/component-a/gulpfile.js
    • executes build task from /components/component-b/gulpfile.js
    • concats /components/component-a/dist/build.js and /components/component-b/dist/build.js (I know how to do this)

我不知道如何从/components/component-?/gulpfile.js 执行构建任务.是否有可能或者我应该以其他方式处理这种情况?

What I don't know is how to execute the build task from /components/component-?/gulpfile.js. Is it even possible or I should deal with this situation otherwise?

推荐答案

require('child_process').spawn;

使用 Node 的 child_process#spawn<从不同的目录运行 Gulpfile 非常简单/code> 模块.

尝试根据您的需要调整以下内容:

Try adapting the following to your needs:

// Use `spawn` to execute shell commands with Node
const { spawn } = require('child_process')
const { join } = require('path')

/*
  Set the working directory of your current process as
  the directory where the target Gulpfile exists.
*/
process.chdir(join('tasks', 'foo'))

// Gulp tasks that will be run.
const tasks = ['js:uglify', 'js:lint']

// Run the `gulp` executable
const child = spawn('gulp', tasks)

// Print output from Gulpfile
child.stdout.on('data', function(data) {
    if (data) console.log(data.toString())
})

咕噜咕噜

虽然使用 gulp-chug 是解决此问题的一种方法，但 它已被 gulp 的维护者列入黑名单 因为...

gulp-chug

Although using gulp-chug is one way to go about this, it has been blacklisted by gulp's maintainers for being...

执行，太复杂，只是将 gulp 用作 globber"

"execing, too complex and is just using gulp as a globber"

官方黑名单声明...

没有理由存在，使用 require-all 模块或节点的 require"

"no reason for this to exist, use the require-all module or node's require"

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