从另一个gulpfile.js的gulpfile.js中运行gulp任务 [英] Running gulp task from one gulpfile.js from another gulpfile.js

问题描述

也许这是我的方法有问题，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 component-一个有一个大文件的。它的一个任务（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 component -b 有一个大文件。它的一个任务（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。这个项目也有一个大文件，我想写一个任务：
   
   
   
   
   • 从/ components / component-a / gulpfile执行构建任务。 js
   
   
   • 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（我知道如何做到这一点）
   
   

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，或者我应该处理这种情况，否则？

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;

通过从其他目录运行Gulp文件非常简单节点的 child_process＃spawn 模块。

尝试根据需要调整以下内容：

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())
})

巨嘴鸟chug

尽管使用 gulp-chug 是解决这个问题的方法之一，它已被 gulp 的维护人员列入黑名单，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...

过于复杂，只是将gulp当作一个globber使用

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

a href =https://github.com/gulpjs/plugins/blob/master/src/blackList.json#L101 =noreferrer>官方黑名单指出......

The official blacklist states...

没有理由存在，请使用require-all模块或节点require

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

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