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:
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component-a that has a gulpfile. One of its tasks (eg. build) builds the component and creates a combined js file in dist foldercomponent-b that has a gulpfile. One of its tasks (eg. build) builds the component and creates a combined js file in dist folder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?
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())
})
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...
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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