問題描述
也許我的方法有問題，但我有以下情況:
Perhaps it's something wrong with my approach but I have a following situation:
- 我有一個包含 gulpfile 的
component-a.它的一項任務(wù)(例如構(gòu)建)構(gòu)建組件并在 dist 文件夾中創(chuàng)建一個組合的 js 文件 - 我有一個包含 gulpfile 的
component-b.它的一項任務(wù)(例如構(gòu)建)構(gòu)建組件并在 dist 文件夾中創(chuàng)建一個組合的 js 文件 - 我有一個使用這兩個組件的項目.這個項目也有一個 gulpfile，我想在其中編寫一個任務(wù):
- 從/components/component-a/gulpfile.js 執(zhí)行構(gòu)建任務(wù)
- 從/components/component-b/gulpfile.js 執(zhí)行構(gòu)建任務(wù)
- concats/components/component-a/dist/build.js 和/components/component-b/dist/build.js(我知道怎么做)
- I have a
component-athat has a gulpfile. One of its tasks (eg. build) builds the component and creates a combined js file in dist folder - I have a
component-bthat has a gulpfile. One of its tasks (eg. build) builds the component and creates a combined js file in dist folder - 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)
我不知道如何從/components/component-?/gulpfile.js 執(zhí)行構(gòu)建任務(wù).是否有可能或者我應(yīng)該以其他方式處理這種情況?
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?
推薦答案
require('child_process').spawn;
使用 Node 的 child_process#spawn<從不同的目錄運行 Gulpfile 非常簡單/code> 模塊.
嘗試根據(jù)您的需要調(diào)整以下內(nèi)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())
})
咕嚕咕嚕
雖然使用 gulp-chug 是解決此問題的一種方法，但 它已被 gulp 的維護者列入黑名單 因為...
gulp-chug
Although using gulp-chug is one way to go about this, it has been blacklisted by gulp's maintainers for being...
執(zhí)行，太復(fù)雜，只是將 gulp 用作 globber"
"execing, too complex and is just using gulp as a globber"
官方黑名單聲明...
沒有理由存在，使用 require-all 模塊或節(jié)點的 require"
"no reason for this to exist, use the require-all module or node's require"
這篇關(guān)于從一個 gulpfile.js 從另一個 gulpfile.js 運行 gulp 任務(wù)的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網(wǎng)！