Perhaps it's something wrong with my approach but I have a following situation:
component-a
that has a gulpfile. One of its tasks (eg. build) builds the component and creates a combined js file in dist folder component-b
that has a gulpfile. One of its tasks (eg. build) builds the component and creates a combined js file in dist folder 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?
Running a Gulpfile from a different directory is quite simple with Node's child_process#spawn
module .
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())
})
Although using gulp-chug
is one way to go about this, it has been blacklisted by gulp
's maintainers for being...
"execing, too complex and is just using gulp as a globber"
The official blacklist states...
"no reason for this to exist, use the require-all module or node's require"
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