I want to add some bash commands at the end of gulp.watch
function to accelerate my development speed. So, I am wondering if it is possible. Thanks!
I would go with:
var spawn = require('child_process').spawn; var fancyLog = require('fancy-log'); var beeper = require('beeper'); gulp.task('default', function(){ gulp.watch('*.js', function(e) { // Do run some gulp tasks here // ... // Finally execute your script below - here "ls -lA" var child = spawn("ls", ["-lA"], {cwd: process.cwd()}), stdout = '', stderr = ''; child.stdout.setEncoding('utf8'); child.stdout.on('data', function (data) { stdout += data; fancyLog(data); }); child.stderr.setEncoding('utf8'); child.stderr.on('data', function (data) { stderr += data; fancyLog.error(data)); beeper(); }); child.on('close', function(code) { fancyLog("Done with exit code", code); fancyLog("You access complete stdout and stderr from here"); // stdout, stderr }); }); });
Nothing really "gulp" in here - mainly using child processes http://nodejs.org/api/child_process.html and spoofing the result into fancy-log
Use https://www.npmjs.org/package/gulp-shell.
A handy command line interface for gulp
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