Running a shell command from gulp

Question

I would like to run a shell command from gulp, using gulp-shell. I see the following idiom being used the gulpfile.

Is this the idiomatic way to run a command from a gulp task?

var cmd = 'ls';
gulp.src('', {read: false})
    .pipe(shell(cmd, {quiet: true}))
    .on('error', function (err) {
       gutil.log(err);
});

score 116 · Accepted Answer · answered Apr 08 '15 at 12:21

116

gulp-shell has been blacklisted. You should use gulp-exec instead, which has also a better documentation.

For your case it actually states:

Note: If you just want to run a command, just run the command, don't use this plugin:

var exec = require('child_process').exec;

gulp.task('task', function (cb) {
  exec('ping localhost', function (err, stdout, stderr) {
    console.log(stdout);
    console.log(stderr);
    cb(err);
  });
})

answered Apr 08 '15 at 12:21

ddprrt

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`gulp-shell has been blacklisted`: What do you mean by that? – Laoujin Sep 08 '15 at 11:52
29

Gulp features a blacklist of packages that aren't written in "the Gulp way". Check out the list here: https://github.com/gulpjs/plugins/blob/master/src/blackList.json – ddprrt Sep 08 '15 at 12:13
9

The function passed to `exec` doesn't log anything to Gulp. Why isn't that happening? – IIllIIll Jan 18 '16 at 16:13
6

what did you mean in your answer "just run the command".. how can I run a command through a gulp task without such a plugin? (must be a task because gulp.watch is calling it on every file change – vsync Feb 15 '16 at 16:49
Task: Yes. Plugin: No. You don't have to run gulp plugins inside a gulp task. See the task definition below my Note. This is everything you need. – ddprrt Feb 16 '16 at 13:34
4

Update: Still doesn't log anything. – IIllIIll May 27 '16 at 16:20
@Arcrammer got the same issue as you – Maria Ines Parnisari Aug 13 '16 at 22:43
You can't use plain 'child_process' exec() with src(). So if you'd like to use pipe(), you may still need this plugin. – kervin Jan 05 '17 at 22:46
I can't see how gulp-exec's documentation is better... Is there some place I'm not finding? – martixy Jan 18 '17 at 16:47
Is there a way to automatically "flag" blacklisted plugins? – jchook Apr 04 '17 at 03:01
1

@jchook try this: https://gist.github.com/ddprrt/6178d3ce2ae66f97c3689622c4834bdd (untested) – ddprrt Apr 13 '17 at 09:22
@TomaszWaszczyk Try [exec-chainable](https://www.npmjs.com/package/exec-chainable) – cjsimon Apr 08 '18 at 02:24

score 44 · Answer 2 · answered Oct 14 '15 at 18:23

44

The new way to do this that keeps console output the same (e.g., with colors):

see: https://nodejs.org/api/child_process.html#child_process_child_process_spawn_command_args_options

var gulp = require('gulp');
var spawn = require('child_process').spawn;

gulp.task('my-task', function (cb) {
  var cmd = spawn('cmd', ['arg1', 'agr2'], {stdio: 'inherit'});
  cmd.on('close', function (code) {
    console.log('my-task exited with code ' + code);
    cb(code);
  });
});

answered Oct 14 '15 at 18:23

AutoSponge

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But be careful if you are on Windows, as spawn() doesn't work reliably, see [this Node issue](https://github.com/nodejs/node-v0.x-archive/issues/2318) – Rich Tebb Jan 28 '16 at 16:27
1

This is the best answer imo +1 – chukkwagon Nov 20 '17 at 14:48
2

this solution doesn't work for me on windows, but found another one here - https://stackoverflow.com/questions/45841902/how-to-run-ng-build-from-gulp-using-child-process-spawn-or-disable-all-output-in – Sergey Apr 03 '18 at 13:27
Excellent answer! I found & created a more verbose method (https://gist.github.com/rmckeel/b4e60922f5098ced9c50bdd96731b34a/ccca5069825096cb286c46c090312f1b5e0d5594), but I updated that code sample with this inherit stdio concept (https://gist.github.com/rmckeel/b4e60922f5098ced9c50bdd96731b34a), because it is elegant and should work for most purposes. Thanks! – ryanm May 07 '18 at 14:46

score 17 · Answer 3 · edited Jun 05 '19 at 13:36

17

With gulp 4 your tasks can directly return a child process to signal task completion:

'use strict';

var cp = require('child_process');
var gulp = require('gulp');

gulp.task('reset', function() {
  return cp.execFile('git checkout -- .');
});

gulp-v4-running-shell-commands.md

edited Jun 05 '19 at 13:36

elad.chen

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answered Jul 18 '16 at 14:38

birnbaum

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Gulp 4 is the modern incarnation of Zeno's Paradox. – Ken Smith Dec 09 '16 at 07:31
1

Newer Gulp versions don't use `gulp.task` like this. If it doesn't work for you, try `cp.exec` instead of `cp.execFile` – Flimm Apr 28 '21 at 10:06

score 1 · Answer 4 · answered Oct 24 '19 at 11:17

You could simply do this:

const { spawn } = require('child_process');
const gulp = require('gulp');

gulp.task('list', function() {
    const cmd = spawn('ls');
    cmd.stdout.on('data', (data) => {
        console.log(`stdout: ${data}`);
    });
    return cmd;
});

Flimm · Answer 5 · 2021-04-28T10:40:58.743

According to Gulp's documentation, you can run a command like echo like this:

const cp = require('child_process');

function childProcessTask() {
  return cp.exec('echo *.js');
}

exports.default = childProcessTask;

exec takes one string that will be parsed by the shell, and it silences output by default.

I personally like using child_process.spawn instead. spawn takes the name of the command, and then a list of arguments. If you use the option stdio: 'inherit' it won't swallow its output.

function childProcessTask() {
  return cp.spawn('echo', ['one', 'two'], {stdio: 'inherit'});
}

References:

Gulp documentation: Returning a child process
Node documentation: child_process.exec(command, options, callback)
Node documentation: child_process.spawn(command, args, options)

Running a shell command from gulp

5 Answers5

References:

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