gulp-run doesn't work when child process changes working directory

[mirik123]

The problem occurs because the child gulp script requires working directory change: gulp --gulpfile "<...>\gulpfile.js" build [13:40:01] Working directory changed to <...> [13:40:03] Using gulpfile <...>\gulpfile.js [13:40:03] Starting 'build'... [13:40:03] Finished 'build' after 6.42 μs

This works:

const cp = require('cp');
gulp.task('gg2', function (done) {
    var gulpfile = path.join(gulpcwd, 'gulpfile.js'); 
    var subShell = cp.spawn('cmd.exe', ['/C', 'gulp', '--gulpfile', gulpfile, '--cwd', gulpcwd, 'build'], {
        env: process.env,
        cwd: process.cwd()
    });

    subShell.stdout.pipe(process.stdout);
    subShell.stderr.pipe(process.stderr);
    subShell.once('exit', function (code) {
            done(code);       
    });
});

This doesn’t work:

const run = require('gulp-run');
gulp.task('gg2', function (done) {
    var gulpfile = path.join(gulpcwd, 'gulpfile.js');
    return run('gulp --gulpfile "' + gulpfile + '" --cwd "' + gulpcwd + '" build', { verbosity: 3 })
        .exec(function (err) {
            done(err);
        }); 
});

In latter case the child process sends a standard gulp error that the gulp file is not found.