Gulp execute commands without plugin

solution for executing shell commands within gulp in NodeJS.

Using child_process.exec in gulp with gulp.watch

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

Below codes for watching files and run shell commands every file changes.

const { exec } = require('child_process');
const gulp = require('gulp');
const { join } = require('path');

gulp.task('default', function () {
  gulp.watch(join(__dirname, 'src/**/*.ts'), function (done) {
    const child = exec('cross-env NODE_ENV=development DEBUG=prerender-it* run-s build test', function () {
      done();
    });
    child.stdout.pipe(process.stdout);
    child.stderr.pipe(process.stderr);
  });
  gulp.watch(join(__dirname, 'test/src/**/*.{ts,tsx}'), function (done) {
    const child = exec('cross-env NODE_ENV=development DEBUG=prerender-it* run-s test-build', function () {
      done();
    });
    child.stdout.pipe(process.stdout);
    child.stderr.pipe(process.stderr);
  });
});

Result using exec

image

References: