Node.js Gulp 能否向 Docker 发出增量构建的第一次迭代完成的信号

例如，在下面的 case 中，只有在构建了FrontServerEntryPoint之后，才必须启动 node .

这意味着您需要两个步骤:

• 一个用于建筑(不含watcher个)
• 一个用于并行运行

即:

gulp.task('initialBuild', ['css-libs', 'scripts']);

gulp.task('developmentBuild', ['initialBuild', 'browser-sync'], function() {
    gulp.watch('app/sass/**/*.sass', ['sass']);
    gulp.watch('app/*.html', browserSync.reload);
    gulp.watch(['app/js/common.js', 'app/libs/**/*.js'], browserSync.reload);
});

你可以用gulp initialBuild && (gulp developmentBuild & docker-compose up)来运行它

The 'developmentBuild' task has 'initialBuild' and 'browser-sync' as its dependencies.
This means that 'initialBuild' and 'browser-sync' tasks will run before the 'developmentBuild' task starts executing the function containing the watchers.

作为替代方法，您可以try 在Node.js中使用child_process.exec(npm install child_process之后)来运行初始构建，然后并行运行Docker Compose和Gulp Watcher.

node run-build-and-compose.js:

const { exec } = require('child_process');

exec('gulp initialBuild', (error, stdout, stderr) => {
  if (error) {
    console.error(`initialBuild error: ${error}`);
    return;
  }
  console.log(`initialBuild output: ${stdout}`);
  console.error(`initialBuild stderr: ${stderr}`);

  // Run the Gulp developmentBuild task and Docker Compose in parallel
  const devBuild = exec('gulp developmentBuild');
  devBuild.stdout.on('data', (data) => {
    console.log(`developmentBuild output: ${data}`);
  });
  devBuild.stderr.on('data', (data) => {
    console.error(`developmentBuild stderr: ${data}`);
  });

  const dockerCompose = exec('docker-compose up');
  dockerCompose.stdout.on('data', (data) => {
    console.log(`Docker Compose output: ${data}`);
  });
  dockerCompose.stderr.on('data', (data) => {
    console.error(`Docker Compose stderr: ${data}`);
  });
});