gulp任务完成所有文件后运行代码

Question

因此，我一直在尝试Gulp，看看它在速度上与Grunt相比如何，我对结果印象深刻，但是我有一件事我不知道如何在Gulp中进行。

因此，我有这个任务来缩小HTML：

gulp.task('html-minify', function() {
  var files = [
    relativePaths.webPath + '/*.html',
    relativePaths.webPath + '/components/**/*.html',
    relativePaths.webPath + '/' + relativePaths.appPath + '/components/**/*.html'
  ];

  var changedFiles = buildMetaData.getChangedFiles(files);

  //TODO: needs to execute only after successful run of the task
  buildMetaData.addBuildMetaDataFiles(changedFiles);
  buildMetaData.writeFile();
  return gulp.src(changedFiles, {
      base: relativePaths.webPath
    })
    .pipe(filelog())
    .pipe(minifyHtml({
      empty: true,
      quotes: true,
      conditionals: true,
      comments: true
    }))
    .pipe(gulp.dest(relativePaths.webPath + '/' + relativePaths.appPath +  '/' + relativePaths.buildPath));
});

buildMetaData对象具有我需要的自定义功能，以及为什么不能使用像gulp-
changed这样的插件。我要弄清楚的是，在缩小完成后如何（如果可能）运行代码块来处理所有文件并成功运行。gulp有可能发生这种情况吗？

Answer 1

您可以根据以下条件完成一项任务html-minify：

gulp.task('other-task', ['html-minify'], function() {
  //stuff
});

您还可以end在html-minify任务中监听流事件：

gulp.task('html-minify', function(done) {
  var files = [
    relativePaths.webPath + '/*.html',
    relativePaths.webPath + '/components/**/*.html',
    relativePaths.webPath + '/' + relativePaths.appPath + '/components/**/*.html'
  ];

  var changedFiles = buildMetaData.getChangedFiles(files);

  //TODO: needs to execute only after successful run of the task
  buildMetaData.addBuildMetaDataFiles(changedFiles);
  buildMetaData.writeFile();
  var stream = gulp.src(changedFiles, {
      base: relativePaths.webPath
    })
    .pipe(filelog())
    .pipe(minifyHtml({
      empty: true,
      quotes: true,
      conditionals: true,
      comments: true
    }))
    .pipe(gulp.dest(relativePaths.webPath + '/' + relativePaths.appPath +  '/' + relativePaths.buildPath));

  stream.on('end', function() {
    //run some code here
    done();
  });
  stream.on('error', function(err) {
    done(err);
  });
});