The very simple and efficient solution that I found out to perform tasks one after the other(when one task gets completed then second task will be initiated) (providing just an example) is : gulp.task('watch', () => gulp.watch(['src/**/*.css', 'src/**/*.pcss'], gulp.series('build',['copy'])) ); This means when you need to run the first-task before second-task, you need to write the second task(copy in this case) in square brackets. NOTE There should be round parenthesis externally for the tasks(until you want them to occur simultaneously)

默认情况下，gulp同时运行任务，除非它们有显式的依赖关系。这对于像clean这样的任务不是很有用，在这些任务中，您不希望依赖它们，但您需要在所有其他任务之前运行它们。

我专门用gulp编写了run-sequence插件来解决这个问题。安装后，像这样使用它:

var runSequence = require('run-sequence');

gulp.task('develop', function(done) {
    runSequence('clean', 'coffee', function() {
        console.log('Run something else');
        done();
    });
});

您可以在README包上阅读完整的说明-它还支持同时运行一些任务集。

请注意，这将在gulp的下一个主要版本中(有效地)修复，因为他们完全消除了自动依赖顺序，并提供类似于run-sequence的工具，允许您手动指定您想要的运行顺序。

然而，这是一个重大的突破性变化，所以当您现在可以使用run-sequence时，没有理由等待。

这个问题的唯一好的解决方案可以在gulp文档中找到:

var gulp = require('gulp');

// takes in a callback so the engine knows when it'll be done
gulp.task('one', function(cb) {
  // do stuff -- async or otherwise
  cb(err); // if err is not null and not undefined, the orchestration will stop, and 'two' will not run
});

// identifies a dependent task must be complete before this one begins
gulp.task('two', ['one'], function() {
  // task 'one' is done now
});

gulp.task('default', ['one', 'two']);
// alternatively: gulp.task('default', ['two']);

Gulp和Node使用承诺。

所以你可以这样做:

// ... require gulp, del, etc

function cleanTask() {
  return del('./dist/');
}

function bundleVendorsTask() {
  return gulp.src([...])
    .pipe(...)
    .pipe(gulp.dest('...'));
}

function bundleAppTask() {
  return gulp.src([...])
    .pipe(...)
    .pipe(gulp.dest('...'));
}

function tarTask() {
  return gulp.src([...])
    .pipe(...)
    .pipe(gulp.dest('...'));
}

gulp.task('deploy', function deployTask() {
  // 1. Run the clean task
  cleanTask().then(function () {
    // 2. Clean is complete. Now run two tasks in parallel
    Promise.all([
      bundleVendorsTask(),
      bundleAppTask()
    ]).then(function () {
      // 3. Two tasks are complete, now run the final task.
      tarTask();
    });
  });
});

如果返回gulp流，则可以使用then()方法添加回调。或者，您可以使用Node的本机Promise创建自己的Promise。在这里，我使用Promise.all()来获得一个回调，当所有promise都解决时触发。

试试这个技巧:-) 吞咽v3。x针对异步错误的Hack

我在Readme中尝试了所有的“官方”方法，它们都不适合我，但是这个方法管用。你也可以升级到gulp 4。x，但是我强烈建议你不要这样做，这样会弄坏很多东西。你可以使用一个真正的js承诺，但嘿，这是快速，肮脏，简单:-) 基本上你可以使用:

var wait = 0; // flag to signal thread that task is done
if(wait == 0) setTimeout(... // sleep and let nodejs schedule other threads

看看这个帖子!

The very simple and efficient solution that I found out to perform tasks one after the other(when one task gets completed then second task will be initiated) (providing just an example) is : gulp.task('watch', () => gulp.watch(['src/**/*.css', 'src/**/*.pcss'], gulp.series('build',['copy'])) ); This means when you need to run the first-task before second-task, you need to write the second task(copy in this case) in square brackets. NOTE There should be round parenthesis externally for the tasks(until you want them to occur simultaneously)