Gulp 任务中`return` 语句和`callback` 参考的用途

我无法让 gulp run-sequence 执行我给它的两个函数(它只会执行其中一个.我有这样的事情:

gulp.task('wire', function(){gulp.src('./index.html').pipe(wiredep()).pipe(gulp.dest('.'));});var filesToInject = ['./app/**/*.js', './app/**/*.css'];gulp.task('注入', function (){var target = gulp.src('./index.html');var sources = gulp.src(filesToInject, {read: false});返回target.pipe(注入(来源)).pipe(gulp.dest('.'));});gulp.task('wire-inject', function(callback){序列('wire'，'inject'，回调)；});

那只会运行 inject 任务.只有当我在 wire 任务中的 gulp.src 之前添加一个 return 时，它才会执行.我还在两个函数参数中添加了回调，并将其传递给每个 文档.

gulp.task('wire', function(callback){gulp.src('./index.html').pipe(wiredep()).pipe(gulp.dest('.'), 回调);});var filesToInject = ['./app/**/*.js', './app/**/*.css'];gulp.task('inject', 函数(回调){var target = gulp.src('./index.html');var sources = gulp.src(filesToInject, {read: false});返回target.pipe(注入(来源)).pipe(gulp.dest('.'), 回调);});

这并没有引发任何错误，但是如果我把它拿出来它不会改变任何东西.这个 return 语句做了什么神奇地使我的序列完全运行.文档中的这些回调是什么，我只是通过不带括号的引用来执行它们?即使它们看似什么都不做，我是否仍然需要它们?

Gulp 需要知道异步任务何时完成.您有 3 种方法可以做到这一点:

1. 返回一个事件流，这就是当你的任务函数中有 return gulp.src(...).pipe(...); 时你所做的.p>

2. 返回一个承诺.

3. 让定义你任务的函数带一个参数.Gulp 将使用回调函数调用您的函数，您应该在任务结束时调用该回调函数.

如果您不这样做，那么 Gulp 将不会知道该任务是异步的.因此，它会在其函数返回后立即认为任务已完成.这会带来各种后果.Gulp 可以在任务完成之前退出.或者它可以在任务 A 完成之前启动依赖于任务 A 的任务 B.在某些情况下，您可能立即看不到问题，但随着您的 gulpfile 变得越来越复杂，您出现不良行为的机会也越来越多.run-sequence 并不特殊:它需要知道任务何时完成才能正确完成工作.

请注意，您在问题中显示的此代码中的 callback 是无用的:

return target.pipe(inject(sources)).pipe(gulp.dest('.'), 回调);

callback 不会被调用.将其替换为 function () { console.log("Hello!");打回来();}.您不会在控制台上看到 Hello!，因为未调用回调.但是，代码确实可以正常工作，因为您返回了流.Gulp 使用您返回的流来确定任务结束，而不是回调(无论如何都不会调用).

I was having trouble getting gulp run-sequence to execute both functions that i gave it (it would only execute one of them. I had something like this:

gulp.task('wire', function(){
  gulp.src('./index.html')
    .pipe(wiredep())
    .pipe(gulp.dest('.'));
});

var filesToInject = ['./app/**/*.js', './app/**/*.css'];
gulp.task('inject', function (){
  var target = gulp.src('./index.html');
  var sources = gulp.src(filesToInject, {read: false});
  return target.pipe(inject(sources))
    .pipe(gulp.dest('.'));
});

gulp.task('wire-inject',  function(callback){
    sequence('wire', 'inject', callback);
});

That would only run the inject task. Only when I added a return before gulp.src in the wire task would it execute. I also added callback in both function params and passed it to the last pipe in both tasks per the documentation.

gulp.task('wire', function(callback){
  gulp.src('./index.html')
    .pipe(wiredep())
    .pipe(gulp.dest('.'), callback);
});

var filesToInject = ['./app/**/*.js', './app/**/*.css'];
gulp.task('inject', function (callback){
  var target = gulp.src('./index.html');
  var sources = gulp.src(filesToInject, {read: false});
  return target.pipe(inject(sources))
    .pipe(gulp.dest('.'), callback);
});

That did not throw any errors, however it doesn't change anything if I take it out. What does this return statement do that magically makes my sequence run completely. What are these callbacks in the documentation that I just pass a reference to with out parentheses to execute them? Do I still need them even though they seemingly do nothing?

return target.pipe(inject(sources))
    .pipe(gulp.dest('.'), callback);