跳过任务列为饮料中的依赖项 [英] Skipping Tasks Listed as Dependencies in Gulp

问题描述

我已经浏览了所有的文档和NPM，试图找到解决办法，但我没有运气。我希望可以选择跳过在运行特定任务时列为依赖关系的任务。例如，如果我有以下内容：

I have looked all over the documentation and NPM to try to find a solution to this, but I have had no luck. I would like to have the option to skip the tasks that I list as dependencies when running a specific task. For example, if I have the following:

gulp.task('prerun', function(){
  // do cleaning, installation, etc.
});

gulp.task('run', ['prerun'], function(){
  // do stuff
});

gulp.task('watch', function(){
  gulp.watch('glob/glob/**', ['run']);
});

我希望能够让我的 gulp.watch 执行运行，而无需触及 prerun 中涉及的开销。这是所有可能的Gulp？

I would like to be able to have my gulp.watch execute run without having to touch the overhead involved in prerun. Is this at all possible in Gulp?

推荐答案

关于助手任务是什么？我使用这种方法来消除我的监视任务中的任何依赖关系。你的例子可以看起来像这样：

What's about a helper task? I use this approach to eliminate any dependencies in my watch tasks. Your example can look like this:

gulp.task('prerun', function(){
    // do cleaning, installation, etc.
});

gulp.task('run', ['prerun'], function(){
    gulp.start('run-dev');
});

gulp.task('run-dev', function() {
    // do the run stuff
});

gulp.task('watch', function(){
    gulp.watch('glob/glob/**', ['run-dev']);
});

如果需要，还可以将预运行任务用作监视任务的依赖项：

The prerun task you can also use as dependency for your watch task if needed:

gulp.task('watch', ['prerun'], function(){
    gulp.watch('glob/glob/**', ['run-dev']);
});

ciao
Ralf

Ciao Ralf

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