Gulp - 使用参数启动任务 [英] Gulp - start task with an argument

问题描述

所以我正在尝试创建一个 gulp 工作流程，并且我想为一些任务实现选项，例如 gulp copy-images --changed.现在，我创建了一个监视任务，它显然监视所有图像文件，它应该使用 --changed 标志启动 copy-images.

So I'm trying to create a gulp workflow and I'd like to implement options for some tasks, like gulp copy-images --changed. Now, I've created a watch task that obviously watches all image files and it should start the copy-images with the --changed flag.

理想情况下，我想做这样的事情:

Ideally, I want to do something like this:

gulp.task('copy-images', function(){
    // some code
});

gulp.task('watch', function(){
    gulp.watch(config.images, ['copy-images --changed']);
});

我也很清楚我可以做到:

I'm also very aware that I could do:

gulp.task('copy-images', function(){
    // some code
});

gulp.task('copy-images-changed', function(){
    // some code
});

gulp.task('watch', function(){
    gulp.watch(config.images, ['copy-images']);
});

但这意味着重复的代码.
任何人有解决方案或一些建议?

but this means duplicate code.
Anyone with a solution or maybe some advice?

提前致谢！

推荐答案

Gulp 不提供指定任务选项的内置方式.您必须使用外部选项解析器模块，例如 yargs.请参阅 this question 了解有关该主题的更多信息.

Gulp does not provide a built-in way of specifying options for tasks. You have to use an external options parser module like yargs. See this question for more on that topic.

这也意味着将 ['copy-images --changed'] 之类的内容传递给 gulp.watch() 将不起作用.整个字符串将被解释为任务名称.

This also means that passing something like ['copy-images --changed'] to gulp.watch() will not work. The entire string will just be interpreted as a task name.

对您来说最好的方法是将您的任务代码分解为一个函数，然后从您的任务和手表中调用该函数:

The best approach for you would be to factor out the code of your task into a function and then call this function from both your task and your watch:

var argv = require('yargs').argv;

function copyImages(opts) {
  if (opts.changed) {
    // some code
  } else {
    // some other code
  }
}

gulp.task('copy-images', function() {
  copyImages(argv);
});

gulp.task('watch', function(){
    gulp.watch(config.images, function() {
      copyImages({changed:true});
    });
});

以上内容应涵盖您的所有基础:

The above should cover all of your bases:

• gulp copy-images 将执行 //一些其他代码.
• gulp copy-images --changed 将执行//一些代码.
• gulp watch 将执行 //some code 任何时候监视的文件发生变化.

• gulp copy-images will execute //some other code.
• gulp copy-images --changed will execute //some code.
• gulp watch will execute //some code any time a watched file is changed.

这篇关于Gulp - 使用参数启动任务的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！