我如何将吞咽任务合并为一个？

Question

My project uses two different code bases - one for desktop and one for mobile. I'd like to use the same gulp "build" task for both. My gulp "build" task for desktop looks like this:

gulp.task('build', function() {
  return gulp.src([
    'file1.js',
    'file2.js',
    'file3.js',
    'file4.js',
    'file5.js',
    'file6.js',
    'file7.js',
    'file8.js',
    'file9.js',
    'file10.js',
  ])
  .pipe(concat('bundle.js'))
  .pipe(uglify())
  .pipe(gulp.dest('desktop/dist'));
});

The only differences between these tasks is:

1) the list of files being concatenated into bundle.js

2) the prefixing dist/bundle.js with "desktop/" or "mweb/" based on which platform I'm compiling code for

Ideally, the task would be generic enough to where I could pass in a different array of files (perhaps using a function argument and/or variables), but am struggling with how to achieve this using gulp.

What I've tried so far:

My first approach was to create two separate build tasks, "desktop-build" and "mobile-build", but this approach feels bloated and redundant.

Answer 1

I recommend looking at this question Pass Parameter to Gulp Task . You can define, for example, an environment variable object which contains an environment name, prefix, file list, and any other useful information, then check for a flag like, gulp build --desktop to use the corresponding environment attributes in the build process, with the most common use case as a default environment. Seems the easiest and most flexible option.