Get the current file name in gulp.src()

Question

In my gulp.js file I\'m streaming all HTML files from the examples folder into the build folder.

To create the gulp task is not difficult:

Answer 1

You can use the gulp-filenames module to get the array of paths. You can even group them by namespaces:

var filenames = require("gulp-filenames");

gulp.src("./src/*.coffee")
    .pipe(filenames("coffeescript"))
    .pipe(gulp.dest("./dist"));

gulp.src("./src/*.js")
  .pipe(filenames("javascript"))
  .pipe(gulp.dest("./dist"));

filenames.get("coffeescript") // ["a.coffee","b.coffee"]  
                              // Do Something With it