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

If you want to use @OverZealous' answer (https://stackoverflow.com/a/21806974/1019307) in Typescript, you need to import instead of require:

import * as debug from 'gulp-debug';

...

    return gulp.src('./examples/*.html')
        .pipe(debug({title: 'example src:'}))
        .pipe(gulp.dest('./build'));

(I also added a title).