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:

var gulp = require('gulp');  gulp.task('examples', function() {     return gulp.src('./examples/*.html')         .pipe(gulp.dest('./build')); }); 

But I can't figure out how retrieve the file names found (and processed) in the task, or I can't find the right plugin.

Answer 1

I'm not sure how you want to use the file names, but one of these should help:

• If you just want to see the names, you can use something like gulp-debug, which lists the details of the vinyl file. Insert this anywhere you want a list, like so:

  var gulp = require('gulp'),     debug = require('gulp-debug');  gulp.task('examples', function() {     return gulp.src('./examples/*.html')         .pipe(debug())         .pipe(gulp.dest('./build')); }); 

• Another option is gulp-filelog, which I haven't used, but sounds similar (it might be a bit cleaner).

• Another options is gulp-filesize, which outputs both the file and it's size.

• If you want more control, you can use something like gulp-tap, which lets you provide your own function and look at the files in the pipe.