Get the current file name in gulp.src()

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.

I found this plugin to be doing what I was expecting: gulp-using

Simple usage example: Search all files in project with .jsx extension

gulp.task('reactify', function(){
        gulp.src(['../**/*.jsx']) 
            .pipe(using({}));
        ....
    });

Output:

[gulp] Using gulpfile /app/build/gulpfile.js
[gulp] Starting 'reactify'...
[gulp] Finished 'reactify' after 2.92 ms
[gulp] Using file /app/staging/web/content/view/logon.jsx
[gulp] Using file /app/staging/web/content/view/components/rauth.jsx