[gulp] Get the current file name in gulp.src()

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.

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

Here is another simple way.

var es, log, logFile;

es = require('event-stream');

log = require('gulp-util').log;

logFile = function(es) {
  return es.map(function(file, cb) {
    log(file.path);
    return cb();
  });
};

gulp.task("do", function() {
 return gulp.src('./examples/*.html')
   .pipe(logFile(es))
   .pipe(gulp.dest('./build'));
});

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).

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 

For my case gulp-ignore was perfect. As option you may pass a function there:

function condition(file) {
 // do whatever with file.path
 // return boolean true if needed to exclude file 
}

And the task would look like this:

var gulpIgnore = require('gulp-ignore');

gulp.task('task', function() {
  gulp.src('./**/*.js')
    .pipe(gulpIgnore.exclude(condition))
    .pipe(gulp.dest('./dist/'));
});