Get the current file name in gulp.src()

gulp

Solution

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.

Problem

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.

Original source