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

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/'));
});