javascript - Gulp.js get filename from .src() -

i'm trying use gulp-proceesshtml (https://github.com/julien/gulp-processhtml) remove unwanted code in build version, problem task requires filename given.

gulp.src('test.html').pipe(processhtml('test.html'));

but can't figure out how work when i'm processing html files in folder

gulp.src('*.html).pipe(processhtml('filename here'));

personally, sounds that's wrong plugin trying accomplish. see below.

however, because it's not clear using for, can able use node-glob process each file one-by-one:

var glob = require('glob')     // need event-stream merging streams     es = require('event-stream');  gulp.task('mytask', function() {     var files = glob.sync('*.html'),         streams;     streams = files.map(function(file) {                 // add *base* option if files stored in                 // multiple subdirectories         return gulp.src(file, {base: 'relative/base/path'})                 // may need require('path').filename(file)                 .pipe(processhtml(file));     });      return es.merge.apply(es, streams); }); 

this create single asynchronous stream out of every file matches initial pattern.

---

for removing text files, can use gulp-replace, so:

var replace = require('gulp-replace');  gulp.src('*.html')    // replaces text between    // <!-- remove-this --> , <!-- /remove-this -->    .pipe(replace(/<!--\s*remove-this\s*-->[\s\s]*?<!--\s*\/remove-this\s*-->/g, ''));