Gulp replace only relevant files

Gulp-Replace-Only-Relevant-Files.md

gulp-replace only relevant files

Piping to gulp.dest would re-write all input files. This checks for the pattern first and only passes along files that actually need to be processed, avoiding unnecessary overwriting with the same content. A benefit is that your files' timestamps stay meaningful.

gulpfile.js

const gulp = require('gulp');
const log = require('gulp/node_modules/fancy-log');
const replace = require('gulp-replace');
const only = require('gulp-ignore').include;
const path = require('path');

const SRC = 'data/*.csv';
const DEST = 'data/';

gulp.task('clean:diacritics', (done) => {
  const REPLACEMENTS = {
    'Ş': 'Ș',
    'ş': 'ș',
    'Ţ': 'Ț',
    'ţ': 'ț',
    'Ã': 'Ă',
    'ã': 'ă',
  };
  const dirty = new RegExp(`[${Object.keys(REPLACEMENTS).join('')}]`, 'g');

  gulp.src(SRC)
    .pipe(only((file) => {
      let result = dirty.test(String(file.contents));
      file.basename = path.parse(file.history[0]).base;
      if (result) log(`Cleaning '${file.basename}'`);
      return result;
    }))
    .pipe(replace(dirty, (match) => REPLACEMENTS[match]))
    .pipe(gulp.dest(DEST))
    .on('end', () => {
      log('Done. Wrong diacritics have been smitten!');
      done();
    });
});

gulp.task('clean', ['clean:diacritics']);

A more generic solution would be better
Maybe something like:

gulp.src(SRC)
.pipe(hash)
.pipe(replace(dirty, match => REPLACEMENTS[match]))
.pipe(onlyHashChanged)
.pipe(gulp.dest(DEST))

Or using gulp-snapshot

gulp.src(SRC)
.pipe(snapshot.take())
.pipe(replace(dirty, match => REPLACEMENTS[match]))
.pipe(snapshot.take())
.pipe(snapshot.compare( difference => {
  // OK now WTF?! How to NOT passthrough if difference.noChanges?!
}
.pipe(gulp.dest(DEST))

.pipe(snapshot.compare( difference => {
  // OK now WTF?! How to NOT passthrough if difference.noChanges?!
}

hi is there any way, how to passthrough only changed files to gulp dest.