How to do that would in gulp copied only the new pictures?

The build copies all of the pictures first every time. And the fact that I had
gulp.task('assets', function() {
 return gulp.src(['frontend/assets/{fonts,images}/**', 'frontend/assets/index.html'], {since: gulp.lastRun('assets')})
.pipe(newer('public'))
.pipe(gulp.dest('public'));
});

had to copy only what has changed. But for some reason did not work.

Perhaps we can do that would be adding the pictures themselves were copied now, but not all and only the new?
Here is my task now:
gulp.task('assets', function() {
 return gulp.src('frontend/assets/{fonts,images}/**/*.*')
.pipe(gulp.dest('public'));
});
July 9th 19 at 11:05
1 answer
July 9th 19 at 11:07
Try gulp-changed
Works well

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