Reputation: 466
Im new to Gulp but managed to create the following gulpfile.js to minify images that reside in an /image/ folder and output to my /images/optimised/ folder:
const gulp = require('gulp');
const imagemin = require('gulp-imagemin');
function minifyimages() {
return gulp.src('./images/*')
.pipe(imagemin())
.pipe(gulp.dest('images/optimised'))
verbose: true
}
exports.minifyimages = minifyimages;
If i run the script for the first time it generates the /images/optimised/ folder on its own, which is fine. But if i run it again then it generates an /images/optimised/optimised/ folder.
Is there a way for it to skip the creation of the /optimised/ folder if it already exists?
Thanks
Upvotes: 2
Views: 322
Reputation: 180945
Try this task:
function minifyimages() {
return gulp.src('./images/*.+(gif|png)') // change here
.pipe(imagemin())
.pipe(gulp.dest('./images/optimised'))
verbose: true
}
I added the file extensions to gulp.src
(you may have more image types) so that gulp is not looking into the optimized
folder at all. And thus optimizing those as well, with their folder structure.
Upvotes: 1