How to exclude a folder from processing gulp-imagemin?

The question in the title. The method is given below, leads to the fact that all the images are not processed. Created issue on github, the author was sent to stackoverflow.

gulp.task('default', function () {
 return gulp.src(['src/images/**', '!src/images/sprites/**'])
.pipe(imagemin())
.pipe(gulp.dest('dist'));
});
September 26th 19 at 07:31
2 answers
September 26th 19 at 07:33
Solution
Which version is used?
Try it so
gulp.task('default', function () {
 return gulp.src(['./src/images/**/*', '!./src/images/sprites/**/*'])
.pipe(imagemin())
.pipe(gulp.dest('./dist/images'));
});


So I recommend along with madamp to use gulp-cache
Thank you, the recommendation gulp-cache prompted the idea that it can be gulp-newer. Turned off, it worked. - ethelyn_Schuppe commented on September 26th 19 at 07:36
September 26th 19 at 07:35
Try to look to the side: gulp-ignore
Thanks for the useful plugin. - ethelyn_Schuppe commented on September 26th 19 at 07:38

Find more questions by tags JavaScriptNode.js