How to set gulp.dest() in same directory as pipe inputs?

Question

I need all the found images in each of the directories to be optimized and recorded into them without setting the path to the each folder separately. I don\'t understand how

Answer 1

Here you go:

gulp.task('optimizeJpg', function () {

    return gulp.src('./images/**/**/*.jpg')
        .pipe(imageminJpegtran({ progressive: true })())
        .pipe(gulp.dest('./images/'));
});

Gulp takes everything that's a wildcard or a globstar into its virtual file name. So all the parts you know you want to select (like ./images/) have to be in the destination directory.