javascript - 用 gulp/imagemin 替换子文件夹中的文件

Question

我的文件夹结构为

assets/images/foo/abc.jpg
assets/images/bar/def.jpg 
..

我想用 gulp imagemin 它们并将图像放在相同的子文件夹中，怎么办？

var paths = {
    images: 'assets/images/**/*'
};

gulp.task('images', function () {
    return gulp.src(paths.images)
        .pipe(imagemin({optimizationLevel: 5}))
        .pipe(gulp.dest(???));
});

我应该在 gulp.dest 中放什么？

Answer 1

看看这个 future 的食谱:

https://github.com/arvindr21/gulp/blob/master/docs/recipes/maintain-directory-structure-while-globbing.md

If you want to maintain the structure, you need to pass {base: '.'} to gulp.src().