寻找方式来复制文件,并根据父目录进行重命名 [英] Looking for way to copy files in gulp and rename based on parent directory
问题描述
对于每个模块,我都需要将一些文件复制到构建目录中,并且正在寻找一种方法来最小化重复代码:
gulp.src('./ client / src / modules / signup / index.js')
.pipe(gulp.dest('./ build / public / js / signup ));
gulp.src('./ client / src / modules / admin / index.js')
.pipe(gulp.dest('./ build / public / js / admin' ));
类似这样:
gulp.src('./ client / src / modules /(.*)/ index.js')
.pipe(gulp.dest('./ build / public / js / $ 1' ));
很明显,上述方法无效,所以有办法做到这一点,或者是一个npm已经这样做了吗?
谢谢
配置您的 base
,如下所示:
gulp.src('./ client / src / modules / ** / index.js',{base:'./client/src/modules'})
.pipe(gulp.dest './build/public/js/'));
这告诉 gulp
使用modules目录作为确定相对路径的起点。
(另外,您可以使用 / ** / *。js
如果你想包括所有的JS文件...)
For each module I have some files that need to be copied over to the build directory, and am looking for a way to minimize the repeated code from this:
gulp.src('./client/src/modules/signup/index.js')
.pipe(gulp.dest('./build/public/js/signup'));
gulp.src('./client/src/modules/admin/index.js')
.pipe(gulp.dest('./build/public/js/admin'));
to something like this:
gulp.src('./client/src/modules/(.*)/index.js')
.pipe(gulp.dest('./build/public/js/$1'));
Obviously the above doesn't work, so is there a way to do this, or an npm that already does this?
Thanks
The best way is to configure your base
when sourcing files, like so:
gulp.src('./client/src/modules/**/index.js', {base: './client/src/modules'})
.pipe(gulp.dest('./build/public/js/'));
This tells gulp
to use the modules directory as the starting point for determining relative paths.
(Also, you can use /**/*.js
if you want to include all JS files...)
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