Closed mrowdy closed 9 years ago
This plugin just gathers up the files in the stream and zips them up. Keep in mind that gulp is just JS and you can do whatever you want with it. You could just loop through the packages
directory, and create a new gulp.src
stream for every subfolder. Then you already have the folder name that you could pass to zip
.
Is it possible to use the filename from gulp.src as filename for the generated archive?
I have a list of folders like this
and want to generate:
First problem:
if i provide *gulp.src('packages/')** it generates one zip file which contains all 3 folders. How can i split this into 3 separate zipfiles?
second problem:
zip('archive.zip') takes the filename as string. is there a way to provide the filename from gulp.src?