My folder structure looks like this:
- /projects
- /projects/proj1
- /projects/proj2
- /projects/proj3
- /zips
For each folder in projects (proj1, proj2 and proj3) I want to zip contents of each of those folders and generate proj1.zip, proj2.zip and proj3.zip in /zips folder.
Following example function generates single zip file from proj1 folder
zip = require('gulp-zip');
gulp.task('default', function () {
return gulp.src('./projects/proj1/*')
.pipe(zip('proj1.zip'))
.pipe(gulp.dest('./zips'));
});
But how I can execute such task for each folder in projects? I can get all folders to zip by gulp.src('./projects/*') but what then?
Old question I know and I am pretty new to gulp so this might be considered a hack but I have just been trying to do what you are after and I ended up with this.
My file structure is this:
proj
proj/dist
proj/dist/sub01
proj/dist/sub02
proj/dist/sub03
proj/zipped
And my task ended up like this:
var gulp = require("gulp");
var foreach = require("gulp-foreach");
var zip = require("gulp-zip");
gulp.task("zip-dist", function(){
return gulp.src("./dist/*")
.pipe(foreach(function(stream, file){
var fileName = file.path.substr(file.path.lastIndexOf("/")+1);
gulp.src("./dist/"+fileName+"/**/*")
.pipe(zip(fileName+".zip"))
.pipe(gulp.dest("./zipped"));
return stream;
}));
});
It grabs all the first level contents of ./dist as its source and then pipes it to gulp-foreach.
gulp-foreach looks at each item and I use a plain javascript substr() to get the name of the current item which I store as a variable.
Finally I set a new src using the stored fileName var and pipe the result to gulp-zip using the stored var again as the name of the zipped file.
The result is a structure that looks like this:
proj
proj/dist
proj/dist/sub01
proj/dist/sub02
proj/dist/sub03
proj/zipped
proj/zipped/sub01.zip
proj/zipped/sub02.zip
proj/zipped/sub03.zip
Again, I am a million miles from being an expert but this worked for me and if I understand the question might work for you as well or at least give you some ideas.
