Configure the gulp bundling destination folder outside the project folder - gulp

Question

We have bundling our application using gulp process(using aurelia-bundler). As of now the bundle process destination folder resides inside the project folder named "dist". We need to place the destination folder outside the project folder
gulp.task('bundle', function (callback) {
runSequence('unbundle',
'bundle-config',
'copy-app-files',
'compress',
function () {
});
});

Answer 1

In theory, you just have to change the exportSrvRoot variable of the build/paths.js file. For instance:
//var exportSrvRoot = 'export/';
var exportSrvRoot = '../'; //this is outside of the project folder.
Now, the gulp export command will export the files to the folder above the project folder. The solution is the same for build output (gulp build/bundle), but you'd have to change the outputRoot variable instead of exportSrvRoot.
However, there is a problem in this approach. Since your export folder is outside of the project folder, if you run gulp export, you will get an error saying that gulp.del cannot delete a folder that is outside of the project folder. This could be solved by passing additional parameters to gulp.del, but the task uses vinyl-paths to call gulp.del, preventing you to send any additional parameters for it =/.
One of the ways to solve the above problem is deleting the line 36 of the export-release.js:
// use after prepare-release
gulp.task('export', function(callback) {
return runSequence(
'bundle',
//'clean-export', <---- this line
'export-copy',
callback
);
});
In this way, gulp export won't try to delete the folder, preventing the error. But now, you must delete the export folder manually before each time you run gulp export.
Another way to solve this is rewriting the clean-export task in order to remove its dependency from vinyl-paths.
Hope this helps!