I have written some code and I want to provide it in a package, but I want to also expose it to package consumers as a worker. For this purpose I have created a wrapper class, that runs an isolate internally and using the send command and listeners communicate with the Isolate to provide the functionality.
The problem arises when I want to use this wrapper class from
web directory: the
Uri provided is interpolated from the directory of the running/main
Isolate instead of from the package root. For
bin it is
packagename|bin/ and for
web it is
I would like to export this class to the consumers so they can chose an easier approach than to construct their own
Isolate, but I am not sure how to specify the main file that will be used in spawnUri.
Is there a way to specify the file so it will always be resolved to the correct file regardless of where the main Isolate is run from.
// Exports the next file so the class in it will be package visible packageroot -> lib/package_exports_code_that_spawns_isolate.dart // This file should contain URI that always resolve to the next file packageroot -> lib/code_that_spawns_isolate.dart // The main worker/Isolate file packageroot -> lib/src/worker/worker.dart
To refer to a library in your package, you should use a
var workerUri = Uri.parse("package:myPackage/src/worker/worker.dart"); var isolate = await Isolate.spawnUri(workerUri,...);
It’s not perfect because it requires you to hard-wire your package name into the code, but I believe it’s the best option currently available.
Isolate.spawnUri function doesn’t (and can’t) resolve a relative URI reference wrt. the source file that called it – nothing in the Dart libraries depends on where it’s called from, that’s simply too fragile – so a relative URI isn’t going to work. The only absolute URI referencing your worker is a
package: URI, so that’s what you have to use.
Answered By – lrn
Answer Checked By – Mildred Charles (FlutterFixes Admin)