In an attempt to separate my .ts source files from the .js output, I'm trying to implement a formal file-watcher in TypeScript and it seems as though the ability to specify an output path for a single file does not exist.
The flow, just to be clear: I begin watching my entire src
directory, make a change to, say, src/views/HomeView.ts
, and i want node to pick up that the file has been changed, and move the compiled version to public/js/views/HomeView.js
.
Using tsc myfile.ts --out myfile.js
it travels through all of the modules and compiles each in the same path that the .ts file exists, without placing the final file(s) in the properly specified path. It does however create an empty file where I would like it to end up.
I'm wondering:
1) Is it possible to use the --out
parameter and only compile that one file? I do not want it to traverse imports and compile every single file, but merely do syntax / error checking at compile-time (and that's only a secondary requirement, not necessary); and
2) Is there a bug in the compiler that prevents it from properly combining / creating files? Again, the final output in the --out
path directive is empty, but a file is indeed created.
Any help would be appreciated.
* Update *
While I don't want to close this question as I do believe it's still an issue, I went ahead and implemented the core TypeScript compiler to achieve what I needed to do, bypassing tsc
altogether. Please see https://github.com/damassi/TypeScript-Watcher for more information and usage.