Upload a file to Convex file storage via a generated upload URL, store the storage ID on a document, and serve the file back

Documented steps

1. Define a mutation generateUploadUrl that calls await ctx.storage.generateUploadUrl() and returns the short-lived URL.
2. From the client, call that mutation to obtain postUrl.
3. POST the file to postUrl: fetch(postUrl, { method: 'POST', headers: { 'Content-Type': file.type }, body: file }).
4. Parse the response JSON to get { storageId }.
5. Call a second mutation (e.g. sendImage) with args { storageId: v.id('_storage'), author: v.string() } that persists the ID, e.g. await ctx.db.insert('messages', { body: args.storageId, author: args.author, format: 'image' }).
6. To serve the file from a query or mutation, call await ctx.storage.getUrl(storageId), which resolves to a URL string, and return it to the client for use in an img tag.
7. Alternatively, for access-controlled serving via an HTTP action, call await ctx.storage.get(storageId) to obtain a Blob and return it inside a Response.
8. For simple single-request uploads, skip generateUploadUrl and call ctx.storage.store(blob) inside an HTTP action instead; it returns an Id<'_storage'>.
9. Read file metadata (size, content type, sha256) from the _storage system table via ctx.db.system.get(storageId) rather than a deprecated metadata API.
10. Official docs: https://docs.convex.dev/file-storage/upload-files and https://docs.convex.dev/file-storage/serve-files