Skip to content
New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Only use last line as fileName when packing #115

Merged
merged 5 commits into from
Nov 28, 2024
Merged
Changes from 1 commit
Commits
File filter

Filter by extension

Filter by extension

Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
3 changes: 2 additions & 1 deletion src/lib/utils/pack.ts
Original file line number Diff line number Diff line change
Expand Up @@ -48,7 +48,8 @@ export async function pack(srcDir: string, dstDir: string) {
);
});

const fileName = path.basename(stdout.trim());
const lastLine = stdout.trim().split("\n").pop() || "";
Copy link
Owner

@0x80 0x80 Nov 26, 2024

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

I think it would be better to throw a descriptive error if lastLine is undefined, don't you think? Maybe even add the stdout to the message, like Failed to parse last line from stdout: ${stdout.trim()}

If basename can't parse the empty string, or if fileName becomes an empty string, the resulting error might not be so obvious.

Also, I have a slight preference for at(-1) instead of pop(), but that's nitpicking :)

Copy link
Contributor Author

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

Agreed, updated now!

const fileName = path.basename(lastLine);

const filePath = path.join(dstDir, fileName);

Expand Down