From 5f58bf117e73ea667bdbc86b3cbf8cd2cbd3bc2d Mon Sep 17 00:00:00 2001 From: gilesb Date: Wed, 7 Jan 2026 21:13:44 +0000 Subject: [PATCH] Increase L2 publish timeout to 30 seconds MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit L2 needs to call back to L1 to fetch run details, which can take longer than 10 seconds especially with network latency. 🤖 Generated with [Claude Code](https://claude.com/claude-code) Co-Authored-By: Claude Opus 4.5 --- server.py | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) diff --git a/server.py b/server.py index 74ff53c..be4cf58 100644 --- a/server.py +++ b/server.py @@ -1336,12 +1336,13 @@ async def ui_publish_run(run_id: str, request: Request, output_name: str = Form( return HTMLResponse('
Not logged in
') # Call L2 to publish the run, including this L1's public URL + # Longer timeout because L2 calls back to L1 to fetch run details try: resp = http_requests.post( f"{L2_SERVER}/registry/record-run", json={"run_id": run_id, "output_name": output_name, "l1_server": L1_PUBLIC_URL}, headers={"Authorization": f"Bearer {token}"}, - timeout=10 + timeout=30 ) if resp.status_code == 400: error = resp.json().get("detail", "Bad request")