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Increase L2 publish timeout to 30 seconds

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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 <noreply@anthropic.com>
This commit is contained in:
gilesb
2026-01-07 21:13:44 +00:00
parent 3436a22a94
commit 5f58bf117e
1 changed files with 2 additions and 1 deletions
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3
server.py
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@@ -1336,12 +1336,13 @@ async def ui_publish_run(run_id: str, request: Request, output_name: str = Form(
return HTMLResponse('<div class="error">Not logged in</div>') return HTMLResponse('<div class="error">Not logged in</div>')
# Call L2 to publish the run, including this L1's public URL # 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: try:
resp = http_requests.post( resp = http_requests.post(
f"{L2_SERVER}/registry/record-run", f"{L2_SERVER}/registry/record-run",
json={"run_id": run_id, "output_name": output_name, "l1_server": L1_PUBLIC_URL}, json={"run_id": run_id, "output_name": output_name, "l1_server": L1_PUBLIC_URL},
headers={"Authorization": f"Bearer {token}"}, headers={"Authorization": f"Bearer {token}"},
timeout=10 timeout=30
) )
if resp.status_code == 400: if resp.status_code == 400:
error = resp.json().get("detail", "Bad request") error = resp.json().get("detail", "Bad request")