Closed ramilexe closed 8 years ago
Why not send (or not send) a 444 status code to reduce the load on NGINX further, because it doesn't have to send a reponse at all to the bad bots? Maybe it would make sense to add a new option "testcookie.status = XYZ".
On 22.08.2014 01:33, Ramil wrote:
When module is enable it modify content but send 200 ok. It is bad for SEO. I suggest send 503 temporary unavailable for search bot come back later and didn't index bad content.
You can merge this Pull Request by running
git pull https://github.com/ramilexe/testcookie-nginx-module master
Or view, comment on, or merge it at:
https://github.com/kyprizel/testcookie-nginx-module/pull/19
Commit Summary
- Update src/ngx_http_testcookie_filter_module.c
Update src/ngx_http_testcookie_filter_module.c
File Changes
M src/ngx_http_testcookie_filter_module.c https://github.com/kyprizel/testcookie-nginx-module/pull/19/files#diff-0 (8)
Patch Links:
- https://github.com/kyprizel/testcookie-nginx-module/pull/19.patch
- https://github.com/kyprizel/testcookie-nginx-module/pull/19.diff
— Reply to this email directly or view it on GitHub https://github.com/kyprizel/testcookie-nginx-module/pull/19.
When occur first request you don't know bot it or not. If you send 200 ok it will be bad for SEO because you content only to check bot () So I suggest return 503 to say search engines not process this response.
@kyprizel it took long time)
sorry :) added testcookie_refresh_status instead
When module is enable it modify content but send 200 ok. It is bad for SEO. I suggest send 503 temporary unavailable for search bot come back later and didn't index bad content.