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I have two python apps I'd like to run on the same server. One requires Python 2.7, the other requires Python 3.3. I'm running CentOS 6.6 with httpd 2.4 and the python 2.7 and 3.3 SCLs. The Python 2.7 app is currently deployed through apache. I'm running the python 3.3 app through the flask dev server on another port, but want to run in in the same apache instance.

When I start apache, and it tries to load the second mod_wsgi instance it tells me :

AH01574: module wsgi_module is already loaded, skipping

So I change: LoadModule wsgi_module modules/mod_python33-wsgi.so to:

LoadModule wsgi_module33 modules/mod_python33-wsgi.so

And then I get Can't locate API module structure `wsgi_module33' in file /opt/rh/httpd24/root/etc/httpd/modules/mod_python33-wsgi.so: /opt/rh/httpd24/root/etc/httpd/modules/mod_python33-wsgi.so: undefined symbol: wsgi_module33

Is this possible?

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  • Yes there is an indirect way of doing but don't have time to provide details right now. You are also better off asking on the mod_wsgi mailing list as it will involve some discussion and ServerFault is not a discussion forum. So go go ask on the mod_wsgi mailing list and I will explain. Mar 31, 2015 at 22:45
  • @GrahamDumpleton ok I will ask. Apr 2, 2015 at 12:04
  • Topic on Google Groups is groups.google.com/forum/#!topic/modwsgi/tq2TI76-q3g I will respond. This week hasn't been a normal week and haven't had a chance yet, plus about to be travelling for a week as well. So please be a little patient with me. Apr 4, 2015 at 9:48

1 Answer 1

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Apparently, it's not possible.

All the customization accepted by mod_wsgi is to use a specific version using virtualenv.

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