Hi,
I tried for the last 2 hours now to somehow grasp how to use ZSI over
mod_python. It's nice that they have a code-example in the ZSI docs,
but it's incomplete and you are left guessing.
The ZSI docs I used as reference can be found on http://
pywebsvcs.sourc eforge.net/zsi.html#SECTIO N00313000000000 0000000
Here is where I am at so far:
First I created the two files as described in the ZSI docs. Namely
"MyHandler. py" and "ws.py". The first name is obvious to someone who
has coded at least a little bit of python, but it could be explicitly
stated in the docs, so the beginners don't get stumped by a silly
filename to start with ;)
The second filename is my own concoction. I suppose it does not matter
how it's named as you specify it in the apache conf anyhow.
So. Now to the apache conf. I put the scripts into the document-root
of my server. In this case: "/var/www/localhost". So this is what I
get (ommitting the unnecessary cruft -- marked it with "[...]"):
---- /etc/apache2/sites-enables/000-default.conf
-----------------------------------
<VirtualHost *>
[...]
DocumentRoot /var/www/localhost
AddHandler mod_python .py
[...]
<Directory /var/www/localhost/>
[...]
PythonHandler ws
PythonDebug On
</Directory>
[...]
--------------------------------------------------------------------------------------------------------
This is as far as I got. Now how do I call a service from this
setting? What's the URI I need to specify in the client?
I was trying "http://server/foo.py" in the browser, and it was
obviously doing *something* with the "ws.py" file. But I have no clue
what. Here's the output:
---- Browser output for http://server/foo.py
---------------------------------------------
Mod_python error: "PythonHand ler ws"
Traceback (most recent call last):
File "/usr/lib/python2.4/site-packages/mod_python/apache.py", line
299, in HandlerDispatch
result = object(req)
File "/var/www/localhost/ws.py", line 9, in handler
dispatch.AsHand ler(modules=(My Handler,), request=req)
File "/usr/lib/python2.4/site-packages/ZSI-2.0_rc3-py2.4.egg/ZSI/
dispatch.py", line 263, in AsHandler
File "/usr/lib/python2.4/site-packages/ZSI-2.0_rc3-py2.4.egg/ZSI/
parse.py", line 61, in __init__
File "/usr/lib/python2.4/site-packages/PyXML-0.8.4-py2.4-linux-
i686.egg/_xmlplus/dom/ext/reader/PyExpat.py", line 65, in fromStream
success = self.parser.Par seFile(stream)
ExpatError: no element found: line 1, column 0
--------------------------------------------------------------------------------------------------------
No matter what I append to the URI, it does not make a difference in
the output. And *why* am I forced to add a "abracadabra.py " to the
URI? Could I not set Apache up in a way that it would use the ZSI
handler for everything that calls -- let's say -- http://
webservices.ser ver.tld/ instead of http://webservices.server.tld/
booh.py ?
I'm off to a little bit more of testing. But any hints/input would be
kindly appreciated ;)
I tried for the last 2 hours now to somehow grasp how to use ZSI over
mod_python. It's nice that they have a code-example in the ZSI docs,
but it's incomplete and you are left guessing.
The ZSI docs I used as reference can be found on http://
pywebsvcs.sourc eforge.net/zsi.html#SECTIO N00313000000000 0000000
Here is where I am at so far:
First I created the two files as described in the ZSI docs. Namely
"MyHandler. py" and "ws.py". The first name is obvious to someone who
has coded at least a little bit of python, but it could be explicitly
stated in the docs, so the beginners don't get stumped by a silly
filename to start with ;)
The second filename is my own concoction. I suppose it does not matter
how it's named as you specify it in the apache conf anyhow.
So. Now to the apache conf. I put the scripts into the document-root
of my server. In this case: "/var/www/localhost". So this is what I
get (ommitting the unnecessary cruft -- marked it with "[...]"):
---- /etc/apache2/sites-enables/000-default.conf
-----------------------------------
<VirtualHost *>
[...]
DocumentRoot /var/www/localhost
AddHandler mod_python .py
[...]
<Directory /var/www/localhost/>
[...]
PythonHandler ws
PythonDebug On
</Directory>
[...]
--------------------------------------------------------------------------------------------------------
This is as far as I got. Now how do I call a service from this
setting? What's the URI I need to specify in the client?
I was trying "http://server/foo.py" in the browser, and it was
obviously doing *something* with the "ws.py" file. But I have no clue
what. Here's the output:
---- Browser output for http://server/foo.py
---------------------------------------------
Mod_python error: "PythonHand ler ws"
Traceback (most recent call last):
File "/usr/lib/python2.4/site-packages/mod_python/apache.py", line
299, in HandlerDispatch
result = object(req)
File "/var/www/localhost/ws.py", line 9, in handler
dispatch.AsHand ler(modules=(My Handler,), request=req)
File "/usr/lib/python2.4/site-packages/ZSI-2.0_rc3-py2.4.egg/ZSI/
dispatch.py", line 263, in AsHandler
File "/usr/lib/python2.4/site-packages/ZSI-2.0_rc3-py2.4.egg/ZSI/
parse.py", line 61, in __init__
File "/usr/lib/python2.4/site-packages/PyXML-0.8.4-py2.4-linux-
i686.egg/_xmlplus/dom/ext/reader/PyExpat.py", line 65, in fromStream
success = self.parser.Par seFile(stream)
ExpatError: no element found: line 1, column 0
--------------------------------------------------------------------------------------------------------
No matter what I append to the URI, it does not make a difference in
the output. And *why* am I forced to add a "abracadabra.py " to the
URI? Could I not set Apache up in a way that it would use the ZSI
handler for everything that calls -- let's say -- http://
webservices.ser ver.tld/ instead of http://webservices.server.tld/
booh.py ?
I'm off to a little bit more of testing. But any hints/input would be
kindly appreciated ;)