SOAPpy small bug

Installing SOAPpy 0.12.0 on python 2.5 I got the following error:

Traceback (most recent call last):
File "setup.py", line 8, in
from SOAPpy.version import __version__
File "/usr/local/src/SOAPpy-0.12.0/SOAPpy/__init__.py", line 5, in
from Client import *
File "/usr/local/src/SOAPpy-0.12.0/SOAPpy/Client.py", line 46
from __future__ import nested_scopes
SyntaxError: from __future__ imports must occur at the beginning of the file

The solution is to open file Client.py and copy/past the "from __future__ imports" line to be the first import.

The same error will appear in several files and also during installation, but the solution is always the same

Easier way to check python

Like we say in Portuguese "Who knows, know it" here is an easier way to get the mod_python info (posted by Grahan Dumpleton in the comments)


It is easier to go:

<Location /mpinfo>
SetHandler mod_python
PythonHandler mod_python.testhandler
</Location>

and then use '/mpinfo' as URL.

mod_python info server request

This is how information of the server and options can be made with mod_python 3.3.1 (like to do phpinfo() in PHP)

1. Make a /test directory in the htdocs tree

2. Create a directory structure in the httpd.conf as follow:

<Directory "/usr/local/apache2/htdocs/test">
Options Indexes FollowSymLinks
AllowOverride All
AddHandler mod_python .py
PythonHandler mod_python.testhandler
PythonDebug On
Order allow,deny
Allow from all
</Directory>

3. Restart Apache

4. Point to some hypothetical python script: http://localhost/test/ImaginaryScript.py

And the information of mod_python and apache should appear!!!!