simple cgi 'system'?

[gabor]

hi,

i'm developing an application which many times downloads differents
stuff from http.

for this i'd need a quick way to create a cgi to for example return
always error 503. so that i can test my program if it handles this error
correctly or not.

because the program is in java, i usually create a small servlet, but i
thought that maybe python would be easier.

what is the easiest way to create a cgi?
i know about the cgi module, but most of times i need something a little
higher level.

usually i only want to specify for example that the cgi should fail with
error 503, i don't want to print all the header stuff manually.

any ideas?

gabor

[Andrew Clover]

Re: simple cgi 'system'?

gabor <gabor@z10n.net > wrote:
[color=blue]
> i know about the cgi module, but most of times i need something a little
> higher level.[/color]

The cgi module is slightly mis-named. It's for reading in HTML form
submissions. You don't need to use it to write a CGI script if you're
just going to return a 503 anyway.

The 'print all the header stuff manually' solution:

#!/path/to/python
print 'Status: 503 Service Unavailable'
print

There isn't a particular module to simplify writing this sort of script
as it is already completely trivial!

--
Andrew Clover
mailto:and@doxd esk.com

http://www.doxdesk.com/