python script will not be executed
Question
I have following script
#!/usr/bin/python
import os
import cgi
import sys
def header():
header = """
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<html>
<body bgcolor="#B7B7B7">
"""
print header
def footer():
footer = """
</body>
</html>
"""
def main():
try:
header()
footer()
except:
print "<!-- --><hr><h1>Oops. An error occured.</h1>"
cgi.print_exception()
main()
This script is called from another script - problem is that is not working. The browser just simply write whole code as plain text beside HTML tags:
#!/usr/bin/python
import os
import cgi
import sys
def header():
header = """ """
print header
def footer():
footer = """ """
def main():
try:
header()
footer()
except:
print "Oops. An error occured."
cgi.print_exception()
main()
I checked the apache error log, there is nothing about it.
I tried chmod 777, still the same
this is how my config.py looks like:
<Directory "{1}">
AddHandler cgi-script .cgi
Options -Indexes FollowSymLinks ExecCGI
AllowOverride All
Order allow,deny
Allow from all
Python version is 2.7
Do you have any idea how to run this script? Thanks a lot!
2 answers
solution1
0 2015-11-09 14:11:59
What output are you expecting from running this script (which I agree with ergonaut) is not clear how you are doing it.
Running the script independently, it works, but it does not have any natural output...so potentially it is working but you have no means of verifying.
solution2
0 2015-11-09 15:17:15
Only other suggestions are calling the python module and breakpointing to see where it fails from an ide (eg eclipse). Also capturing logging error may help, but I presume you have tried both of those.
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