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Configuraion to run pyhton script on ubuntu

+1 vote
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I want to run a python script which contains simple form of html on firefox browser , but don't know what should be the configuration on ubuntu 12.04 to run this script i.e CGI configuration.

My code is

ubder 
in /var/www/cgi-bin/forms__.py

#!/usr/bin/env python
import webapp2

form ="""

 """

class MainPage(webapp2.RequestHandler):
 def get(self):
 #self.response.headers['Content-Type'] = 'text/plain'
 self.response.out.write(form)

app = webapp2.WSGIApplication([('/', MainPage)], debug=True)
posted Jul 28, 2013 by Sonu Jindal

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2 Answers

0 votes
 
Best answer

The first step is to update your server:

sudo apt-get update

Then install your Apache2 web server:

sudo apt-get install apache2

Create your CGI directory
This is where all script you wish to execute will go

sudo mkdir /var/www/cgi-bin

The next step tells Apache which directories it can execute in and how to handle a .py file when it sees one. With an editor open /etc/apache2/sites-available/default

sudo vi /etc/apache2/sites-available/default

Inside is a long list of configurations. We want to look at the block of text which should start around line 16. It looks like this:

ScriptAlias /cgi-bin/ /usr/lib/cgi-bin/
<Directory "/usr/lib/cgi-bin/">
    AllowOverride None
    Options +ExecCGI -MultiViews +SymLinksIfOwnerMatch
    Order allow,deny
    Allow from all
</Directory>

Change it to look like this:
(be sure to copy this carefully)

ScriptAlias /cgi-bin/ /var/www/cgi-bin/
<Directory "/var/www/cgi-bin">
    AllowOverride None
    Options +ExecCGI -MultiViews +SymLinksIfOwnerMatch
    Order allow,deny
    Allow from all
    AddHandler cgi-script .py
    AddHandler default-handler .html .htm
</Directory>

Now save and exit.
Now it's time to make your homepage. Apache will automatically display /var/www /index.html, so we will use this file to make our home page.
Open the file with:

sudo vi /var/www/index.html

Delete everything inside.
Replace it with:




First Name:





Last Name:









Now that we have that, we need to make a script to parse it input and return a response.
Go to your cgi-bin directory and with an editor create serverscript.py:

cd /var/www/cgi-bin
sudo vi serverscript.py

Here is the code we will be using for some sample tasks:

!/usr/bin/python

import cgi
import cgitb
cgitb.enable()

Retrieve form fields

form = cgi.FieldStorage() # Get POST data
fname = form.getfirst("fname") # Pull fname field data
lname = form.getfirst("lname") # Pull lname field data
secret = form.getfirst("secretsquirrel") # Pull secretsquirrel field from POST data

Begin HTML generation

print "Content-Type: text/html; charset=UTF-8" # Print headers
print "" # Signal end of headers

HTML is pushed to the page with print

To save time you can use ''' to open and close a print

statement for block printing

print '''


'''
print "First name:",fname
print "
"
print "Last name:",lname
print "
"
print "Whole name:",fname+" "+lname
print "
"
print "The secret was:",secret
print '''


'''

Now allow the script to be executable:

sudo chmod +x serverscript.py

Restart apache2 so that config file changes take effect:
(This does not need to be done when changing site content, only apache2 config files)

sudo /etc/init.d/apache2 restart

answer Jul 29, 2013 by Manish Kumar SIngh
+1 vote

In order for you app to run as cgi, you would have to call the webapp2 run() function. Otherwise, it is going to implement wsgi interfaces.

Have a look at:
http://webapp-improved.appspot.com/api/webapp2.html#webapp2.WSGIApplication.run

As well as:
http://httpd.apache.org/docs/current/mod/mod_alias.html#scriptalias

answer Jul 28, 2013 by Luv Kumar
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