En Tue, 03 Mar 2009 04:13:46 -0200, JohnV <loftmas...@gmail.com> escribió:
Thanks for your suggestion, but I am not able to get it to work for
me.
My original script was:
f = open('C:\Users\Owner\Desktop\mydata.txt', 'r')
read_data = f.read()
f.close()
\ is the escape character in Python. You must double it when you want it
to stand by itself:
'C:\\Users\\Owner\\Desktop\\mydata.txt'
Alternatively, use a raw string: r'C:\Users\Owner\Desktop\mydata.txt'
The same applies to the output file.
I imported urllib2 However, this the scriot does not seem to work for
me for two reasons
First the url that recieves the post is
http://www.thenational.us/pages/start/test/getpost.html so I belive
the "url line" should read:
url = "http://www.thenational.us:80/pages/start/test/getpost.html";
Yes!
I do not know how to capture the text that is displayed in the
interpreter window when I run the program, as the window closes after
the script is run. How do I keep the window open so I can read what
messages are printed there?
Open a command prompt first (Start > Run > type "CMD" and enter), go to
the directory where your script is located (type: CD c:\some\directory)
and then you're ready to execute it using: python scriptname.py
Finally, do I need conn.close() with the new code you suggested?
No, it should be response.close() instead. The whole script is now:
<code>
import httplib, urllib, urllib2
f = open(r'C:\Users\Owner\Desktop\mydata.txt', 'r')
read_data = f.read()
f.close()
params = urllib.urlencode({'textarea1': read_data})
headers = {"Content-type": "application/x-www-form-urlencoded",
"Accept": "text/plain"}
url = "http://www.thenational.us:80/pages/start/test/getpost.html";
req = urllib2.Request(url, params, headers)
response = urllib2.urlopen(req)
data = response.read()
response.close()
print data
f = open(r'C:\Users\Owner\Desktop\pydata.txt', 'a')
f.write(data)
f.close()
</code>
--
Gabriel Genellina
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