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getting a URL

johnvert at my-deja.com johnvert at my-deja.com
Fri Jun 9 21:02:56 EDT 2000 
In article <8hop1b$c1b$1 at nnrp1.deja.com>,
  johnvert at my-deja.com wrote:
> Hello,
>
> As an exercise I'm trying to write a simple cgi script that will get
the
> source code of an html file and echo it to the screen.  Here's what I
> have:
>
> import cgi, urllib
>
> (snip)
>
> 		url = urllib.urlopen("http://www.server.org/foo.html")
> 		print "<pre>"
> 		print cgi.escape(url.read()),
> 		print "</pre>"
>
> But nothing is printed.  The problematic line seems to be:
>
> url = urllib.urlopen("http://www.server.org/foo.html")
>
> But the URL I gave it in the actual program is correct.  I tried a
few.
> If I run this on the shell (it's meant to be a cgi) it raises IOError
> sayin 'network unreachable'.  What could be wrong?  Also, why does
> execution stop at:
>
>       print "<pre>"
>
> Why isn't the rest (print "</pre>" among others) printed?
>
> Thanks,
>  -- John

I changed the code a little bit and ran it on the shell:

print "Content-type: text/html"
print

#form = cgi.FieldStorage()
#if form.has_key("text1"):
#	if form["text1"].value != "":
#		print cgi.escape(form["text1"].value)
url = urllib.urlopen("http://www.tuxedo.org/~esr/sitemap.html")
print "<pre>"
#		print cgi.escape(url.read()),
print "</pre>"

And I get:
$ ./test.py
Content-type: text/html

<pre>
</pre>

Thanks,
 -- John


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