Formatting a column's value output

[Κώστας Παπαδόπουλος]

Τη Κυριακή, 27 Ιανουαρίου 2013 8:36:42 μ.μ. UTC+2, ο χρήστης Joel Goldstick έγραψε:
> On Sun, Jan 27, 2013 at 1:05 PM, Κώστας Παπαδόπουλος <nikos... at gmail.com> wrote:
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> Τη Κυριακή, 27 Ιανουαρίου 2013 6:12:59 μ.μ. UTC+2, ο χρήστης ru... at yahoo.com έγραψε:
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> > On 01/27/2013 02:04 AM, Ferrous Cranus wrote:
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> > >[...]
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> > >             data = cur.fetchall()
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> > >             for row in data:
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> > >                     print ( "<tr>" )
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> > >
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> > >                     for item in row:
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> > >                             print( '''<td>  <a href='http://www.%s?show=log'>%s</a>  </td>''' % (item, item) )
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> > >[...]
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> > > Okey, so far BUT i want the url linking to happen only for the URL column's
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> > > value, and not for the hits column too. How do i apply the url link to the
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> > > URL column's value only?
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> > Ferrous,
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> > 'row' has two items (the url and the hit count) in it, right?
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> > So print each separately rather than in a loop:
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> >     data = cur.fetchall()
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> >     for row in data:
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> >       url = row[0]
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> >       hits = row[1]
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> >
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> >         print ( "<tr>" )
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> >         print( "<td>  <a href='http://www.%s?show=log'>%s</a>  </td>: % (url, hits) )
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> 
> Yes that can work, but i want to ask you if its possible to do the same thing from within the loop as i have it coded it until now.
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> Is there a way to seperate the values within the loop?
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> for row in data:
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>     print( "<td>  <a href='http://www.%s?show=log'>%s</a>  </td>: " % (row[0], row[1]) )

This is not correct.
the <a href> attribute is for url (row[0]) only, not for url and hits too.

i want the following working code:

=============
		data = cur.fetchall()
		for row in data:
			url = row[0]
			hits = row[1]
			print ( "<tr>" )
			print( "<td>  <a href='http://www.%s?show=log'>%s</a>  </td>" % (url, url) )
			print( "<td><b> %s </td>" % (hits) )
=============

inside the loop i posted initially, if its possible.