problems using % in strings with %(var)s replacements
Michael Haggerty
mhagger at alum.mit.edu
Thu May 24 10:19:53 EDT 2001
"Thomas Weholt" <thomas at cintra.no> writes:
> Say I got a string s = "%(number)s among 50% are efficient for
> %(customer_name)s."
> [...]
> I see the % in 50% is the problem but how can I use the %-char in strings
> and still use %(...)s replacements??
Use '%%' to get a single '%':
>>> s = "%(number)s among 50%% are efficient for %(customer_name)s."
>>> s % {'number':42,'customer_name':'Acme Inc.'}
'42 among 50% are efficient for Acme Inc..'
Michael
--
Michael Haggerty
mhagger at alum.mit.edu
More information about the Python-list
mailing list