On 11/22/2011 09:20 AM, Cranky Frankie wrote:
OK, but this is still not working:
class Qb:
def __init__(self, first_name='', last_name='', phone='',
email='', stadium=''):
self.first_name = first_name
self.last_name = last_name
self.phone = phone
self.email = email
self.stadium = stadium
Qb_list = [["Joe", "Montana", "415-123-4567",
"joe.mont...@gmail.com","Candlestick Park"],
["Fran", "Tarkington","651-321-7657",
"frank.tarking...@gmail.com", "Metropolitan Stadidum"],
["Joe", "Namath", "212-222-7777", "joe.nam...@gmail.com", "Shea Stadium"],
["John", "Elway", "303-9876-333", "john.el...@gmai.com", "Mile
High Stadium"],
["Archie", "Manning", "504-888-1234", "archie.mann...@gmail.com",
"Louisiana Superdome"],
["Roger", "Staubach", "214-765-8989", "roger.staub...@gmail.com",
"Cowboy Stadium"]]
len_Qb_list = len(Qb_list)
for i in range(0, len_Qb_list):
quarterbacks = Qb(*Qb_list[i])
i = i + 1
print(quarterbacks.last_name(2))
You'll generally get better responses by saying in what way it's not
working. In this case, I get an exception in the print statement at the
end. So your message ought to state that, and show the traceback.
The specific error that causes that exception is you're trying to call a
string. What's that (2) all about anyway? quarterbacks is an objext
of type Qb, and quarterbacks.last_name is a string.
Your variable quarterbacks currently contains an object representing the
last line of the list, the one for Roger Staubach. You kept replacing
the value in quarterbacks with the next item from the original list. If
you actually want a list of the objects, you need to say so.
quarterbacks = []
for ....
quarterbacks.append( )
Now that you really have a list, then you can print a particular one with:
print (quarterbacks[2].last_name)
(I tested my own variant using python 2.7. And i'm not saying your
code can't be improved. You're learning, and getting it correct is much
more important than getting it "optimal.")
--
DaveA
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